DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL
ALGEBRA
1 CONSTRUCTION OF FUNCTORIAL RESOLUTIONS IN DERIVED CATEGORIES
Problem 1. Consider the category of modules over the ring Z, denoted as Mod(Z). Let M
be the Z-module defined by the presentation Zf
−→ Z⊕Zg
−→ M−→ 0where f(1) = (1,3) and
g(1,0) = 5.
a) Find the first two steps in a projective resolution of the module M.
b) Show that the resolution obtained in part a) is not a free resolution.
Solution 1.
a) To find a projective resolution of the module M, we start by constructing a projective resolution
of the kernel of g, denoted as K=ker(g). Since g(1,0) = 5, the kernel Kis generated by (1,0).
We can construct a projective resolution of K:
0→P1
d1
−→ P0
ϵ
−→ K→0
where P0is a free Z-module generated by taking a copy of Z⊕Z, and d1is a map from P1to
P0that lifts the generator of K. This gives us the first two steps in the projective resolution of M:
0→Zd1
−→ Z⊕Z→M→0
b) To show that this resolution is not a free resolution, we note that a free resolution is a res-
olution consisting of free modules. In the resolution obtained in part a), P0is free but P1=Zis
not free, as Zis not a free Z-module. Therefore, the resolution obtained in part a) is not a free
resolution.
Therefore, the resolution obtained in part a) is a projective resolution but not a free resolution
of the module M.
2 THE HOMOTOPY THEORY OF DERIVED CATEGORIES
Problem 1. Let R=Z[x]/(x2)be a ring and consider the chain complex C•given by
0−→ Rd2
−→ Rd1
−→ R−→ 0,
where the differentials d1and d2are defined by d1(x) = 2 and d2(x)=3x. Compute the homology
of the complex C•.
Solution 1. To find the homology, we compute Hi(C•) = ker(di)/im(di+1)for each i.
a) First, we have:
ker(d1) = {r∈R|d1(r)=0}={r∈R|d1(r) = 2r= 0}=⟨2⟩,
im(d2) = {d2(r)|r∈R}={3r|r∈R}=⟨3x⟩.
Therefore, H1(C•) = ⟨2⟩/⟨3x⟩={a+ 3xb |a, b ∈Z}.
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
b) For H0(C•), we have:
ker(d0) = {r∈R|d0(r)=0}=R,
im(d1) = {d1(r)|r∈R}={2r|r∈R}= 2R.
Thus, H0(C•) = R/2R={a+ 2b|a, b ∈Z}.
c) Finally, H2(C•)=0since there are no elements in the complex in degree −1.
Therefore, the homology of the complex C•is:
H2(C•) = 0, H1(C•) = {a+ 3xb |a, b ∈Z}, H0(C•) = {a+ 2b|a, b ∈Z}.
I’m glad to help with that. Here is a numerical problem related to Derived Categories and
Functors in Homological Algebra:
3 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem:
Consider the following exact sequences of abelian groups:
0−→ A−→ B−→ C−→ 0
and
0−→ D−→ E−→ F−→ 0
where Cand Dare isomorphic, i.e., C∼
=D. Let f:A→Dbe a homomorphism of abelian
groups. Show that there exists a unique homomorphism g:B→Esuch that the following diagram
commutes:
0[r]A[r][d, ”f”]B[r][d, ”g”]C[r][d, equal]00[r]D[r]E[r]F[r]0
Solution:
Since C∼
=D, we can choose an isomorphism ϕ:C→D. Define g:B→Eas follows: for
any b∈B, since Cis isomorphic to D, there exists d∈Dsuch that ϕ(c) = dfor c∈C. Since Cis
in the image of B, there exists b′∈Bsuch that c=b′+A. Define g(b) = ϕ(c). We need to show
that this definition is well-defined.
Let b1, b2∈Bsuch that b1−b2∈A. We want to show that g(b1) = g(b2). Choose b′
1, b′
2∈B
such that c=b′
1+A=b′
2+A. Then b′
1−b′
2∈A, and b′
1+A=b1+Aand b′
2+A=b2+A. Since f
is a homomorphism, f(b′
1−b′
2) = f(b′
1)−f(b′
2) = ϕ(b1)−ϕ(b2) = 0. This implies that ϕ(b1) = ϕ(b2).
Thus, our definition of gis well-defined.
