advance topology homework

MA500-2: Topological Structures 2016

Aisling McCluskey, Daron Anderson

[email protected], [email protected]

Contents

0 Preliminaries 2

1 Topological Groups 8

2 Morphisms and Isomorphisms 15

3 The Second Isomorphism Theorem 27

4 Topological Vector Spaces 42

5 The Cayley-Hamilton Theorem 43

6 The Arzelà-Ascoli theorem 44

7 Tychonoff ’s Theorem if Time Permits 45

Continuous assessment 30%; final examination 70%. There will be a weekly

workshop led by Daron during which there will be an opportunity to boost

continuous assessment marks based upon workshop participation as outlined in

class.

This module is self-contained; the notes provided shall form the module text.

Due to the broad range of topics introduced, there is no recommended text.

However General Topology by R. Engelking is a graduate-level text which has

relevant sections within it. Also Undergraduate Topology: a working textbook by

McCluskey and McMaster is a useful revision text. As usual, in-class discussion

will supplement the formal notes.

1

0 PRELIMINARIES

0 Preliminaries

Reminder 0.1. A topology τ on the set X is a family of subsets of X, called

the τ-open sets, satisfying the three axioms.

(1) Both sets X and ∅ are τ-open

(2) The union of any subfamily is again a τ-open set

(3) The intersection of any two τ-open sets is again a τ-open set

We refer to (X,τ) as a topological space. Where there is no danger of ambi-

guity, we suppress reference to the symbol denoting the topology (in this case,

τ) and simply refer to X as a topological space and to the elements of τ as its

open sets. By a closed set we mean one whose complement is open.

Reminder 0.2. A metric on the set X is a function d: X×X → R satisfying the five axioms.

(1) d(x,y) ≥ 0 for all x,y ∈ X

(2) d(x,y) = d(y,x) for x,y ∈ X

(3) d(x,x) = 0 for every x ∈ X

(4) d(x,y) = 0 implies x = y

(5) d(x,z) ≤ d(x,y) + d(y,z) for all x,y,z ∈ X

Axiom (5) is often called the triangle inequality.

Definition 0.3. If d′ : X × X → R satisfies axioms (1), (2), (3) and (5) but maybe not (4) then we call it a pseudo-metric.

Reminder 0.4. Every metric on X induces a topology on X, called the metric

topology. We define an open ball to be a set of the form

B(x,r) = {y ∈ X : d(x,y) < r}

for any x ∈ X and r > 0. Then a subset G of X is defined to be open (wrt the metric topology) if for each x ∈ G, there is r > 0 such that B(x,r) ⊂ G. Thus open sets are arbitrary unions of open balls.

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The definition of the metric topology makes just as much sense when we are

working with a pseudo-metric. Open balls are defined in the same manner, and

the open sets are exactly the unions of open balls. Pseudo-metric topologies are

often neglected because they do not have the nice property of being Hausdorff.

Reminder 0.5. Suppose f : X → Y is a function between the topological spaces X and Y . We say f is continuous to mean that whenever U is open in

Y , the preimage f−1(U) = {x ∈ X : f(x) ∈ U} is open in X.

Reminder 0.6. The following are equivalent for a function f : X → Y between metric spaces (X,dX) and (Y,dY ).

(1) The function f is continuous

(2) ∀x ∈ X ∀ε > 0 ∃δ > 0 : dX(x,y) < δ =⇒ dY (f(x),f(y)) < ε

(3) Whenever (xn) satisfies dX(xn,x) → 0 we have also dY (f(xn),f(x)) → 0

Definition 0.7. Suppose (X,τ) and (Y,σ) are topological spaces. We define

the product topology on the set X×Y as follows. A point x is called an interior point of the set G ⊂ X × Y if there exists open sets U ∈ τ and V ∈ σ such that x ∈ U ×V ⊂ G. A set is declared open in the product topology if each of its points is an interior point.

Lemma 0.8. This defines a topology on X ×Y .

Proof. We need only check the three axioms for a topology are satisfied.

(1) Since ∅ has no points, each of its zero-many points is an interior point. To show X ×Y is open we recall that X is open in itself and likewise for Y . The let U = X and V = Y . Then for each (x,y) ∈ X × Y we have (x,y) ∈ U ×V ⊂ X ×Y . Therefore each (x,y) is an interior point.

(2) Suppose W is a family of open sets. We must show ⋃ W is open. Let

z ∈ ⋃ W. Then z ∈ W for some open set W in the family W. Since W is

open there exist U ∈ τ and V ∈ σ such that z ∈ U ×V ⊂ W . But since W ⊂

⋃ W this implies z ∈ U ×V ⊂

⋃ W . Therefore each z ∈

⋃ W is an

interior point.

(3) Let W and W ′ be open in X × Y and let z ∈ W ∩ W ′. Since z ∈ W there exist open sets U ∈ τ and V ∈ σ such that z ∈ U × V ⊂ W . Likewise since z ∈ W ′ there exist open sets U′ ∈ τ and V ′ ∈ σ such that

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z ∈ U′ × V ′ ⊂ W . Recall that U ∩ U′ is open in X and V ∩ V ′ is open in Y . Taking intersections gives that z ∈ (U ∩U′) × (V ∩V ′) ⊂ W ∩W ′. Therefore z is an interior point of W ∩W ′.

Exercise 0.9. Call a basic open set one of the form U ×V where U is open in X and V is open in Y . The open sets in the product topology are exactly

the unions of basic open sets.

Reminder 0.10. The Euclidean metric:

d2(x,y) = √

(x1 −y1)2 + . . . + (xn −yn)2

and the uniform metric:

d3(x,y) = max{|x1 −y1|, . . . , |xn −yn|}

generate the same topology on Rn.

Lemma 0.11. The Euclidean topology on R2 agrees with the product topology on R×R.

Proof. Recall the Euclidean metric on R2 generates the same topology as the uniform metric:

d3((x,y), (x ′,y′)) = max{|x−x′|, |y −y′|}.

With respect to the uniform metric the open ball with radius r and centre (x,y)

looks like (x−r,x+r)×(y−r,y +r). This is the open square with centre (x,y) and side 2r. Observe that (x−r,x + r) and (y −r,y + r) are open in R. If we let U = (x−r,x + r) and V = (y −r,y + r) we see the open uniform ball is of the form U ×V and hence is also open in the product topology. By extension every union of uniform-open balls, and hence every uniform-open set, is open in

the product topology.

Now we will prove every product-open set is uniform-open. Suppose the set

W ⊂ X×Y is open with respect to the product topology, and let (x,y) ∈ W be any point. Then we have (x,y) ∈ U ×V ⊂ W where U and V are open. Then x ∈ U and y ∈ V . So there exist r1,r2 > 0 such that x ∈ (x− r1,x + r1) ⊂ U and y ∈ (y − r2,y + r2) ⊂ V . Let r be the lesser of r1 and r2. Then (x,y) ∈ (x − r,x + r) × (y − r,y + r) ⊂ U × V ⊂ W . Thus (x,y) is an element of a uniform-open ball contained in W . We conclude W is uniform-open.

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Definition 0.12. Suppose (xn) is a sequence in the topological space X. We

say (xn) converges to x and write xn → x to mean that for every open set U containing x there exists (a sufficiently large) N ∈ N such that xm ∈ U whenever m > N.

Homework 0.13. Let X be a metric space and (xn) ⊂ X a sequence. Show that (xn) converges to the point x ∈ X if and only if the sequence of real numbers d(xn,x) converges to zero.

Asides like the following are not directly examinable, but are valuable to

challenge your understanding of the course-content, and avoid mistakes.

Now that we have defined convergence in spaces without a metric, we might

hope to generalise Reminder 0.6 into something like this.

Non-Lemma 0.14. The following are equivalent for a function f : X → Y between topological spaces.

(1) The function f is continuous

(2) Whenever (xn) ⊂ X is a sequence with xn → x then f(xn) → f(x) for the sequence (f(xn)) ⊂ Y .

It is true that (1) =⇒ (2) in general. However the reverse implication does not hold for non-metric spaces. To see this, consider the following

example.

Example 0.15. Let R0 be the real line with the topology whose proper closed sets are the countable ones. Let R be the real line with the Euclidean topology. We claim the identity map f : R0 → R satisfies condition (2) above but is not continuous. This is because every convergent sequence

in R0 is eventually constant. That means xn → x if and only if (xn) = {x1,x2, . . .xm,x,x, . . .} for some m ∈ N.

To see this suppose that xn → x is not eventually constant. Then we can find a subsequence (ym) of (xn) such that all ym 6= x. Since (ym) is a subsequence of (xn) we have as well that ym → x. Then since {y1,y2, . . .} is countable its complement is open. Moreover the complement is an open

neighbourhood of x that contains no ym. It follows that ym may not con-

verge to x in R0, a contradiction.

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It is easy to see that f maps the sequence {x1,x2, . . . ,xm,x,x, . . .} to the sequence {f(x1),f(x2), . . .f(xm),f(x),f(x), . . .} which converges to f(x). Therefore condition (2) is met. To see f : R0 → R is not continuous consider the preimage of an uncountable closed set, for example a closed

interval.

The main point to take away is that, while convergent sequences make

sense in non-metric spaces, they do not control the topology to the same

extent as when there is a metric present.

Lemma 0.16. Let X and Y be topological spaces with x ∈ X, y ∈ Y . The sequence (xn,yn) in X×Y converges to (x,y) if and only if xn → x and yn → y.

