Chapter 2. Topological Spaces - wj32 · PDF fileChapter 2. Topological Spaces ... Let (M;d) be...

Chapter 2. Topological Spaces

Example 1. [Exercise 2.2] Show that each of the following is a topological space.

(1) Let X denote the set {1, 2, 3}, and declare the open sets to be {1}, {2, 3},{1, 2, 3}, and the empty set.

(2) Any set X whatsoever, with T = {all subsets of X}. This is called the discretetopology on X, and (X, T ) is called a discrete space.

(3) Any set X, with T = {∅, X}. This is called the trivial topology on X.(4) Any metric space (M,d), with T equal to the collection of all subsets of M

that are open in the metric space sense. This topology is called the metrictopology on M .

We only check (4). It is clear that ∅ and M are in T . Let U1, . . . , Un ∈ T and letp ∈ U1 ∩ · · · ∩ Un. There exist open balls B1 ⊆ U1, . . . , Bn ⊆ Un that contain p, so theintersection B1 ∩ · · · ∩Bn ⊆ U1 ∩ · · · ∩Un is an open ball containing p. This shows thatU1 ∩ · · · ∩ Un ∈ T . Now let {Uα}α∈A be some collection of elements in T which we canassume to be nonempty. If p ∈

⋃α∈A Uα we can choose some α ∈ A, and there exists an

open ball B ⊆ Uα containing p. But B ⊆⋃α∈A Uα, so this shows that

⋃α∈A Uα ∈ T .

Theorem 2. [Exercise 2.9] Let X be a topological space and let A ⊆ X be any subset.

(1) A point q is in the interior of A if and only if q has a neighborhood containedin A.

(2) A point q is in the exterior of A if and only if q has a neighborhood containedin X \ A.

(3) A point q is in the boundary of A if and only if every neighborhood of q containsboth a point of A and a point of X \ A.

(4) IntA and ExtA are open in X, while ∂A is closed in X.(5) A is open if and only if A = IntA, and A is closed if and only if A = A.(6) A is closed if and only if it contains all its boundary points, which is true if and

only if A = IntA ∪ ∂A.(7) A = A ∪ ∂A = IntA ∪ ∂A.

Proof. (1) and (2) are obvious. Let q ∈ ∂A and let N be a neighborhood of q. By (1)and (2), N is not contained in A or X \A, i.e. N contains a point of A and a point ofX \ A. Conversely, if every neighborhood of q contains both a point of A and a pointof X \ A then q /∈ (IntA ∪ ExtA) by (1) and (2). Parts (4) and (5) are obvious. If Ais closed and p ∈ ∂A then

p ∈ X \ (IntA ∪ ExtA) = X \ (IntA ∪ (X \ A)) ⊆ A.

If A contains all its boundary points then IntA ∪ ∂A ⊆ A, so A = IntA ∪ ∂A sinceit is always true that A ⊆ IntA ∪ ∂A. If A = IntA ∪ ∂A then A is closed since

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A = IntA∪ (X \ (IntA∪ExtA)) = X \ExtA and ExtA is open. This proves (6). For(7),

IntA ∪ ∂A = IntA ∪ (X \ (IntA ∪ ExtA)) = X \ ExtA = A

and A ∪ ∂A = IntA ∪ ∂A since IntA ∪ ∂A ⊆ A ∪ ∂A and p ∈ A ∪ ∂A implies thatp ∈ ∂A or p ∈ A \ ∂A = IntA. �

Theorem 3. [Exercise 2.10] A set A in a topological space X is closed if and only if itcontains all of its limit points.

Proof. Suppose A is closed and let q ∈ X \ A be a limit point of A. Since everyneighborhood of q contains both a point of X \A and a point of A, we have q ∈ ∂A ⊆ A.Conversely, suppose that A contains all of its limit points. If q ∈ ∂A \ A and N is aneighborhood of q then N contains a point of A which cannot be equal to q since q /∈ A.Therefore q is a limit point of A, and is contained in A. This shows that A contains allits boundary points, i.e. A is closed. �

Theorem 4. [Exercise 2.11] A subset A ⊆ X is dense if and only if every nonemptyopen set in X contains a point of A.