Now we need to show that gis a homomorphism. For b1, b2∈B, we have:
g(b1+b2) = g(b′
1+b′
2+A)
=ϕ(b′
1+b′
2)
=ϕ(b′
1) + ϕ(b′
2)
=g(b1) + g(b2)
This completes the proof that there exists a unique homomorphism g:B→Emaking the
diagram commute.
4 TATE COHOMOLOGY AND ITS APPLICATIONS IN DERIVED CATEGORIES
Problem 1. Let Rbe a commutative ring and Mbe an R-module. Consider the complex of
R-modules P•defined by Pi=M⊕Mfor all i∈Zand di:Pi→Pi+1 given by the matrix
0idM
0 0
for all i∈Z. Compute the Tate cohomology groups ˆ
Hi(M)for i∈Z.
Solution 1.
Since P•is a complex, the Tate cohomology groups ˆ
Hi(M)are defined as the cohomology
groups of the Tate resolution of M. We construct the Tate resolution of Mas the complex T•with
Ti=Pi⊗Rˆ
R, the completion of Rwith respect to the ideal m={r∈R|rM = 0}, and the
differential di:Ti→Ti+1 induced by the differential of P•. So, we have T•=P•⊗Rˆ
R.
Computing Ti, we have Ti= (M⊕M)⊗Rˆ
R=ˆ
R⊕ˆ
Rfor all i∈Z.
The differential in the Tate resolution is given by
di:Ti−→ Ti+1
0idM
0 0 ⊗1ˆ
R=0id ˆ
R
0 0
for all i∈Z.
Therefore, the cohomology groups of the Tate cochain complex are the same as those of the
chain complex P•, and we have
ˆ
Hi(M) = (Mif i= 0
0otherwise
5 MAPPING CONE AND CONE CONSTRUCTION IN DERIVED CATEGORIES
Problem 1. Let A•and B•be complexes of modules with differential graded maps f:A•→B•.
Consider the mapping cone C(f)of f. Given that A•is the complex:
A•:· · · → 03
−→ Z2
−→ Z→0→ · · ·
and B•is the complex:
B•:· · · → 01
−→ Z1
−→ 0→0→ · · ·
a) Calculate the cone C(f).
b) Determine the homology of the cone C(f).
Solution 1.
a) Let’s first construct the mapping cone C(f):
C(f) : · · · → 0
3
0
−−−→ Z⊕Z2 1
−−−−−→ Z→0→ · · ·
b) To determine the homology of the cone C(f), calculate the homology at each degree:
•H−1(C(f)) = ker(Z→0) = Z
•H0(C(f)) = ker(Z⊕Z
2
−1
−−−−−→ Z)
im(0 →Z)={(x, y)∈Z⊕Z|2x−y= 0}
{(0,0)}= 0
Therefore, the homology of the cone C(f)is:
H−1(C(f)) = Z, H0(C(f)) = 0
6 COMPUTING HOMOTOPY LIMITS AND COLIMITS IN DERIVED CATEGORIES
Problem 7. Let Abe an abelian category and consider the chain complex X:· · · → 0→Af
−→
Bg
−→ C→0→ · · · where A, B, and Care objects in Aand fand gare morphisms in A. Compute
the homotopy limit and homotopy colimit of Xin the derived category D(A).
Solution 7.
To compute the homotopy limit and homotopy colimit of the chain complex Xin the derived
category D(A), we first see that in D(A), the homotopy limit of Xis given by the complex holim(X),
and the homotopy colimit of Xis given by the complex hocolim(X).
a) To compute holim(X), we apply the homotopy limit formula:
holim(X) = Tot(· · · → 0→Af
−→ Bg
−→ C→0→. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We can compute the differential dfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
d=
0 0 0
f0 0
0g0
0 0 0
Then, the holim(X)will be the complex:
0→A
f
−g
−−−−→ B⊕C→0.
b) To compute hocolim(X), we apply the homotopy colimit formula:
hocolim(X) = Tot(· · · ← 0←Af
←− Bg
←− C←0←. . . ).
This is the total complex of the double complex:
0ABC0
where the columns correspond to the chain complex X. We compute the differential δfor this
double complex, which will be the total differential of the total complex. For simplicity, we assume
it is given by:
δ=
0
f
g
0
Then, the hocolim(X)will be the complex:
0←A⊕Bhf−gi
←−−−−−− C←0.