Proof. First suppose (xn,yn) → (x,y). We will show xn → x. The case for yn is identical. Let U be open with x ∈ U ⊂ X. Then U × Y is open in the product topology. Since (xn,yn) → (x,n) with respect to the product topology there exists N ∈ N such that (xm,ym) ∈ U × Y ∀m > N. But this implies xm ∈ U ∀m > N. Hence xn → x in X.

Now suppose xn → x and yn → y and let W ⊂ X × Y be open with (x,y) ∈ W . Then we can find U and V open such that (x,y) ∈ U × V ⊂ W . Since xn → x there exists N1 ∈ N such that xm ∈ U ∀m > N1. Likewise since yn → y there exists N2 ∈ N such that ym ∈ V ∀m > N2. Define N = max{N1,N2}. Then whenever m > N we have (xm,ym) ∈ U × V ⊂ W . Therefore (xn,yn) → (x,y).

Homework 0.17. Given topological spaces X and Y , show that the (first)

projection map π1 : X ×Y → X defined by (x,y) 7→ x is continuous

The second projection map π2 : X×Y → Y defined by (x,y) 7→ y is similarly continuous. These facts readily establish the following result:

Exercise 0.18. A function h into a product space Y × Z is continuous if for each i, πi ◦h is continuous.

Homework 0.19. Given topological spaces X, Y and Z, suppose the functions

f : X → Y and g : X → Z are continuous. Prove the following function is continuous:

h: X → Y ×Z x 7→ (f(x),g(x))

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when the codomain carries the product topology.

Homework 0.20. Let (X,dX) and (Y,dY ) be metric spaces. We can define a

function (X ×Y )2 → R +

as follows:

d((x,y), (x′,y′)) = dX(x,x ′) + dY (y,y

′).

This is called the product metric. Prove it is indeed a metric, and that it

generates the product topology. Thus show every product of metric spaces is a

metric space.

Assignment 1: Please return solutions to all four Homework problems at

class on Wednesday January 27th.

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1 Topological Groups

Definition 1.1. By a topological group we mean a set that is simultaneously

a group and a topological space. Formally it is a triple (G,∗,τ) such that ∗ is a binary operation on the set G satisfying the group axioms, and τ is a topology on

G. We require the binary operation and topology to interact in such a way that

the two maps (group operations), corresponding to multiplication and inversion

m : G×G → G inv : G → G

(g,h) 7→ g ∗h g 7→ g−1

are both continuous.

Henceforth when we say group, we mean a topological group. We will usually

write just G instead of (G,∗,τ) and write the product of a,b ∈ G as ab instead of a∗b. The identity will be denoted e. When two or more groups are involved we will make sure to be more precise.

The topology τ on the set X is called a group topology to mean there

exists some binary operation ∗ on X that makes (X,∗,τ) into a topological group.

Remark. There are, of course, many ways of recognising continuity − through open sets, closed sets, neighbourhoods, closure et cetera − and consequently many equivalent versions for the continuity of m and inv. One that may be

useful is:

(i) m is continuous iff for all x,y ∈ X, for all open U such that xy ∈ U, there is open V , W such that x ∈ V,y ∈ W,V W ⊂ U

(ii) inv is continuous: for all x ∈ X, for all open U such that x ∈ U, there is open V such that x ∈ V and V −1 ⊂ U.

Here we are using V W and V −1 in the group-theoretic sense i.e.

V W = {vw: v ∈ V,w ∈ W}, V −1 = {v−1 : v ∈ V}.

Lemma 1.2. Let q denote the “division” operation in a group X. i.e. the map

q : X × X → X is defined by q((x,y)) = xy−1; then (i) and (ii) above are together equivalent to

(iii) q is continuous.

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Proof. First suppose both (i) and (ii) hold. Consider the map θ: G×G → G×G defined by θ((x,y)) = (x,y−1). To prove θ is continuous we must check both

projections π1 ◦θ and π2 ◦θ are continuous. The second projection is the map G × G → G given by (x,y) 7→ y−1.

That is to say π2 ◦ θ finds the second coordinate and then takes the inverse. So we can write π2 ◦ θ = inv ◦ π2. We proved in Homework 0.17 that π2 is always continuous. Moreover the map inv is continuous by (ii). Therefore as a

composition of continuous functions inv◦π2 = π2 ◦θ is continuous as required. Continuity of π1 ◦θ is proved similarly.

By (ii) the product map m is continuous. Therefore so is m◦ θ. We claim that m ◦ θ = q and hence q is also continuous as required. To prove this take the input (x,y) and compute (m ◦ θ)((x,y)) = m(θ((x,y))) = m((x,y−1)) = xy−1 = q((x,y)).

Now suppose (iii) holds. That means the map q : G×G → G is continuous. Consider the function ϕ: G → G×G given by ϕ(x) = (e,x). It can be shown this map is continuous (check the projections!). By taking an input as above,

we can prove inv = q ◦ ϕ and hence inv is continuous, as a composition of continuous functions. This proves (ii).

Now we know both inv and q are continuous. By considering the projections

we can show the map θ((x,y)) = (x,y−1) is continuous in this context as well.

To show m is continuous we represent it as m = q ◦θ. This proves (i)

Exercise 1.3. Prove the following are topological groups. That means prove

the group operations are continuous − you need not prove the axioms for a topology are satisfied.

(1) Q, R and C under addition and the Euclidean topology

(2) Q 6=0, R 6=0 and C 6=0 under multiplication and the Euclidean topology

(3) The circle group T = {z ∈ C: |z| = 1} of complex numbers of modulus 1, taken with multiplication and the subspace topology

(4) Any group with the discrete or trivial topology.

(5) The additive group of vectors of any normed vector space

(6) Homework: The set of bounded continuous functions X → R where X is any fixed topological space, under addition of functions and either the

topology of pointwise convergence or the topology of uniform convergence.

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(7) Let Ω be the collection of Lebesgue-measurable subsets of [0, 1], and let

µ denote the Lebesgue measure. Define the symmetric product on Ω

by A ∗ B = A∆B = (A − B) ∪ (B − A), and the pseudo-metric d′ by d′(A,B) = µ(A∆B).

(8) For a simpler example, let Ω be the collection of all subsets of some finite

set, and let µ(A) = |A| be the counting measure. Unlike the previous example, this defines a metric.

The topology of uniform convergence on the group of bounded continuous

functions X → R is generated by the uniform norm or sup norm.

‖f‖ = sup{|f(x)|: x ∈ X}

The fact that f is bounded ensures its norm is finite. In fact if we allow

unbounded functions and permit ‖f‖ = ∞ as a possible value, then this topology still makes sense, but is no longer metrisable.

The pointwise topology gives each f : X → R the neighbourhood basis of sets

(f,F,ε) = {g : X → R : |g(x) −f(x)| < ε ∀x ∈ F} Where F ⊂ X is any finite set and ε > 0 any positive number.

We say f is an interior element of the set U if there is some (f,F,ε)

such that f ∈ (f,F,ε) ⊂ U. We say U is open if each element is an interior element. Equivalently the open sets are arbitrary unions

⋃ i(fi,Fi,εi) of

the basic open sets defined above.

The pointwise topology is not metrisable unless X is countable. So

for example letting X be any of R, C, [0, 1] or T will give a group whose topology is not generated by any metric.

Exercise 1.4. Prove that example (7) above only defines a pseudo-metric space,

and not a metric space.

Exercise 1.5. Prove the group in example (8) has the discrete topology.

Exercise 1.6. Let Ω be the collection of all subsets of N = {1, 2, . . .} and let a1,a2, . . . be positive constants such that

∑ n∈N an is finite. Define a weighted

counting measure on Ω by µ(A) = ∑ n∈A an. Prove that d(A,B) = µ(A∆B)

defines a metric, and that the metric topology and symmetric product make Ω

into a topological group.

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Exercise 1.7. Prove that, for any choice of positive constants a1,a2, . . . , the

group described in Exercise 1.6 never has the discrete topology.

Homework 1.8. By treating the set of complex n×n matrices as a subspace of Cn×n we can give it the Euclidean topology. Show the group Matn(C) of invert- ible matrices forms a topological group under multiplication and the Euclidean

topology. (Hint: Homework 0.19)

Exercise 1.9. Show that that R is not an additive topological group under any of the following topologies.

(1) Homework: The upper-limit topology, whose nonempty open sets are of

the form {(a,∞) : a ∈ R}.

(2) The Sorgenfrey topology, whose nonempty open sets are finite unions of

right-open intervals [a,b) = {x ∈ R: a ≤ x < b}.

(3) The cofinite-measure topology, whose nonempty closed sets are those with

finite Lebesgue measure.

(4) Homework: The cofinite topology whose proper closed sets are the

finite ones.

(5) Homework: The particular point topology, whose nonempty open sets

are those that contain the identity.

Exercise 1.10. (Bonus Points!) Let Z2 be set of binary integers that extend infinitely to the left. For example (. . . 010101) and (. . . 111111) are elements of

Z2. Formally Z2 = {(. . .a3a2a1) : an = 0, 1} where each an is the nth digit of the integer. We can add these numbers together in the same way normal integers

are added, just repeating the process ad infinitum. Formally if a = (. . .a3a2a1)

and b = (. . .b3b2b1) are elements, we define the first n digits of a + b to be the

first n digits of (an . . .a3a2a1) + (bn . . .b3b2b1). Define a metric on Z2 as follows:

d(a,b) = 1

2n where n is the first digit that an 6= bn

Prove this is indeed a metric and a topological group, called the 2-adic

integers.