Proof. Suppose that A = X. Let U be a nonempty open set in X and let p be anypoint in U . Since ExtA is empty, no neighborhood of p is contained in X \ A. ButU is a neighborhood of p, so U contains a point of A. Conversely, suppose that everynonempty open set in X contains a point of A. Then every p ∈ X \ A is a limit pointof A, so p ∈ A and therefore X ⊆ A. �

Theorem 5. [Exercise 2.12] Let (M,d) be a metric space with the usual topology. Thefollowing are equivalent for a sequence {qi} and a point q in M :

(1) For every neighborhood U of q there exists an integer N such that qi ∈ U for alli ≥ N .

(2) For every ε > 0 there exists an integer N such that d(qi, q) < ε for all i ≥ N .

Proof. The direction (1)⇒ (2) is clear. Suppose (2) holds and let U be a neighborhoodof q. Since U is open, there exists an open ball B ⊆ U of radius r around q, so thereexists an integer N such that d(qi, q) < ε for all i ≥ N . But then qi ∈ B ⊆ U for alli ≥ N , which proves (1). �

Theorem 6. [Exercise 2.13] Let X be a discrete topological space. Then the onlyconvergent sequences in X are the ones that are “eventually constant”, that is, sequences{qi} such that qi = q for all i greater than some N .

Proof. Suppose that qi → q in X. Since {q} is open, there exists an integer N suchthat qi ∈ {q}, i.e. qi = q, for all i ≥ N . �

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Theorem 7. [Exercise 2.16] A map f : X → Y between topological spaces is continuousif and only if the inverse image of every closed set is closed.

Proof. Suppose that f is continuous and let A ⊆ Y be a closed set. Since Y \A is open,f−1(Y \ A) is open and therefore f−1(A) = X \ f−1(Y \ A) is closed. The converse issimilar. �

Theorem 8. [Exercise 2.18] Let X, Y , and Z be topological spaces.

(1) Any constant map f : X → Y is continuous.(2) The identity map Id : X → X is continuous.(3) If f : X → Y is continuous, so is the restriction of f to any open subset of X.(4) If f : X → Y and g : Y → Z are continuous, so is their composition g ◦ f :

X → Z.

Proof. For (1), suppose that f(x) = y for all x ∈ X. Let U ⊆ Y be an open set. Ify ∈ U then f−1(U) = X and if y /∈ U then f−1(U) = ∅; in either case, f−1(U) is open.Part (2) is obvious. Part (3) follows from the fact that the topology is closed underfinite intersections. For (4), let U ⊆ Z be an open set. Since g is continuous g−1(U) isopen, and since f is continuous f−1(g−1(U)) is open. But (g ◦ f)−1(U) = f−1(g−1(U)),so this proves that g ◦ f is continuous. �

Theorem 9. [Exercise 2.20] “Homeomorphic” is an equivalence relation.

Proof. For any space X the identity map Id : X → X is a homeomorphism betweenX and itself. If ϕ : X → Y is a homeomorphism then ϕ−1 : Y → X is also ahomeomorphism. Finally, if ϕ : X → Y and ψ : Y → Z are homeomorphisms thenψ ◦ ϕ : X → Z is a homeomorphism since (ψ ◦ ϕ)−1 = ϕ−1 ◦ ψ−1 is continuous. �

Theorem 10. [Exercise 2.21] Let (X1, T1) and (X2, T2) be topological spaces and letf : X1 → X2 be a bijective map. Then f is a homeomorphism if and only if f(T1) = T2in the sense that U ∈ T1 if and only if f(U) ∈ T2.

Proof. This is clear from the definition. �

Theorem 11. [Exercise 2.27] Let C = {(x, y, z) | max(|x| , |y| , |z|) = 1} and let ϕ :C → S2 be given by

ϕ(x, y, z) =(x, y, z)√x2 + y2 + z2

.

Then ϕ is a homeomorphism with inverse

ϕ−1(x, y, z) =(x, y, z)

max(|x| , |y| , |z|).