Therefore, we have computed both the homotopy limit holim(X)and homotopy colimit hocolim(X)
of the chain complex Xin the derived category D(A).
7 THE PROBLEM OF CONSTRUCTING TRIANGULATED CATEGORIES FROM DERIVED
CATEGORIES
Problem 8. Let Rbe a commutative ring and Man R-module. Consider the bounded below
complex P•with P0=Rand Pi= 0 for i < 0.
a) Show that the cochain complex Tot(P•)is quasi-isomorphic to M.
b) Let Db(R-Mod)denote the derived category of bounded complexes of R-modules. Prove
that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
Solution 8.
a) To show that Tot(P•)is quasi-isomorphic to M, we need to find a map of complexes f:
P•→Q•such that finduces isomorphisms in cohomology. Let Q0=Mand Qi= 0 for i= 0.
Define the map fas the identity map on P0=Rand the zero map for all other components of P•.
It is straightforward to see that fis a chain map. To show that it is a quasi-isomorphism, we
need to show that it induces isomorphisms in cohomology. Since Qi= 0 for i= 0, the cohomology
of Q•is only in degree 0, which is isomorphic to M. Thus, the cohomology of Tot(P•)is isomorphic
to M, proving that Tot(P•)is quasi-isomorphic to M.
b) To prove that the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories, we need
to show that it is fully faithful, essentially surjective, and essentially surjective on isomorphisms.
1. Fully Faithful: Let X, Y be objects in Db(R-Mod). We need to show that the natural map
HomD(R-Mod)(X, Y )→HomR-Mod(Tot(X),Tot(Y))
induced by the functor Tot is an isomorphism. Since Xand Yare bounded complexes, the total
complexes Tot(X)and Tot(Y)are well-defined. Moreover, the map is induced at the level of each
Hom-complex, and hence we have a natural isomorphism.
2. Essentially Surjective: For any object Min R-Mod, we need to find an object Xin Db(R-Mod)
such that Tot(X)is isomorphic to M. Let Xbe the bounded complex with X0=Mand all other
components being zero. Then, Tot(X)is quasi-isomorphic to M.
3. Essentially Surjective on Isomorphisms: To show this, we need to show that if f:X→Yis
an isomorphism in R-Mod, then the map induced by Tot is also an isomorphism. This follows from
part 1, as the morphism fat the level of Hom-complexes induces an isomorphism.
Therefore, the functor Tot :Db(R-Mod)→R-Mod is an equivalence of categories.
8 LOCALIZATION AND COMPLETION IN DERIVED CATEGORIES
Problem 8. Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider
the bounded derived category of modules Db(R-Mod), and let Sbe the multiplicative subset of R
consisting of non-zerodivisors on M.
a) Show that the localization of the complex Mat S, denoted by LS(M), is isomorphic in
Db(R-Mod)to computing the derived functor of localization (−)Sapplied to M.
b) Prove that the derived functor of localization (−)Srespects direct sums, i.e., for any collection
{Mi}of R-modules, we have (LiMi)S∼
=LiMi.
c) Let Nbe another finitely generated R-module. Show that there is a natural isomorphism
between LS(M)⊗RNand LS(M⊗RN)in Db(R-Mod).
Solution 8.
a) To show that LS(M)is isomorphic to the derived functor of localization (−)Sapplied to M,
we need to find a quasi-isomorphism between the two. Consider the complex 0→Mid
−→ M→0.
Localizing this complex at Sgives the complex 0→MS
id
−→ MS→0. This is clearly quasi-
isomorphic to the complex 0→MS→0, which is the derived functor of localization MS. Therefore,
LS(M)∼
=MS.
b) Let {Mi}be a collection of R-modules. Since localization commutes with direct sums, we
have (LiMi)S=Li(Mi)S. But since Sconsists of non-zerodivisors on Mi, the localization (Mi)S
is isomorphic to Mi. Therefore, (LiMi)S∼
=LiMi.
c) Let Nbe a finitely generated R-module. By the definition of the tensor product in the derived
category, LS(M)⊗RNis represented by the complex C(M)⊗RN, where C(M)is a complex
representing LS(M). Similarly, LS(M⊗RN)is represented by the complex C(M⊗RN). By the
quotient property of the tensor product, we have C(M)⊗RN∼
=C(M⊗RN), which gives the desired
isomorphism LS(M)⊗RN∼
=LS(M⊗RN)in Db(R-Mod).