Definition 1.11. Suppose G and H are topological groups with operations ∗G and ∗H respectively. Define their product to be the set G×H under the product topology and group operation (g,h) ∗ (g′,h′) = (g ∗G g′,h∗H h′).

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Exercise 1.12. Show this defines a topological group.

Definition 1.13. A left translation on G is a function of the form x 7→ ax for some a ∈ G. A right translation is a function of the form x 7→ xb for some b ∈ G.

There is a rich supply of homeomorphisms in any topological group; in par-

ticular, left or right translations by any fixed element are homeomorphisms.

Lemma 1.14. For a topological group X, and a ∈ X, the following are home- omorphisms of X onto X:

(i) λa, defined by λa(x) = ax (left translation);

(ii) ρa, defined by ρa(x) = xa (right translation);

(iii) the map x 7→ axa−1 (inner automorphism);

(iv) inv.

Proof. The map Ψ : X → X×X given by Ψ(x) = (a,x) is continuous (consider its composites with projections); therefore so is m ◦ Ψ which is λa; and the inverse of λa is λa−1 which is likewise continuous; hence (i). Proof of (ii) is

similar. The map in (iii) is λa ◦ρa−1 , so (iii) follows from (i) and (ii). Also (iv) is immediate because (inv)−1 = inv.

Definition 1.15. A topological space (X,τ) is said to be homogeneous if,

for every x and y in X, we can find a homeomorphism h : X → X such that h(x) = y.

Corollary 1.16. A topological group is homogeneous.

Proof. Given x and y, take h to be λyx−1 .

Remark. So, topologically speaking, a topological group “looks the same” at

each point. Here are a few of the many immediate consequences:

(i) It is discrete if any one point (e.g. {e}) is open;

(ii) It is locally compact if any one point (e.g. e) has a compact neighbour-

hood;

(iii) It is T1 (coming up!) if any one point (e.g. {e} is closed.

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Definition 1.17. For any two subsets A,B ⊂ G and element g ∈ G define the following sets.

A−1 = {a−1 : a ∈ A}

AB = {ab: a ∈ A,b ∈ B}

An = {a1a2 . . .an : a1,a2, . . . ,an ∈ A}

gB = {gb: b ∈ B} Bg = {bg : b ∈ B}

The set gB is called the left-translate of B by g. The set gB is called the right-

translate of B by g. A translate of a set B means either a left or right translate

by some group element.

Lemma 1.18. The nonempty open sets of a group are exactly the translates

of the open sets about the identity.

Proof. First suppose U ⊂ G is an open set containing the identity. Since trans- lations are homeomorphisms, every translate of U is open. Now suppose V ⊂ G is a nonempty open set. V must contain some g and hence we can write

V = g(g−1V ). Since g ∈ V we have e ∈ g−1V . Thus we have expressed V as a left-translate of an open set about the identity. Observe we could have

done the same for a right-translation as well.

Lemma 1.19. (The Half Lemma) For any open neighborhood U of the identity

there is an neighborhood V of the identity such that V 2 ⊂ U.

Proof. Consider the preimage of U ⊂ G under the product map. We have m−1(U) = {(g,h) : g ∗h ∈ U} is open by continuity of the product. Moreover (e,e) ∈ m−1(U) since e∗e = e ∈ U. By the definition of the product topology there are open sets U1,U2 such that (e,e) ∈ U1 × U2 ⊂ m−1(U). Define V = U1 ∩U2. Since V ×V ⊂ m−1(U) we have m(V ×V ) = V 2 ⊂ U as required.

Definition 1.20. The topological space X is called regular to mean that,

given any point x ∈ X outside the closed set B ⊂ X, there exist disjoint open sets U ⊂ X and V ⊂ X such that x ∈ U and B ⊂ V .

Theorem 1.21. Topological groups are regular.

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Proof. In the notation of Definition 3.26 suppose B ⊂ G is closed and x /∈ B. Then G − B is an open neighborhood of x. So G − B = xW for some open neighborhood W of the identity. The Half Lemma ensures there is some open

neighborhood Y of the identity such that Y 2 ⊂ W . Let U = xY and V = BY −1. We claim U and V are disjoint. For otherwise we have xy1 = by

−1 2 for some

y1,y2 ∈ Y and b ∈ B. But this implies b = xy1y2 ∈ xY 2 ⊂ xW = G−B and so b ∈ G−B, a contradiction. Therefore the open sets U and V are disjoint, as required.

We want to show how easy it is for a topological group to be Hausdorff. As

it turns out, being Hausdorff is equivalent to the following − seemingly-weaker − condition.

Definition 1.22. The space X is said to be T0 when, for every two points

x,y ∈ X, there exists an open set U ⊂ X such that either x ∈ U ⊂ (X −y) or y ∈ U ⊂ (X −x)

Theorem 1.23. A group is T0 if and only if it is Hausdorff.

Proof. That every Hausdorff space is T0 is left as an exercise. Now suppose the

group G is T0 and let x,y ∈ G be arbitrary. Without loss of generality there exists an open set U ⊂ X such that x ∈ U ⊂ (X −y). Then there is some open V at the identity such that U = xV . By the Half Lemma we can find an open

set W at the identity such that W2 ⊂ V . We claim the open sets xW and yW−1 are disjoint. For otherwise we have

w1,w2 ∈ W such that xw1 = yw−12 and so y = xw1w2. This implies y ∈ xW 2 ⊂

xV = U, contrary to our assumption that U ⊂ (G−y). We conclude that for every two points x,y ∈ G there are disjoint open sets A,B ⊂ G such that x ∈ A and y ∈ B. This means G is Hausdorff.

Homework 1.24. A T1 topological space is one whose every singleton is closed.

Prove the following implications for the space X.

X is Hausdorff =⇒ X is T1 =⇒ X is T0

Homework 1.25. Suppose the group G has identity element e. Prove that G

is Hausdorff if and only if the singleton {e} is closed.

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2 Morphisms and Isomorphisms

Remember that in (non topological) Group Theory there is a special class of

functions that preserve the group structure.

Reminder 2.1. Suppose G and F are two groups and φ: G → F is a function. We call φ a group-homomorphism to mean that φ(gh) = φ(g)φ(h) for any

g,h ∈ G. Suppose φ is an invertible map and the inverse is also a group- homomorphism. Then we call φ a group-isomorphism.

The group-homomorphisms between G and F are the functions that ‘care

about’ how G and F are groups and not just sets. Obviously if we change the

group operations on G and F, then we will get a different set of possible group-

homomorphisms. A group-isomorphism is a function that witnesses how two

groups are ‘the same’. Some easy properties of group-homomorphisms are as

follows.

Reminder 2.2. Suppose φ: G → F is a group-homomorphism. Then φ satisfies the following identities for any g,g1,g2, . . . ,gn ∈ G and n ∈ N.

(1) φ(g1g2 . . .gn) = φ(g1)φ(g2) . . .φ(gn)

(2) φ(gn) = φ(g)n

(3) φ(g−1) = φ(g)−1

(4) φ(eG) = eF

Recall from Group Theory that for the group-homomorphism φ to be a

group-isomorphism, it is enough that φ be bijective.

Reminder 2.3. Suppose G and F are two groups and φ: G → F is a bijective group-homomorphism. Then the inverse function φ−1 : F → G is also a group- homomorphism and hence φ is a group-isomorphism.

Likewise in topology there is a special class of functions that preserve the

topological (or physical) structure of a space. These are called continuous func-

tions. A continuous function whose inverse function exists and is continuous is

called a homeomorphism.

We already know a lot about continuous functions and homeomorphisms.

For example we know the analogue of Reminder 2.3 fails if we replace groups

with topological spaces and group-homomorphism with continuous maps.

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Exercise 2.4. Give an example of two topological spaces X and Y and a

bijective function f : X → Y whose inverse function f−1 : Y → X fails to be continuous.

Now suppose G and F are topological groups. That means they have both

a group structure and a topological structure. Requiring a map G → F to preserve both structures leads to the following definition.

Definition 2.5. A morphism of topological groups is a continuous group-

homomorphism between topological groups. An isomorphism of topological

groups is a bijective morphism whose inverse is also a morphism.

Exercise 2.6. Suppose G and F are topological groups. Prove the function

φ: F → G is an isomorphism of topological groups if and only if it is simulta- neously a group-isomorphism and an homeomorphism.

Exercise 2.7. Prove the following are morphisms:

(1) The determinant function Matn(R) → R 6=0 is a morphism from the group of invertible matrices to the multiplicative group of nonzero reals.

(2) The transpose function Matn(R) → Matn(R) which sends each matrix A = (a

j i ) to its transpose A

t = (aij) is continuous but is not a group-

homomorphism under the usual matrix multiplication. But we can use

the ‘reverse product’ A∗B = BA on the codomain to make the transpose into an isomorphism.

(3) Similar to the above, except we define the group Matn(C) of complex invertible n × n matrices. Then the conjugate transpose function which sends B = (b

j i ) to B

∗ = (bij) is a morphism when the codomain has the

reverse product. Like the previous example this map is self-inverse and is

thus an isomorphism.

(4) The logarithm function x 7→ log(x) is an isomorphism from the multiplica- tive group of positive reals to the additive group of all reals. The inverse

morphism is the exponential function.