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Proof. Note that ϕ and ϕ−1 are continuous; it suffices to check

ϕ(ϕ−1(x, y, z)) =(x, y, z)√x2 + y2 + z2

= (x, y, z)

since (x, y, z) ∈ S2 implies that√x2 + y2 + z2 = 1 and

ϕ−1(ϕ(x, y, z)) =(x, y, z)

max(|x| , |y| , |z|)= (x, y, z)

since (x, y, z) ∈ C. �

Theorem 12. [Exercise 2.28] Let X denote the half-open interval [0, 1) ⊆ R, and letS1 denote the unit circle in R2. Define a map a : X → S1 by a(t) = (cos 2πt, sin 2πt).Then a is continuous and bijective but not a homeomorphism.

Proof. Let I = [0, 1/2), which is open in X. Since a(I) is not open in S1, the inverseof a cannot be continuous. �

Theorem 13. [Exercise 2.31]

(1) Every local homeomorphism is an open map.(2) Every homeomorphism is a local homeomorphism.(3) Every bijective continuous open map is a homeomorphism.(4) Every bijective local homeomorphism is a homeomorphism.

Proof. Let f : X → Y be a local homeomorphism and let U ⊆ X be an open set.Let y ∈ f(U) so that y = f(x) for some x ∈ U . There exists a neighborhood V ofx such that f(V ) is open and f |V : V → f(V ) is a homeomorphism; in particular,Ny = f(V ∩ U) ⊆ f(U) is open. Therefore f(U) =

⋃y∈f(U)Ny is open, which proves

(1). Parts (2) and (3) are obvious, and (4) follows from (1) and (3). �

Theorem 14. [Exercise 2.33] Let Y be a trivial topological space (that is, a set withthe trivial topology). Then every sequence Y converges to every point of Y .

Proof. Let {ri} be a sequence in Y and let r be any element of Y . If U is a neighborhoodof r then U = Y , so it is trivial that ri → r. �

Theorem 15. [Exercise 2.35] Suppose X is a topological space, and for every p ∈ Xthere exists a continuous function f : X → R such that f−1({0}) = {p}. Then X isHausdorff.

Proof. Let p, q be distinct points of X and let f : X → R be a continuous function suchthat f−1({0}) = {p}. Then f(q) 6= 0, and we can choose disjoint neighborhoods U of 0and V of f(q) in R. We have f−1(U)∩ f−1(V ) = ∅, while f−1(U) is a neighborhood ofp and f−1(V ) is a neighborhood of q. �

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Theorem 16. [Exercise 2.38] The only Hausdorff topology on a finite set is the discretetopology.

Proof. Let X be a finite set with a Hausdorff topology T . By Proposition 2.37, everyone-point set in X is closed. But every subset of X can be written as a finite union ofone-point sets, so every subset of X is closed and T must be the discrete topology. �

Example 17. In each of the following cases, the given set B is a basis for the giventopology.

(1) M is a metric space with the metric topology, and B is the collection of all openballs in M .

(2) X is a set with the discrete topology, and B is the collection of all one-pointsubsets of X.

(3) X is a set with the trivial topology, and B = {X}.

(1) follows trivially from the definition of the metric topology. For (2), let U be an openset in X. If x ∈ U then x ∈ {x} ⊆ U , so B is a basis. For (3), if U is an open set thenit is either empty or equal to X. In the latter case, for any x ∈ U we have x ∈ X ⊆ U ,so B is a basis.

Theorem 18. [Exercise 2.42] The following collections are bases for the Euclideantopology on Rn:

(1) B1 = {Cs(x) : x ∈ Rn and s > 0}, where Cs(x) is the open cube of side s aroundx:

Cs(x) = {y = (y1, . . . , yn) : |xi − yi| < s/2, i = 1, . . . , n}(2) B2 = {Br(x) : r is rational and x has rational coordinates}.

Proof. Let U be an open set and let x ∈ U . There exists some open ball B ⊆ U ofradius r around x. For (1), the open cube Cr/

√2(x) is an element of B1 that is contained

in B. For (2), choose any rational r′ with 0 < r′ ≤ r/2 and any point x′ ∈ Br′(x) withrational coordinates; then Br′(x

′) ⊆ B is an element of B2 that contains x. �


(1) Every second countable space has a countable dense subset.(2) A metric space is second countable if and only if it has a countable dense subset.