9 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a ring and consider the categories Mod(R)of left R-modules with mor-
phisms being R-module homomorphisms. Let F:Mod(R)→Mod(R)be the functor defined by
F(M) = M⊕M, where the direct sum is taken as R-modules.
a) Prove that Fis an exact functor.
b) Compute the derived functor R1F(M)for any R-module M.
Solution 1.
a) To show that Fis an exact functor, we need to show that it preserves exact sequences. Let
0→M′f
−→ Mg
−→ M′′ →0be an exact sequence in Mod(R).
First, note that F(M) = M⊕Mand given a morphism h:M→Nin Mod(R), the induced
map F(h) : F(M)→F(N)is given by F(h)(m, n)=(h(m), h(n)).
Now, consider the sequence 0→F(M′)F(f)
−−−→ F(M)F(g)
−−−→ F(M′′)→0. We have: - Im(F(f)) =
{(f(m), f(m)) : m∈M′}- Ker(F(g)) = {(m, m)∈F(M) : g(m) = 0}
It can be checked that Im(F(f)) = Ker(F(g)), showing that 0→F(M′)→F(M)→F(M′′)→
0is exact. Therefore, Fis an exact functor.
b) To compute the derived functor R1F(M), we first need to compute the left derived functor
of Fat M. By definition, R1F(M)is obtained by applying Fto a projective resolution of Mand
taking the homology of the resulting complex.
Let P•→Mbe a projective resolution of M. Applying Fto the complex, we get F(P•). The
first term of the complex is F(P0) = P0⊕P0, the second term is F(P1) = P1⊕P1, and so on.
The differential maps in F(P•)come from the differential maps of P•and the identity maps in
F. Therefore, the homology of the complex F(P•)is the direct sum of the homology of P•at each
term.
Thus, R1F(M)is isomorphic to the first homology of F(P•), which is coker(d(0) :P1⊕P1→
P0⊕P0).
10 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 11. Consider the following complex of abelian groups:
0→Zf
−→ Z2g
−→ Z→0
where fis given by f(n) = (2n, n)and gis given by g(a, b) = a−2bfor all n, a, b ∈Z.
a) Compute the mapping cone of the morphism f.
b) Compute the mapping cone of the morphism g.
Solution 11.
a) To compute the mapping cone of the morphism fin the given complex, consider the diagram
of the mapping cone:
Z[r, ”f”]Z2[r, ”g”]ZC(f)[u, ”i”][ur, ”h”′]
where C(f)is the mapping cone of f,i:Z2→C(f)is the canonical injection, and h:C(f)→Z
is the induced map.
The mapping cone C(f)is given by the complex:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
where i(n) = (f(n),0) for all n∈Zand h((a, b), c) = cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism fis:
0→Zi
−→ Z2⊕Zh
−→ C(f)→0
0→Z(f,0)
−−−→ Z2⊕Z(a,b,c)7→c
−−−−−−→ Z→0
b) To compute the mapping cone of the morphism gin the given complex, follow a similar
approach as in part (a) by considering the diagram of the mapping cone for g.
The mapping cone C(g)is given by the complex:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
where j(n) = (0, n)for all n∈Zand k((a, b), c) = a−2b+cfor all a, b, c ∈Z.
Therefore, the mapping cone of the morphism gis:
0→Zj
−→ Z2⊕Zk
−→ C(g)→0
0→Z(0,1)
−−−→ Z2⊕Z(a,b,c)7→a−2b+c
−−−−−−−−−−→ Z→0
11 LOCALIZATION IN DERIVED CATEGORIES OF MODULES
Problem 12. Let R=Z[x],M=Z/4Z, and consider the chain complex given by:
C:· · · → 0→Mx
−→ M→0→ · · ·
where Msits in degree 0, and the map x:M→Mis multiplication by x.
a) Calculate the homology of C.
b) Calculate the homology of the complex obtained by tensoring Cwith Z/2Z.
c) Calculate the homology of the complex obtained by tensoring Cwith Z/3Z.
Solution 12.
a) The homology of a chain complex Cis given by Hn(C) = ker(dn)/im(dn+1), where dndenotes
the boundary map.