(5) The exponential function C → C 6=0 given by z 7→ ez is a morphism from the complex plane under addition, to the punctured complex plane under

multiplication. This morphism has no inverse because it is impossible to

define a global logarithm function.

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(6) The function z 7→ z/|z| which takes the multiplicative group C 6=0 into the circle group T.

(7) Suppose F is an additive group of real-valued functions on the space X,

under pointwise or uniform convergence. For each x ∈ X the function f 7→ f(x) defines a morphism F → R.

(8) The function R → T which ‘wraps the real line around the circle’. This means each x 7→ eiθ where θ smallest positive element of {x+2πn: n ∈ Z}. This is an example of a quotient morphism. We will see more on these

morphisms later.

(9) When G is Abelian the function g 7→ gn is a morphism for each n ∈ Z. In particular inversion is an isomorphism.

(10) Suppose G and F are two groups and define their product G × F as Definition 1.12. Then the projection G×F → G given by (g,f) 7→ g is a morphism.

(11) For any group G and x ∈ G a fixed element, the conjugation function g 7→ xgx−1 is an isomorphism from G to itself. This isomorphism is the identity whenever G is Abelian.

Exercise 2.6 tells us we should not expect every bijective morphism to be

an isomorphism. If a morphism is bijective its inverse will always be a group-

isomorphism but may fail to be continuous. The next exercise provides an

example.

Exercise 2.8. Let Cd denote the complex plane under addition and the discrete topology. Let C denote the complex plane under addition and the standard Eu- clidean topology. Prove the identity map Cd → C is a continuous and bijective group-homomorphism, but the inverse function is not continuous. Hence this is

a bijective morphism but not an isomorphism.

The First, Second and Third Isomorphism Theorems from Group Theory

have analogues for topological groups. Indeed the non-topological versions can

be thought of as the special case when all groups are discrete.

When proving the Isomorphism Theorems, we are given one or more group-

homomorphisms, and from these we construct the desired group-isomorphisms.

When proving the Topological Isomorphism Theorems, we are given one or more

morphisms and from these we construct the relevant isomorphisms.

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But since not every bijective morphism is an isomorphism, we will require

extra hypotheses on the morphisms we start with, in order to show the desired

functions are continuous. These take the form of requiring certain morphisms

map open sets to open sets. Not all continuous maps have this property, and

we will do a lot of work to find conditions when this holds.

Definition 2.9. The function f : G → F is called an open mapping to mean that f(U) ⊂ F is open whenever U ⊂ G is open.

Homework 2.10. Suppose G and F are topological groups. Show that for the

group-homomorphism f : G → F to be open, it is enough that eF be an interior point of f(U) for each open neighborhood U ⊂ G of eG. (Hint: Use translations to reposition a general element at eG.)

Next we will discuss what subgroups and quotient groups mean in a topo-

logical setting.

Reminder 2.11. Suppose (G,∗) is a group and F ⊂ G is a subset such that (F,∗) is a group. Then we say F is a subgroup of G.

Exercise 2.12. Suppose F ⊂ G is a subgroup. Prove F becomes a topological group when equipped with the subspace topology. Prove moreover the inclusion

map F ↪→ G is a morphism.

Reminder 2.13. The subgroup N ⊂ G is called normal to mean that gNg−1 = N for every g ∈ G.

Exercise 2.14. Suppose G is a topological group with identity element e. Prove

the closure {e} of the singleton {e} is a normal subgroup of G.

Exercise 2.15. Suppose G is a Hausdorff group and F ⊂ G is an Abelian subgroup. Prove that F is also an Abelian subgroup.

Whenever F ⊂ G is a subgroup and g ∈ G is an element, the subset gF ⊂ G is called the left coset of g. The subset Fg ⊂ G is likewise called the right coset of g. The group G is partitioned into cosets. What does that mean?

Reminder 2.16. Suppose X is a set and P is a family of subsets of X. Then P is called a partition of X if it has the two properties.

(1) The elements of P cover X. That means X = ⋃ {P : P ∈P}.

(2) Any two distinct elements of P are disjoint. That means if P,Q ∈P then either P = Q or P ∩Q = ∅.

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It can be helpful to think of X as something like a house. Each element

x ∈ X is one of the atoms that make up the house. Each element of P ∈P can be thought of as a brick, and the elements p ∈ P are the atoms that make up that brick. Condition (1) says that every atom is contained in some brick, and

condition (2) says that no two bricks share atoms.

Reminder 2.17. Suppose G is a group and N ⊂ G is a normal subgroup. Then G/N = {gN : g ∈ G} is a partition of G.

When N is normal the set G/N can be made into a group in its own right.

We define the product by (gN ∗hN) = (gh)N. The normality of N ensures this is a well-defined group operation.

When G carries a compatible (ie group) topology we would like the same for

G/N. That is we would like G/N to be a topological group whenever G is. The

way this topology is constructed works for general spaces and not just groups.

Definition 2.18. Suppose we are given a topological space X and a set Y .

Then for any surjective map π : X → Y we define the quotient topology on Y under π as follows. Each set U ⊂ Y is declared to be open exactly if π−1(U) ⊂ X is open.

The map π should be thought of as ‘classifying’ the elements of X into

different categories. Each category is a subset of X and looks like π−1(y) for

some y ∈ Y . You can think of Y as the set of all category-names. The set π−1(y) is also called the fibre over y. You should check these fibres form a

partition of X.

Exercise 2.19. Prove that, under the notation of Definition 2.18, the quotient

topology is indeed a topology on Y .

Here is a very silly example of what the quotient topology looks like. Suppose

X is the space of all dogs in the world, endowed with some suitable topology.

Perhaps it is a metric topology that compares the colour, size, shape, age,

temperament, and intelligence of two dogs and assigns a number based on how

similar the two dogs are. Two very similar dogs are close together under this

metric and two very different dogs are far apart. Let Y be the set of all possible

dog breeds.

Y = {Labrador, beagle, mastiff, dachshund, collie, greyhound, corgi, wolf, . . .}

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There is an obvious map π : X → Y that takes a dog and classifies it by breed. The fibre π−1(beagle) ⊂ X is just the subset of all beagles. We can use this data to put a topology on the set of breeds of dog.

To find whether the set of breeds U = {beagle, mastiff, collie} is an open subset of Y we simply apply the inverse function. Then π−1(U) is the set of

all beagles, mastiffs, and corgis in the world. We must then consult with the

topology of X to determine if π−1(U), and hence U is an open set.

To intuit what sets of the form π−1(π(B)) look like, suppose B ⊂ X is a set of dogs. To find π(B) we look at all breeds of the dogs in B. Suppose there

are nine Labradors, three collies, one mastiff, and a thousand wolves. Then

π(B) = {Labrador, mastiff, collie, wolf} and π−1(π(B)) ⊂ X is the set of all Labradors, all mastiff, all collies, and all wolves.

In the case of groups π : G → G/N will be the same quotient map we are used to. It will classify the elements of G by sending g ∈ G to whichever coset has g as an element. Henceforth the quotient group G/N will be equipped with

the quotient topology under π. It remains to prove that this makes G/N a

topological group.

Homework 2.20. Prove the quotient map π : G → G/N is a continuous group- homomorphism.

Homework 2.21. Suppose π : G → G/N is the quotient map and B ⊂ G is an arbitrary set. Prove that π−1(π(B)) = BN. Thus conclude π is an open map.

We will prove G/N is a topological group in two steps, corresponding to

the multiplication and inversion maps. The multiplication step relies on the

following exercise.

Homework 2.22. Suppose f : A → X and g : B → Y are open maps. Prove the following is an open map:

(f ×g) : A×B → X ×Y (a,b) 7→ (f(a),g(b))

when the domain and codomain carry the product topology.

Theorem 2.23. Multiplication on G/N is a continuous map.

Proof. Write p: (G/N) × (G/N) → G/N for the multiplication on the quotient group. We know the map m: G × G → G is continuous. We will use this to show p is continuous as well. To that end let U ⊂ G/N be any open set. We must show p−1(U) ⊂ (G/N) × (G/N) is open. We can write

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p−1(U) = {(gN,hN) : p(gN,hN) ∈ U}

= {(gN,hN) : (gN) ∗ (hN) ∈ U}

= {(gN,hN) : (gh)N ∈ U}

= {(gN,hN) : π(gh) ∈ U}

= {(gN,hN) : gh ∈ π−1(U)}

= {(gN,hN) : m(g,h) ∈ π−1(U)}

= {(gN,hN) : (g,h) ∈ m−1(π−1(U))}.

Now consider the map π × π that sends (g,h) to (gN,hN). You proved in Homework 2.22 that, since the quotient map is open, so is the map π ×π. But the above set can be written as

{(π ×π)(g,h) : (g,h) ∈ m−1(π−1(U))}

= (π ×π)(m−1(π−1(U)))

The maps π and m are continuous. Since the set U is open in G/N, the

set π−1(U) is open in G. Then m−1(π−1(U)) is open in G×G. Finally, since (π×π) is an open map, the set (π×π)(m−1(π−1(U))) is open in (G/N)×(G/N) a required.

Theorem 2.24. Inversion on G/N is a continuous map.