Proof. Let X be a second countable space and let B be a countable basis for X. LetE be the countable set formed by choosing a point from each B ∈ B. Let U be anonempty open set in X and let x ∈ U . There exists a B ∈ B with x ∈ B and B ⊆ U ,so U contains a point from E. This shows that E is a countable dense subset of X.

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Now let (M,d) be a metric space with a countable dense subset E. We want to showthat the countable set

B = {Br(p) : (p, r) ∈ E ×Q}is a basis for M . Let U be an open set in M , let x ∈ U , and let B ⊆ U be an open ballof radius r containing x. Since E is dense in M , there exists a point p ∈ E ∩ Br/2(x).Let r′ be a rational number with d(p, x) < r′ < r/2. Then Br′(p) is an element of Bwith x ∈ Br′(p) and Br′(p) ⊆ B, since

d(x, y) ≤ d(x, p) + d(p, y) < r

for every y ∈ Br′(p). �

Theorem 20. [Exercise 2.54] A topological space X is a 0-manifold if and only if it isa countable discrete space.

Proof. Let x ∈ X and let N be a neighborhood of x homeomorphic to an open subsetof R0, i.e. a point. Then N must contain exactly one element, so N = {x}. This showsthat every one-point subset of X is open, and therefore X is discrete. Also, any secondcountable discrete space must also be countable, so X is a countable discrete space. �

Lemma 21. Let X and Y be topological spaces and let f : X → Y be a continuousmap. If A ⊆ X and B ⊆ Y are subsets with f(A) ⊆ B, then f(A) ⊆ B.

Proof. Suppose that f(x) ∈ ExtB for some x ∈ A. Choose a neighborhood U of f(x)such that U ⊆ Y \B. Then f−1(U) is a neighborhood of x such that f−1(U) ⊆ X \A,which contradicts the fact that x ∈ A. �

Lemma 22. Let (X, T1) and (X, T2) be topological spaces with the same underlying set.Let B1 and B2 be bases for T1 and T2 respectively. Then T1 = T2 if and only if B2 ⊆ T1and B1 ⊆ T2.

Proof. If T1 = T2 then it is clear that B2 ⊆ T2 = T1 and B1 ⊆ T1 = T2. Conversely,suppose that B2 ⊆ T1 and B1 ⊆ T2. Every open set in T1 can be written as the unionof elements from B1, which are open in T2 by assumption. Therefore every open set inT1 is also open in T2. Similarly, every open set in T2 is also open in T1. �

Lemma 23. Let (M1, d1) and (M2, d2) be metric spaces with M = M1 = M2 (butdifferent metrics). Then the following are equivalent:

(1) The two spaces have the same topology.(2) Let x ∈M . Every open ball in M1 around x contains an open ball in M2 around

x, and vice versa.

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Proof. Suppose that the two spaces have the same topology. If B is an open ball in M1

around x ∈ M then B is open in M2, so there is an open ball B′ ⊆ B in M2 aroundx. An identical argument holds when B is an open ball in M2. Now suppose that (2)holds and let U be open in M1. If x ∈ U then there exists an open ball B ⊆ U in M1

around x, so there is an open ball B′ ⊆ B in M2 around x. This shows that U is openin M2. Similarly, every set that is open in M2 is also open in M1. This shows that thetwo spaces have the same topology. �

Example 24. [Problem 2-1] Let X be an infinite set. Consider the following collectionsof subsets of X:

T1 = {U ⊆ X : X \ U is finite or is all of X} ;

T2 = {U ⊆ X : X \ U is infinite or is empty} ;

T3 = {U ⊆ X : X \ U is countable or is all of X} .For each collection, determine whether it is a topology.

T1 is a topology; T2 is not a topology; T3 is a topology.

Theorem 25. [Problem 2-3] Let X be a topological space and let B be a subset of X.

(1) X \B = X \ IntB.(2) Int(X \B) = X \B.