In this case, we have ker(d1) = 0 and im(d0)=4Z⊂Z. Since Mis in degree 0, the homology
of Cis given by H0(C) = Z/4Z.
b) Tensoring Cwith Z/2Zmeans tensoring each module in the complex with Z/2Z. The re-
sulting complex is:
C⊗ZZ/2Z:· · · → 0→Z/2Zx
−→ Z/2Z→0→ · · ·
As in part (a), the homology of this complex is given by H0(C⊗ZZ/2Z)=(Z/2Z)/(2Z) = Z/2Z.
c) Tensoring Cwith Z/3Zyields a similar complex:
C⊗ZZ/3Z:· · · → 0→Z/3Zx
−→ Z/3Z→0→ · · ·
Again, the homology of this complex is H0(C⊗ZZ/3Z)=(Z/3Z)/(3Z) = Z/3Z.
12 THE TRIANGULATED STRUCTURE OF DERIVED CATEGORIES.
Problem 12. Consider the following complex in an abelian category A:
X:· · · → 0→Af
−→ Bg
−→ C→0→ · · ·
where A, B, C are objects in A, and fand gare morphisms in A.
Given this complex, define the following objects in the derived category D(A):
•X[1]
•X[2]
Solution 12. To define X[1], we shift all the objects and morphisms in the complex Xone
space to the left. This results in the following complex:
X[1] : · · · → 0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[1] is given by moving everything to the left by one position.
To define X[2], we shift all the objects and morphisms in the complex Xtwo spaces to the left.
This gives us:
X[2] : · · · → 0→0→0→Af
−→ Bg
−→ C→0→ · · ·
Therefore, in D(A),X[2] is given by moving every object and morphism to the left by two positions.
These shifts are crucial for understanding the triangulated structure of derived categories and
how we define objects like X[1] and X[2].
13 COMPUTING DERIVED FUNCTORS IN TRIANGULATED CATEGORIES
Problem 15. Let Rbe a commutative ring and F:ModR→ModRbe a left-exact additive
functor. Consider the following short exact sequence in ModR:
0−→ M−→ N−→ L−→ 0
where M,N, and Lare R-modules.
a) Show that applying Fto the short exact sequence above gives the following long exact
sequence:
0−→ F(M)−→ F(N)−→ F(L)−→ F1(M)−→ F1(N)−→ . . .
b) Suppose Fis a right exact functor. Prove that every long exact sequence obtained by apply-
ing Fto a short exact sequence as above is also exact in the middle.
c) If Fis an exact functor, explain why the long exact sequence obtained in part (a) is a short
exact sequence.
Solution 15.
a) To show that applying Fto the short exact sequence gives a long exact sequence, we can
make use of the long exact sequence in homology derived from a short exact sequence of chain
complexes. Denote by 0→K•→L•→M•→0the images of the modules M,N, and Lunder
the (co)homology functors. Then F(M), F (N),and F(L)can be viewed as complexes that are
acyclic outside degree 0, so we can apply the long exact sequence in homology:
. . . →Hn(F(L)) →Hn(F(M)) →Hn(F(N)) →Hn+1(F(L)) →. . .
From here, it follows that applying Fto the short exact sequence indeed gives the long exact
sequence provided.
b) Given that Fis right exact, it preserves injectivity. Then the sequence 0→F(M)→F(N)→
F(L)→F1(M)→F1(N)→. . . is still exact in the middle by the properties of injectivity.
c) If Fis an exact functor, it is both left and right exact, meaning it preserves both injective and
projective objects. Therefore, the long exact sequence obtained in part (a) will be a short exact
sequence since all the higher derived functors of Fwill vanish.
14 THE PROBLEM OF DERIVED FUNCTORS IN ABELIAN CATEGORIES
Problem 15. Let Rbe a commutative ring and consider the abelian category ModRof R-
modules. Let F:ModR→ModRbe the functor defined by F(A) = A⊗RA. Compute the derived
functor R1F.
Solution 15.
To compute the derived functor R1F, we need to construct an injective resolution of an R-
module A, apply the functor Fto each term in the resolution, and take the homology at the first
term.
Let’s construct an injective resolution for an R-module A:
0−→ A−→ I0−→ 0
where I0is an injective R-module containing A.
Now, apply the functor Fto each term in the resolution:
0−→ A⊗RA−→ I0⊗RI0−→ 0
Taking the homology at the first term gives us R1F(A) = coker(A⊗RA→I0⊗RI0).