Proof. The approach here is similar to before. Write jnv : G/N → G/N for the inversion on the quotient group. We know the map inv : G → G is continuous. We will use this to show jnv is continuous as well. To that end let U ⊂ G/N be any open set. We must show jnv−1(U) ⊂ G/N is open. We can write

jnv−1(U) = {gN : jnv(gN) ∈ U}

= {gN : (gN)−1 ∈ U}

= {gN : (g−1)N ∈ U}

= {gN : π(g−1) ∈ U}

= {gN : g−1 ∈ π−1(U)}

= {gN : inv(g) ∈ π−1(U)}

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= {gN : g ∈ inv−1(π−1(U))}

= π(inv−1(π−1(U)))

Since U is an open set, π and inv are continuous, and π is an open map, the

fact that the above set is open follows from similar reasoning to before.

Corollary 2.25. G/N is a topological group.

Homework 2.26. Suppose G is a topologial group and N ⊂ G a normal subgroup. Prove that G/N is Hausdorff if and only if N is closed. (Hint:

Homework 1.25)

Homework 2.27. Let G and F be groups, N ⊂ G be normal, π the canonical quotient map, and φ any function. Consider the mapping diagram.

G G/N

F

π

φ◦π φ

Prove that φ is a morphism if and only if φ◦π is a morphism.

From here on out we use the symbol ∼= to denote isomorphism of topological groups. For any two topological groups G and F we write G ∼= F to mean some f : G → F is simultaneously a group-isomorphism and a homeomorphism.

Homework 2.28. Prove the quotient group R/Q has the trivial topology.

Definition 2.29. Suppose φ: G → F is a group-homomorphism. Define the kernel of φ to be the subset ker φ = {g ∈ G: φ(g) = eF}.

Exercise 2.30. Prove, under the notation of Definition 2.29, that ker φ is a

normal subgroup of G. Now suppose that N ⊂ G is a normal subgroup and π : G → G/N the quotient map. Show that ker π = N.

Theorem 2.31. (First Isomorphism Theorem) Suppose f : G → F is an open morphism. Then G/N ∼= f(G) where N = ker(f).

Proof. We claim the map θ : G/N → f(G) defined by gN 7→ f(g) is an isomor- phism. We know from the non-topological First Isomorphism Theorem that θ

is a group-isomorphism. We need only show it is a homeomorphism.

To show θ is open take an open subset of U ⊂ G/N and consider the image

θ(U) = {θ(gN) : gN ∈ U}

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= {f(g) : gN ∈ U}

= {f(g) : π(g) ∈ U}

= {f(g) : g ∈ π−1(U)}

= f(π−1(U))

The set above is open, since π is continuous and f is an open map. We conclude

θ is an open map. To show θ is continuous take an arbitrary open V ⊂ f(G) and consider the preimage

θ−1(V ) = {gN ∈ G/N : θ(gN) ∈ V}

= {gN ∈ G/N : f(g) ∈ V}

= {gN ∈ G/N : g ∈ f−1(V )}

= {π(g) : g ∈ f−1(V )}

= π(f−1(V ))

The set above is open, since π is an open map and f is continuous. We conclude

θ is continuous.

Exercise 2.32. Let G be a topological group. Suppose we have two open

morphisms φ: G → F and ϕ: G → E such that ker φ = ker ϕ. Prove the images φ(G) ⊂ F and ϕ(G) ⊂ E are isomorphic as topological groups.

Suppose G is a non-topological group. The upshot of the non-topological

First Isomorphism Theorem is that the normal subgroups of G are in one-

to-one correspondence with the surjective group-homomorphisms on G. Thus

when searching for all possible images of G we need not investigate the ‘world

of all possible groups’. We can restrict our attention to looking at subsets and

subgroups of G.

Now suppose G carries a compatible topology. Theorem 2.31 says that every

open morphism on G is a quotient morphism. In Homework 2.21 you proved

the converse − that quotient morphisms are always open. We conclude that the normal subgroups of G are in one-to-one correspondence with the surjective

morphisms from G.

Exercise 2.33. When G is a topological group consider the correspondence.

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( Normal

subgroups of G

) ←→

( Surjective open

morphisms on G

) The left hand side does not refer to the topology. Whether a subset N ⊂ G is a normal subgroup is a purely group-theoretic notion. However the right-hand

side does refer to the topology, in order to define an open (and continuous)

morphism. How do you justify this to yourself?

Computing quotient topologies directly can be a hassle. The topological

First isomorphism theorem can be used to avoid this. Suppose we want to

compute the topology of R/Z, where Z = {. . . ,−1, 0, 1, . . .} is a normal subgroup of R. We can do this by finding an open morphism whose kernel is Z. The image of R under that morphism will have the same topology as R/Z.

It turns out we already know such a function. Namely the morphism f : R → T that ‘wraps the real line around the circle’. We can write f(x) = e2πix which is clearly continuous. Properties of the exponential ensure it is also a group-

homomorphism.

We will explain why f is open and leave the details as an exercise. Suppose

U ⊂ R is open. Then it is a union of open intervals. Suppose we paint over each interval in red paint and, before the paint dries, wrap the line around the

circle. Then f(U) is the part of the circle that becomes covered in paint. It is

a union of ‘circular intervals’.

Since f is an open morphism, the First Isomorphism Theorem says that

R/Z = R/ ker(f) ∼= f(R) = T. We conclude that R/Z ∼= T. This means the quotient group R/Z has the same topology as T. Observe we avoided working out the topology on R/Z directly.

Exercise 2.34. Prove the morphism f : R → T defined by f(x) = e2πix is open.

Example 2.35. Let Matn(R) be the space of invertible n × n matrices with the Euclidean topology and SMatn(R) = {A ∈ Matn(R) : det(A) = 1} be the subspace of Special Invertible Matrices. If we accept the determinant morphism

is open we get:

Matn(R)/SMatn(R) ∼= R 6=0.

In Theorem 2.31 we need f to be open to ensure θ is open. Otherwise we

have the following kind of counterexample.

Example 2.36. Let Rd be the additive group of real numbers with the discrete topology. The identity morphism id: Rd → R is continuous because the domain

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is discrete. However the morphism is not open. The kernel of id is {0} and so Rd/ ker(id) ∼= Rd which is not homeomorphic to R and so cannot be isomorphic as a topological group.

The Third Isomorphism Theorem is much easier to prove than it is to un-

derstand. The setup is as follows: G is a group with normal subgroups F and

E such that F ⊂ E ⊂ G. This time there are two partitions. G is first partitioned into larger blocks;

that is to say the cosets of E. But, on top of that, each block is partitioned into

smaller bricks; that is to say the cosets of F. The next exercise is to prove this

analogy is accurate.

Exercise 2.37. Suppose G is a group with normal subgroups F and E such

that F ⊂ E ⊂ G. Prove each coset of E is a disjoint union of cosets of F.

To continue our analogy, the blocks partition the atoms of G. But since each

brick is contained in a block, we can equally think of the blocks as forming a

partition of the bricks. In the other direction, we can think of the identity block

− that is to say E − as being composed of bricks rather than atoms. Under this interpretation G is the group of all atoms. G/F is the group of

all bricks. The subgroup E/F of G/F is the group of all bricks that live inside

the identity block. Finally G/E is the group of blocks.

When we form the quotient G/E, both groups G and E are considered as

collections of atoms. What the Third Isomorphism Theorem says, is that it

makes no difference if we consider both groups as collections of bricks instead.

Theorem 2.38. (Third Isomorphism Theorem) Suppose G has normal sub-

groups F and E such that F ⊂ E ⊂ G. Then we have the relation

G/E ∼= G/F

E/F .

Proof. Let πE : G → G/E and πF : G → G/F be the natural quotient maps. Define a group-homomorphism θ: (G/F) → (G/E) by gF 7→ gE. The function θ is well defined since gF = hF =⇒ h−1g ∈ F =⇒ h−1g ∈ E =⇒ gE = hE =⇒ θ(gF) = θ(hF).

That θ is continuous and open follows from the two observations (Think of

them in terms of bricks and blocks!)

θ−1(U) = πF (π −1 E (U)) and θ(W) = πE(π

−1 F (W))

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and the fact that both quotient maps are continuous and open. The result then

follows from the First Isomorphism Theorem since ker(θ) = E/F.

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3 The Second Isomorphism Theorem

We have already proved the First and Third Isomorphism Theorems for topolog-

ical groups. These required few modifications from the non-topological picture.

All we needed was for a certain morphism to be open. The picture for the

Second Isomorphism Theorem is not so simple.

For the Second Isomorphism Theorem, we again have a system of groups and

morphisms, and we will use these morphisms to construct a group-isomorphism.

But the construction is complicated enough that, even requiring the original

morphisms all be open, the group-isomorphism we construct may fail to be

continuous or open.

To solve this problem, we will introduce two classes of ‘well-behaved’ group

topologies. They are well-behaved in the sense that morphisms to or from them

tend to be open. The Second Isomorphism Theorem holds if we restrict our

attention to groups of this sort.

Reminder 3.1. Suppose the topological space X has the following property:

Whenever we are given a collection U of open subsets such that X = ⋃ U, then

there exist finitely many U1,U2, . . .Un ∈ U such that X = U1 ∪U2 ∪ . . .∪Un. Then we say X is compact.

Reminder 3.2. A closed subset of a compact space is itself compact.

Definition 3.3. Let X be a space and x ∈ X a point. We say X is locally- compact at x to mean that x has an open neighborhood with compact closure.

We say X is locally-compact to mean it is locally-compact at each point.