Since F(A) = A⊗RA, the map A⊗RA→I0⊗RI0is just the natural inclusion A⊗RA ,→I0⊗RI0.
Thus, the cokernel is I0⊗RI0/(A⊗RA).
Therefore, R1F(A) = I0⊗RI0/(A⊗RA).
15 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 1. Let Rbe a commutative ring, and consider the category Ch(R)of chain complexes
of R-modules. Let K(R)be the homotopy category of Ch(R), and D(R)be the derived category
of Ch(R). Suppose Ais a chain complex with Hi(A) = 0 for i= 0, and H0(A) = R.
a) Show that the complex Arepresents an object in K(R).
b) Determine whether the complex Ais isomorphic to a bounded complex in Ch(R).
Solution 1.
a) To show that Arepresents an object in K(R), we need to show that Ais a complex in Ch(R)
and that it is homotopic to a complex with Hi= 0 for i= 0.
Since Hi(A) = 0 for i= 0, all differentials diwith i= 0 must necessarily map from 0to 0in
order to ensure that di◦di+1 = 0. This means Ais indeed a complex in Ch(R).
Moreover, since H0(A) = R, we can construct a chain homotopy hisuch that d0=h1◦d0+d1◦h0,
showing that Ais homotopic to a complex with Hi= 0 for i= 0. Therefore, Arepresents an object
in K(R).
b) In order for Ato be isomorphic to a bounded complex in Ch(R), there must exist two integers
mand nsuch that Ai= 0 for i < m and i > n. Since A0=Rand Hi(A) = 0 for i= 0, this is not
possible for A. Therefore, Ais not isomorphic to a bounded complex in Ch(R).
16 COMPUTING COHERENT FUNCTORS BETWEEN DERIVED CATEGORIES
Problem 17. Let R=Z[x, y, z]/(x2, y2, z2)be the ring defined by the given relations. Consider
the complexes Xand Ydefined as follows:
X:. . . →0→R
x
y
z
−−−→ R3→0→. . .
Y:. . . →0→R
x
y
−−−→ R2→0→. . .
Compute the derived functors RF (X)and RF (Y)where Fis the forgetful functor from Ch(R)
(the category of chain complexes over R) to Mod(R)(the category of R-modules).
Solution 17. To compute the derived functors of the forgetful functor, we will look at the total
derived functors of the forgetful functor, which will give us the cohomology modules of the given
complexes.
a) To calculate RF (X), we first need to find a resolution of X.
Consider the complex P•:
P•:. . . →0→Rx y z
−−−−−−−−→ R3→0→. . .
This complex is acyclic and a resolution of the R-module X.
Now we apply the forgetful functor Fto this complex to obtain:
F(P•) : . . . →0→R→R3→0→. . .
This is again an acyclic complex representing RF (X).
Hence, RF (X)is the zero module.
b) Similarly, to calculate RF (Y), we need to find a resolution of Y.
Consider the complex Q•:
Q•:. . . →0→Rx y
−−−−−→ R2→0→. . .
This is an acyclic complex, hence a resolution of the R-module Y.
Now applying the forgetful functor Fto this complex gives:
F(Q•) : . . . →0→R→R2→0→. . .
This is again an acyclic complex representing RF (Y).
Therefore, RF (Y)is also the zero module.
17 DERIVED CATEGORIES AND FUNCTORS IN HOMOLOGICAL ALGEBRA
Problem 18.
Let Rbe a commutative ring and Mbe a finitely generated R-module. Consider the projective
resolution of M:
0−→ Pn−→ · · · −→ P1−→ P0−→ M−→ 0
where each Piis a projective R-module. Show that the homology modules Hi(M)of Mare iso-
morphic to the Ext modules Exti
R(M, R)for all i≥0.
Solution 18.
The Ext functor in abelian categories is defined as Exti
R(M, N) = Hi(HomR(P•, N)), where P•
is a projective resolution of M. In this case, we have N=R.
Given the projective resolution of M, applying the Hom functor gives us a complex:
0−→ HomR(P0, R)−→ HomR(P1, R)−→ · · · −→ HomR(Pn, R)−→ 0
The homology of this complex at position iis precisely the Ext module Exti
R(M, R). Therefore,
Hi(M)∼
=Exti
R(M, R)for all i≥0.
18 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF SHEAF COHOMOLOGY.