The real line will be our motivating example and main building-block for

constructing locally-compact spaces. That R is locally-compact follows from the Heine-Borel theorem, according to which the compact subsets are exactly

the closed and bounded ones. We will build on this fact to show how many

spaces constructed from copies of the real line are locally-compact. We first

need a result about compactness.

Theorem 3.4. The product of two compact spaces is compact.

Proof. Suppose X and Y are compact spaces and W is an open cover of X×Y . Let x0 ∈ X be arbitrary. The space {x0}×Y is a copy of Y sitting inside X×Y . So {x0}×Y is compact.

Each point (x0,y) ∈ {x0} × Y is included in some element of W ∈ W. Without loss of generality W = U × V is a basic open set. Since {x0}× Y is

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compact we can select finitely many cover elements U1×V1,U2×V2, . . . ,Un×Vn that cover {x0}×Y . That means the union V1 ∪V2 ∪ . . .∪Vn = Y . Moreover, since each Ui contains x0, we have U(x0) = U1 ∩ U2 ∩ . . . ∩ Un is a nonempty open neighborhood of x0 in X. Then we have

U(x0) ×Y = U(x0) × (⋃

i

Vi

) = ⋃ i

( U(x0) ×Vi

) ⊂ ⋃ i

( Ui ×Vi

) .

And so the ‘tube’ U(x0) × Y is covered by finitely many elements of W. Now repeat this process for each x ∈ X to get a different ‘tube’ U(x) × Y . As the basepoint x ranges over the elements of X, the opens sets U(x) form

an open cover of X. By compactness of X we can extract a finite subcover

U(x0), . . . ,U(xm) of X. Then we have

X ×Y = (⋃

j

U(xj) ) ×Y =

⋃ j

( U(xj) ×Y

) So X×Y is the union of finitely many ‘tubes’. Since each ‘tube’ is itself covered by finitely many elements of W, we take the union to get a finite subcover of X ×Y .

By iterating the above theorem we can show that if X1,X2, . . .Xn is some

finite collection of topological spaces, then X1 ×X2 × . . .×Xn is compact. For example the Cartesian cube [0, 1]n is compact for any n ∈ N. This fact can be applied in the next exercise.

Exercise 3.5. Prove that if each space X1,X2, . . .Xn is locally-compact then

the product (X1 ×X2 × . . .×Xn) is locally-compact.

The next Homework exercise should be attempted using the fact that groups

are homogeneous.

Homework 3.6. Prove that for a topological group to be locally-compact, it

is enough that it contain at least one open neighborhood with compact closure.

Exercise 3.5 tells us Rn is locally-compact for any n ∈ N. Perhaps you knew this already. Next we want to take subsets of Rn to reach much more complicated spaces.

Lemma 3.7. Suppose X is locally-compact and M ⊂ X is closed. Then M is locally-compact under the subspace topology.

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3 THE SECOND ISOMORPHISM THEOREM

Proof. Let x ∈ M be any point. Then we have x ∈ U ⊂ X for some open subset U ⊂ X such that U is compact. Observe the set M∩U is closed in the compact space U and therefore M ∩U is compact.

By definition of the subspace topology, M ∩U is a neighborhood of x in M. We want to show the closure ClM (M ∩ U) in M is compact. This is because M ∩ U ⊂ M ∩ U. The right-hand-side is closed in M and therefore contains ClM (M ∩ U). As a closed subspace of the compact space M ∩ U we know ClM (M ∩U) is compact in its own right.

We have just shown that local-compactness is closed-hereditary: Every closed

subspace of a locally-compact space is locally-compact in its own right. Local-

compactness is also open-hereditary for Hausdorff spaces. However the proof for

groups is a lot more straightforward, and does not actually need the Hausdorff

property. We first need something like the Half Lemma.

Lemma 3.8. Suppose U ⊂ G is an open neighborhood of the identity. There exists a smaller neighborhood V of the identity such that V ⊂ U.

Proof. By the Half Lemma we can choose W ⊂ G to be open at the identity such that W2 ⊂ U. The set V = W ∩ W−1 is open and nonempty since it contains the identity. Moreover by construction we have V −1 = V .

To see that V ⊂ U choose an arbitrary element g /∈ U. We will find an open neighborhood of g that is disjoint from V . This will show g /∈ V . We claim the neighborhood gV is sufficient.

Suppose otherwise. Then gV ∩ V 6= ∅. That means gw1 = w2 for some w1,w2 ∈ V . Multiplying on both sides we get g = w2w−11 ∈ V V

−1 = V 2 ⊂ U by construction. Therefore g ∈ U, a contradiction.

Exercise 3.9. Suppose U ⊂ G is an open neighborhood of the identity. There exists a neighbourhood V of the identity such that V

−1 V ⊂ U.

We will use the previous lemma to establish local-compactness at the origin.

Then we will use translations to extend the result to general points and open

sets.

Lemma 3.10. Suppose G is a locally-compact group and U ⊂ G is an open neighborhood of the identity. Then the subspace U is locally-compact at the

identity.

Proof. Since G is locally-compact we have e ∈ W ⊂ G for some open subset W ⊂ G such that W is compact. Then U ∩ W is an open neighborhood of e.

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By the previous lemma we can select a further open neighborhood V such that

e ∈ V ⊂ U ∩W. Observe that since V ⊂ W we know V is compact. Now observe V is open in the subspace U and V is closed in the subspace U.

Therefore we have e ∈ V ⊂ ClU (V ) ⊂ V . As a closed subspace of the compact space V , the set ClU (V ) is compact in its own right.

Lemma 3.11. Suppose G is a locally-compact group and W ⊂ G is open. Then the subspace W is locally-compact.

Proof. Let g ∈ W . We may write W = gU for some open U at the identity. The left-translation λg induces a homeomorphism from U to W that sends e to

g. Apply Lemma 3.10 to get an open set A of U such that e ∈ A ⊂ ClU (A) ⊂ U and ClU (A) is compact.

We can apply λg to each part to get g ∈ λg(A) ⊂ λg(ClU (A)). Since homeomorphisms preserve openness λg(A) is an open neighborhood of g in W .

Since homeomorphisms preserve compactness we know λg(ClU (A)) is compact.

But since homeomorphisms also commute with taking the closure we have as

well that λg(ClU (A)) = ClW (λg(A)) is the closure of λg(A) in W . So λg(A)

witnesses now W is locally-compact at g, as required.

So we may construct locally-compact spaces by starting with some finite

power Rn, selecting open or closed subspaces, taking finite powers of those subspaces, then selecting closed or open subsets of those powers, and repeating

the process a finite number of times. This method can be used to approach the

following exercise and in particular parts (4) − (6), where it will be helpful to recall that the determinant function Matn(R) → R is continuous.

Homework 3.12. Prove that the following groups (wrt the usual topologies if

not otherwise specified) are locally-compact:

(1) R and C under addition

(2) R 6=0, C 6=0 and T under multiplication

(3) Any group with a discrete or trivial topology

(4) The group Matn(R) of invertible matrices under the Euclidean topology

(5) The subgroup SMatn(R) ⊂ Matn(R) of matrices of determinant 1

(6) The subgroup OMatn(R) ⊂ Matn(R) of matrices M such that MtM = I

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Homework 3.13. Prove that if G is a locally-compact group and N ⊂ G a closed normal subgroup, that G/N is locally-compact.

At this stage we should provide an example of what non-locally-compact

spaces look like. We know the finite-dimensional vector spaces Rn are locally compact. Our example will be an infinite-dimensional vector space. We will need

the following characterisation of compact spaces. This same characterisation will

also be useful in a later section.

Definition 3.14. The following is called the Cantor property for the topo-

logical space X.

Whenever F is a family of closed subsets of X with F1 ∩F2 ∩ . . .∩Fn 6= ∅ for each choice of finitely many F1,F2, . . . ,Fn ∈F, then we have

⋂ F 6= ∅ as well.

Homework 3.15. Having the Cantor property is equivalent to being compact.

For our example `∞ is the Banach space of bounded real sequences with

norm ‖x‖ = sup{x1,x2, . . .}. Let each en = (0, . . . , 0, 1, 0, . . .) be the ‘coordinate vector’ whose nth entry is 1 and every other entry is zero. We will first show

the closed unit ball B(0, 1) = {x ∈ `∞ : ‖x‖ ≤ 1} is not compact. To that end define the family of subsets Fn = {en,en+1, . . .} of B(0, 1).

It is easy to see that F1 ⊃ F2 ⊃ F3 ⊃ . . . and therefore F = {F1,F2, . . .} satisfies the finite-intersection condition from Definition 3.14. Moreover

⋂ F =

∅. So once we show each Fn is closed, Homework 3.15 will tell us B(0, 1) is not compact.

Observe that each d(en,em) = ‖en − em‖ = 1. By the triangle inequality, that means no ball of radius 1/2 can include more than one en. So let x /∈ Fn and consider the ball B(x, 1/2). If B(x, 1/2) ∩Fn is empty we are done. Otherwise there is a unique em ∈ B(x, 1/2). Then we may shrink the radius of the ball to exclude em by setting the radius equal to d(x,em) for example.

By a similar argument no closed ball B(x,ε) is compact for x ∈ `∞ and ε > 0. Now suppose we had some open set U ⊂ `∞ with compact closure. Then we could choose a point x ∈ U and find an open ball such that x ∈ B(x,ε) ⊂ U. Taking closures would give x ∈ B(x,ε) ⊂ U. Then B(x,ε) would be a closed subset of the compact space U, and so would itself be compact. This contradicts

what we already proved.

Homework 3.16. Prove the group Q of rational numbers with the subspace topology fails to be locally-compact.

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Next we introduce another ‘smallness’ condition for topological spaces. The

notion of a σ-compact space will be needed to formulate the topological version

of the Second Isomorphism theorem.

Definition 3.17. The topological space X is called σ-compact to mean there

are compact subspaces K1,K2, . . . ⊂ X such that X = K1 ∪K2 ∪ . . . .

Often we can prove a given space is σ-compact by explicitly finding the

compact subsets K1,K2, . . . as in the definition. Observe that we can always

require these compact subsets to be nested by redefining L1 = K1,L2 = K1 ∪ K2,L3 = K1 ∪K2 ∪K3, . . .

Homework 3.18. Prove the following groups are σ-compact:

(1) Q, R and C under addition

(2) Q 6=0, R 6=0 and C 6=0 and T under multiplication

(3) Any compact group

(4) Any closed subgroup of a σ-compact group

(5) Any continuous image of a σ-compact group

(6) Any quotient of a σ-compact group

(7) Any finite product of σ-compact groups.

As before the group `∞ fails to be σ-compact. But the proof of that fact

goes beyond the scope of this course. Unlike before, the group Q is σ-compact − can you see why? Another example of a non σ-compact space is provided in the next exercise.

Exercise 3.19. Prove that no uncountable discrete space is σ-compact.

Exercise 3.20. Suppose G is a group and K ⊂ G is a compact subset. Define 〈K〉 ⊂ G to be the smallest subgroup that contains K. Such subgroups are called compactly generated. Prove that compactly generated subgroups are

σ-compact.

Next we introduce a notion of the ‘negligibly small’ subsets of a given topo-

logical space.

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Definition 3.21. Suppose X is a topological space and A ⊂ X is a subset. We say A is nowhere-dense in X to mean that (A)◦ = ∅. That means the closure of A has empty interior.

Homework 3.22. Determine, without proof, which of the following sets are

nowhere-dense in R2.

(1) The rational plane {(x,y) ∈ R2 : x,y ∈ Q}

(2) The rational x-axis {(x, 0) ∈ R2 : x ∈ Q}

(3) The unit circle {(x,y) ∈ R2 : x2 + y2 = 1}

(4) The open disc {(x,y) ∈ R2 : x2 + y2 ≤ 1}

(5) The rational disc {z = reiθ : r,θ ∈ Q and r < 1}

Observe it makes no sense to say ‘the set A is nowhere dense’ without spec-

ifying which space it is nowhere dense in. For example consider the spaces

N ⊂ Z ⊂ R. It can be shown that N is nowhere-dense in R but that N is not-nowhere dense in Z.

Exercise 3.23. Prove the set A ⊂ X is nowhere dense in X if and only if its complement has dense interior.

Definition 3.24. The first category subspaces of the topological space X are

the countable unions of nowhere-dense subsets of X. First category subsets are

sometimes called meagre. Any subspace of X which is not first category in X

is said to be of second category in X.

Warning: The word ‘category’ means something entirely different in other

branches of mathematics. In Category Theory, a category is something like a

family of sets, along with a collection of maps between them, which are closed

under composition.

Just like with nowhere-dense sets, it means nothing to say ‘the set A is

first category’ or ‘the set A is second category’. We must always specify which

ambient space is intended.

Exercise 3.25. Find a topological space which is meagre in itself. Prove that

if a space is meagre in itself, then every subspace is meagre.

The Baire Category Theorem states that certain ‘well-behaved’ spaces can-

not be composed of only countably many negligibly small pieces. To prove it

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we need the following lemma − which can be safely replaced by Lemma 3.8 if we are only interested in groups.

Lemma 3.26. Suppose X is regular and x ∈ U ⊂ X for some point x ∈ X and open set U ⊂ X. Then there exists an open set V ⊂ X such that x ∈ V ⊂ U

Proof. Let x and U be as above. Then the closed set Uc does not contain x. By

regularity there exist disjoint open sets V and W such that x ∈ V and Uc ⊂ W . Then V ⊂ Wc and since W is open, Wc is closed. Thus V ⊂ Wc. Since Wc is disjoint from Uc we know V is too. But this means V ⊂ U as required.

Theorem 3.27. (Baire category theorem): No regular locally-compact topo-

logical space (or group) is meagre in itself.

Proof. Suppose the space X can be written in the form ⋃∞ i=0 Di where each Di

is nowhere-dense. First replace each Di with its closure. Then we still have

X = ⋃∞ i=0 Di. By De Morgan’s law we have ∅ =

(⋃∞ i=0 Di

)c = ⋂∞ i=0 D

c i where

each Dci is open and dense in X. Letting Bi = D c i , we have already shown⋂∞

i=0 Bi = ∅. We will derive a contradiction. Take some x0 in X. This x0 is contained in an open subset U0 such that U0

is compact. Since B0 is dense U0 ∩B0 cannot be empty. So there exists some x1 ∈ U0 ∩B0. Since the open set U0 ∩B0 contains x1 we can use Lemma 3.26 to find an open set U1 such that x1 ∈ U1 ⊂ U0 ∩B0

Again U1∩B1 is nonempty, so we can find a point x2 ∈ U1∩B1 and an open set U2 such that x2 ∈ U2 ⊂ U1 ∩ B1 Proceeding inductively we attain a chain of nonempty closed sets U1 ⊇ U2 ⊇ U3 ⊇ . . . of the compact U0.

Since U0 is compact, by the Cantor property the intersection is nonempty.

Finally note that since each Ui ⊂ Bi we have that ⋂∞ i=0 Bi contains the nonempty⋂∞

i=0 Ui. This contradicts our assumption that ⋂∞ i=0 Bi = ∅.

Theorem 3.28. (Open Mapping Theorem) Suppose the group D is locally-

compact and σ-compact; and the group E is locally-compact and Hausdorff.

Then every surjective morphism φ : D → E is open.

Proof. Let U ⊂ D be any open neighborhood of the identity. By local compact- ness there is some neighborhood V of the identity such that V is compact and

contained in U. Without loss of generality (see exercise 3.9), we may select this

V such that V −1

V ⊂ U. Then the family of sets {xV : x ∈ D} forms an open cover of D.

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Since D is σ-compact we can write D = K1 ∪K2 ∪ . . . where each Kn ⊂ D is compact. Each Kn is covered by finitely many elements of {xV : x ∈ D}. Putting these finite subcovers together gets a countable subcover {xiV : i ∈ N} of D. This means we can write D =

⋃∞ i=1(xiV ).

The morphism φ: D → E is surjective. So we can write

E = φ(D) = φ ( ∞⋃ i=1

(xiV ) )

=

∞⋃ i=1

φ(xiV ) =

∞⋃ i=1

xiφ(V ) =

∞⋃ i=1

xiφ(V )

.

Since V is compact so is φ(V ). Hence φ(V ) ⊂ E is closed because E is Haus- dorff. Since translations are homeomorphisms, either each xiφ(V ) is nowhere-

dense or none are. In the first case we have written E as a countable union

of closed sets with empty interior. This is not allowed by the Baire Category

Theorem.

We conclude that each xiφ(V ) has nonempty interior. In particular φ(V )

has nonempty interior. That means W ⊂ φ(V ) for some nonempty open set W ⊂ E. Choose any g ∈ V such that φ(g) ∈ W . We can write:

φ(g) ∈ W ⊂ φ(V ) We just proved this.

eE ∈ φ(g)−1W ⊂ φ(g)−1φ(V ) Multiplying each part by φ(g)−1.

eE ∈ φ(g−1)W ⊂ φ(g−1)φ(V ) Because φ is a group-homomorphism.

eE ∈ φ(g−1)W ⊂ φ(g−1V ) Again because φ is a group-homomorphism.

eE ∈ φ(g−1)W ⊂ φ(V −1 V ) Since g ∈ V .

Recall how we chose V ⊂ D such that V −1

V ⊂ U. So the final line implies that eE ∈ φ(g−1)W ⊂ φ(U). Since φ(g−1)W is open this says that eE is an interior element of φ(U).

But U ⊂ D was an arbitrary open neighborhood of the identity. Thus we have proved that eE is in the interior of the image of every open neighborhood

of eD. Homework 2.10 completes the proof to show φ: D → E is an open morphism.

Since the hypotheses for the Open Mapping Theorem are rather complicated,

it is helpful to look at an example. Consider the morphism f : R → T defined earlier, which takes each x ∈ R to f(x) = e2πix.

You have already proved the domain is locally-compact and σ-compact. The

range is Hausdorff and not only locally-compact but compact, as a closed and

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bounded subset of the plane. Since all conditions are met, the Open Mapping

Theorem tells us f is an open morphism. This is exactly what we needed earlier

to conclude that R/Z ∼= T. We might also look at the corollary as a ‘reality check’. Do you see why this

follows from the Open Mapping Theorem?

Corollary 3.29. Every surjective morphism between compact Hausdorff groups

is an open morphism.

Thus far, we have developed the topological theory necessary for the Second

Isomorphism Theorem. Before we get started on the algebraic theory, we will

outline what the theorem means when there is no topology present

Suppose we have a group G and a normal subgroup N ⊂ G. We can think of the quotient group G/N as being formed by identifying any pair of elements

g,h ∈ G by saying g ∼ h if and only if gh−1 ∈ N. This defines an equivalence relation on G, and we take the ∼ equivalence classes as the elements of G/N. The fact that N ⊂ G is normal ensures that it makes sense to multiply classes. You should convince yourself this construction of G/N is equivalent to the earlier

one involving cosets.