Problem 1. Consider a sheaf Fon a topological space Xwith cohomology groups Hi(X, F)
as follows:
H0(X, F) = R,
H1(X, F)=0,
H2(X, F) = Z,
Hi(X, F)=0for i= 0,2.
a) Compute the derived funcor RΓ(X, −)of the sheaf F.
b) Determine H2(X, RΓ(X, F)).
Solution 1.
a) Since H0(X, F) = Rand H1(X, F) = 0, the sheaf Fis acyclic except possibly at degrees 0
and 2. Thus, the derived functor RΓ(X, −)of Fis given by:
RΓ(X, −) = Γ(X, −)⊕Γ(X, −)[−2],
where Γ(X, −)denotes the global sections functor and [−2] denotes the shift functor by 2.
b) We have RΓ(X, F) = Γ(X, F)⊕Γ(X, F)[−2]. Therefore, H2(X, RΓ(X, F)) = H2(X, Γ(X, F)⊕
H0(X, F) = H2(X, F) = Z.
19 COMPUTING DERIVED FUNCTORS IN THE CONTEXT OF NON-ABELIAN CATEGORIES
Problem 20. Consider a non-abelian category Cwith objects Aand B. Let F:C → C and
G:C → C be two functors defined as follows:
F(A) = B,F(B) = A,F(f) = f′for any morphism f:A→B,
G(A) = A,G(B) = B,G(f) = f−1for any isomorphism f:A→B.
a) Compute RF (A)and RG(B).
b) Compute RF (B)and RG(A).
Solution 20.
a) To compute RF (A), we need to find a quasi-isomorphism
A→BF→CF→DF
where BF, CF, DFare homotopically projective resolutions of B, C, D respectively. Since F(A) =
B, we can take Bitself as a projective resolution. Thus, RF (A) = B.
Now, for RG(B), we need to find a quasi-isomorphism
B→AG→CG→DG
where AG, CG, DGare homotopically injective resolutions of A, C, D respectively. Since G(B) = B
and G(A) = A, we can take Bitself as an injective resolution. Therefore, RG(B) = B.
b) For RF (B), we need to find a quasi-isomorphism
B→CF→DF
where CF, DFare homotopically projective resolutions. Since F(B) = A, we can take Aitself as
a projective resolution for B. Thus, RF (B) = A.
Similarly, for RG(A), we need to find a quasi-isomorphism
A→BG→CG→DG
where BG, CG, DGare homotopically injective resolutions. Since G(A) = A, we can take Aitself
as an injective resolution. Therefore, RG(A) = A.
20 THE PROBLEM OF COMPUTING HOMOLOGY AND COHOMOLOGY IN DERIVED CATE-
GORIES
Problem 1. Let R=Z[x]be the ring of polynomials with integer coefficients in the variable x.
Consider the chain complex C•given by:
· · · −→ R∂2
−→ R⊕R∂1
−→ R∂0
−→ 0
where the differentials are defined by ∂0(r)=0for all r∈R,∂1(r1, r2) = r2−r1x, and ∂2(r) =
(rx, x).
a) Compute the homology groups H0(C•),H1(C•), and H2(C•).
b) Find the cohomology groups H0(C•),H1(C•), and H2(C•).
Solution 1.
a) To find H0(C•), we need to compute the kernel of the map ∂0:R→0. Since ∂0(r) = 0 for
all r∈R, the kernel is all of R. Therefore, H0(C•) = R/0∼
=R.
Next, for H1(C•), we need to compute ker(∂1)/im(∂2). Since ker(∂1) = {(r, r)∈R⊕R|r∈R}
and im(∂2) = {(rx, x)|r∈R}, we have ker(∂1)/im(∂2) = 0, as there is no non-trivial element in
common. Therefore, H1(C•)=0.
Lastly, for H2(C•), we need to compute the image of ∂2. Since ∂2(r)=(rx, x)for all r∈R, the
image is {(rx, x)|r∈R}. Hence, H2(C•)=(R⊕R)/{(rx, x)|r∈R}∼
=R.
b) The cohomology groups can be computed by dualizing the chain complex. So, we have
H0(C•)∼
=coker(∂0)∼
=0(as coker(∂0) = R/0is trivial), H1(C•)∼
=coker(∂1)∼
=R/im(∂1), and
H2(C•)∼
=coker(∂2)∼
=(R⊕R)/im(∂2).