Now suppose F ⊂ G is a subgroup. We would like to impose the same relation on the elements of F . That is to say identify two elements a,b ∈ F by saying a ∼ b if and only if ab−1 ∈ N. Once we show multiplication of these new classes is well-defined, we can again define a group − call it F∼ − whose elements are the ∼ equivalence classes.

We would like the group F∼ to ‘play nice’ with the machinery of group

theory. So we would like to express it as a quotient group of F. Our first guess

might be F/N. But do you see the problem here? Quotient groups are only

defined when the right-hand-side is a normal subgroup of the left-hand-side.

And we have no reason to believe that N ⊂ F. Therefore, in order to express F∼ as something like F/N, we can do one of

two things: We can either shrink the right-hand-side to be a normal subgroup

contained in F. Or we can expand the left-hand-side to contain N.

If we try expanding, our first idea might be FN/N. You should check that

FN is indeed a group and N ⊂ FN a normal subgroup. You should also check that FN is the smallest subgroup of G that contains N as a normal subgroup.

Thus we have expanded F by as little as possible.

If we try shrinking, our first idea might be F/(F∩N). As before, you should check that F ∩N is indeed a normal subgroup of F. You should also check it is

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the largest normal subgroup of F that is contained in N. Thus we have shrunk

N by as little as possible.

We can check the shrinking approach works by considering the surjective

group-homomorphism φ: F → F∼ that sends every element of F to its ∼ class. According to the The First Isomorphism Theorem we know F∼ ∼= F/ ker φ. Once we compute that ker φ = F ∩N have the result.

What the Second isomorphism theorem says is that the expanding approach

works equally well − and in fact the groups FN/N and F/(F ∩N) are isomor- phic. Thus F∼ can be expressed as a quotient in either of the two ways.

The proof works by defining a group-homomorphism F → FN/N and com- puting the kernel as F ∩N. We then apply the First Isomorphism Theorem to see F/(F ∩N) ∼= FN/N as groups. As you can imagine this becomes a lot more complicated when we require isomorphism as topological groups.

The next result is quite subtle and a little long-winded. It essentially

amounts to a more convenient way to express the group FN/N. However it

does a lot of the work setting up a short proof of the topological Second Iso-

morphism Theorem.

The setup is the same as described above: We have a group G, a subgroup

F ⊂ G, and a normal subgroup N ⊂ G. It follows that FN ⊂ G is also a subgroup. By restricting the quotient map π : G → G/N to FN we can define the quotient group FN/N. By restricting π to F we can define the subgroup

π(F) ⊂ π(G) = G/N. All relevant groups and morphisms are shown in the diagram, where ϕ is just the restriction of π to FN.

FN F

FN/N π(F)

ϕ

ι

π

ψ

We define the map ψ as follows. Whenever g ∈ FN we have ψ(gN) = π(g). We want to show ψ is an isomorphism. To that end we must check ψ is well

defined − that means if gN = hN we must also have π(g) = π(h). We must also check ψ is a group-isomorphism and a homeomorphism.

Lemma 3.30. The map ψ given by gN 7→ π(g) is well-defined.

Proof. Suppose g,h ∈ FN are such that gN = hN.

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gN = hN We assume this.

=⇒ h−1(gN) = h−1(hN) Multiplying by h−1 on both sides.

=⇒ (h−1g)N = N Simplifying the right-hand-side.

=⇒ h−1g ∈ N Using how e ∈ N.

=⇒ π(h−1g) = e Applying π to both sides.

=⇒ π(h)−1π(g) = e Because π is a group-homomorphism.

=⇒ π(g) = π(h) Multiplying by π(h)−1 on both sides.

=⇒ ψ(gN) = ψ(hN) From the definition of ψ.

Therefore the map ψ is well-defined.

Next, we want to construct the inverse map to ψ, namely π(g) 7→ gN. On the face of it, ψ−1 should take π(g) as input, figure out what g is, and then use

g to construct gN. This is, however, impossible if π(g) = π(h) for some distinct

elements g,h ∈ F. But as it turns out, the fact that the inverse returns gN rather than g is important.

Lemma 3.31. The map ψ−1 given by π(g) 7→ gN is well-defined.

Proof. Suppose g,h ∈ F are such that π(g) = π(h).

π(g) = π(h) We assume this.

π(h−1g) = e Because π is a group-homomorphism..

h−1g ∈ N Because π : G → G/N is the quotient map.

h−1g = n For some n ∈ N

h−1gN = nN Multiplying by N on both sides.

h−1gN = N Since N is a group.

gN = hN Multiplying by h on both sides.

Therefore the inverse map is well-defined.

Lemma 3.32. The map ψ defined above is a group-isomorphism.

Proof. Since ψ is bijective it is enough to prove it is a group-homomorphism.

Take gN,hN ∈ FN/N. Then we have ψ(gN ∗ hN) = ψ(ghN) = π(gh) = π(g)π(h) = ψ(gN)ψ(hN), as required.

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We show ψ is a homeomorphism in two almost identical steps, corresponding

to how images and preimages of open sets are open.

Lemma 3.33. The map ψ defined above is continuous

Proof. Consider the same diagram as before, with an extra map shown.

FN F

FN/N π(F)

ϕ ψ◦ϕ

ι

π

ψ

From Homework 2.27 we know ψ ◦ ϕ is a morphism if and only if ψ is. Fortunately, we already know the form of the map ψ ◦ ϕ. For each g ∈ FN we have (ψ ◦ ϕ)(g) = ψ(ϕ(g)) = ψ(gN) = π(g). Therefore (ψ ◦ ϕ)(g) = π(g). It follows the diagonal arrow is just the restriction of the morphism π to FN.

Since π is a morphism so is the restriction. In particular the restriction is

continuous.

Lemma 3.34. The map ψ defined above is open

Proof. Consider the same diagram as before, with an extra map shown, going

in the other diagonal.

FN F

FN/N π(F)

ϕ ψ−1◦π

ι

π

ψ−1

The map ψ being open is equivalent to ψ−1 being continuous. Again Home-

work 2.27 tells us ψ−1 is a morphism if and only if ψ−1 ◦π is. Again we already know the form of the map ψ−1◦π. By definition the map ψ−1 takes π(g) ∈ π(F) to gN ∈ FN/N for each g ∈ F. Therefore each (ψ−1 ◦ π)(g) = gN ∈ FN/N. But this is exactly the morphism ϕ ◦ ι. We conclude ψ−1 is a morphism and hence continuous.

Corollary 3.35. Suppose G is a group with subgroup F ⊂ G and normal subgroup N ⊂ G. The bottom row of the diagram

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3 THE SECOND ISOMORPHISM THEOREM

FN F

FN/N π(F)

π|F N

ι

πF

gN 7→π(g)

is a topological isomorphism.

Theorem 3.36. (Second Isomorphism Theorem) Suppose G is Hausdorff and

the subgroup F ⊂ G is locally-compact and σ-compact. Suppose moreover the normal subgroup N ⊂ G is closed and FN is locally-compact. Then we have the relation

FN

N ∼=

F

(F ∩N)

Proof. Consider the mapping diagram, where φ and π are the obvious quotient

maps; θ(g(F ∩ N)) = gN is the group-isomorphism established in the non- topological Second Isomorphism Theorem; and the isomorphism ψ−1 given by

π(g) 7→ gN has been established in Corollary 3.35.

F π(F)

F

F ∩N FN

N

π

φ ψ−1

θ

We have assumed FN is locally-compact. Since N ⊂ FN is closed, Home- work 3.13 says the quotient FN/N is also locally-compact. Homework 2.26 says

that since the normal subgroup is closed, the quotient is Hausdorff. Therefore

both groups on right-hand column are locally-compact and Hausdorff.

By the same token F ∩N is closed subgroup of F. Therefore F/(F ∩N) is locally-compact. We have also assumed F is σ-compact. By Homework 3.18 so

is the continuous image φ(F) = F/(F ∩N). We already know θ is a group-isomorphism. We have shown the domain

is locally-compact and σ-compact, and the range is locally-compact Hausdorff.

This means that, once we prove θ is continuous, we can invoke the Open Map-

ping Theorem to show it is open as well. This will prove it is a homeomorphism

and hence an isomorphism of topological groups.

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By Homework 2.27 we know θ is continuous if and only if θ◦φ is continuous. We claim that θ ◦φ = ψ−1 ◦π which we already know is continuous. To check these maps are the same take a general element g ∈ F and compute the two outputs

(θ ◦φ)(g) = θ(φ(g)) = θ(g(F ∩N)) = gN

(ψ−1 ◦π)(g) = ψ−1(π(g)) = gN

We conclude the maps are the same and both are continuous. Finally we apply

the Open Mapping Theorem to show θ is open. Hence θ is an open continuous

group-isomorphism. In other words an isomorphism of topological groups.

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4 TOPOLOGICAL VECTOR SPACES

4 Topological Vector Spaces

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5 THE CAYLEY-HAMILTON THEOREM

5 The Cayley-Hamilton Theorem

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6 THE ARZELÀ-ASCOLI THEOREM

6 The Arzelà-Ascoli theorem

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7 TYCHONOFF’S THEOREM IF TIME PERMITS

7 Tychonoff ’s Theorem if Time Permits

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