Proof. Suppose that x ∈ X \B. We can assume that x ∈ B, for otherwise x ∈ X \B ⊆X \ IntB. Then x is a limit point of X \B, so any neighborhood of x contains a pointof X \ B. This shows that x /∈ IntB, i.e. x ∈ X \ IntB. Conversely, suppose that

x /∈ X \B. Then there is a neighborhood N of x that does not contain a point of X \B.Then N ⊆ B, which shows that x ∈ IntB. This proves (1). For (2), apply (1) withX \B to get B = X \ Int(X \B) and take complements to get X \B = Int(X \B). �

Example 26. [Problem 2-4] Let X = {1, 2, 3}. Give a list of topologies on X such thatany topology on X is homeomorphic to exactly one on your list.

The topologies up to homeomorphism are:

T1 = {∅, {1, 2, 3}}T2 = {∅, {1} , {1, 2, 3}}T3 = {∅, {1} , {1, 2} , {1, 2, 3}}T4 = {∅, {1} , {2, 3} , {1, 2, 3}}T5 = {∅, {1} , {1, 2} , {1, 3} , {1, 2, 3}}T6 = {∅, {1} , {2} , {1, 2} , {1, 2, 3}}T7 = {∅, {1} , {2} , {1, 3} , {1, 2, 3}}

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T8 = {∅, {1} , {2} , {1, 2} , {1, 3} , {1, 2, 3}}T9 = {∅, {1} , {2} , {3} , {1, 2} , {1, 3} , {2, 3} , {1, 2, 3}} .

Theorem 27. [Problem 2-7] Suppose X is a Hausdorff space and A ⊆ X. If p ∈ X isa limit point of A, then every neighborhood of p contains infinitely many points of A.

Proof. Construct a sequence of distinct points {qn} in A as follows. Choose any pointq1 ∈ A not equal to p. Suppose we have chosen the distinct points q1, . . . , qn, none ofwhich are equal to p. For each i, there exist neighborhoods Ui of qi and Vi of p suchthat Ui ∩ Vi = ∅. Then V = V1 ∩ · · · ∩ Vn is a neighborhood of p, so V contains somepoint qn+1 ∈ A not equal to p. Furthermore, qn+1 is not equal to any qi for i ≤ n sinceV ∩ Ui = ∅ for every i. �

Theorem 28. [Problem 2-8] Let X be a Hausdorff space, let A ⊆ X, and let A′ denotethe set of limit points of A. Then A′ is closed in X.

Proof. Let x be a limit point of A′ and let E be a neighborhood of x. Then there existssome y ∈ E ∩ A′ not equal to x. Choose neighborhoods U of x and V of y such thatU ∩ V = ∅ and U, V ⊆ E. Since y is a limit point of A′, there exists some z ∈ V ∩ Anot equal to y. Then z 6= x since U and V are disjoint, which proves that x ∈ A′.Therefore A′ is closed in X. �

Theorem 29. [Problem 2-9] Let X be a discrete space, Y be a space with the trivialtopology, and Z be any topological space. Any maps f : X → Z and g : Z → Yare continuous, and if Z is Hausdorff, then the only continuous maps h : Y → Z areconstant maps.

Proof. Suppose that Z is Hausdorff and h : Y → Z is non-constant continuous map.Choose distinct points z1, z2 ∈ Z such that z1 = h(y1) and z2 = h(y2) for some y1, y2 ∈Y , and choose neighborhoods U1 of z1 and U2 of z2 with U1 ∩ U2 = ∅. The set h−1(U1)is nonempty and open in Y , so h−1(U1) = Y . But then z2 ∈ h(Y ) = h(h−1(U1)) = U1,which contradicts the fact that U1 ∩ U2 = ∅. �

Theorem 30. [Problem 2-11] Let f : X → Y be a continuous map between topologicalspaces, and let B be a basis for the topology of X. Let f(B) denote the collection{f(B) : B ∈ B} of subsets of Y . If f is surjective and open, then f(B) is a basis forthe topology of Y .

Proof. Every set in f(B) is open since f is an open map. Let U be an open set inY and let y ∈ U . Since f is surjective we have y = f(x) for some x, so f−1(U) isa neighborhood of x. There exists some B ∈ B with x ∈ B and B ⊆ f−1(U), sof(B) ∈ f(B) with y ∈ f(B) and f(B) ⊆ f(f−1(U)) = U . �

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Theorem 31. [Problem 2-12] Suppose X is a set, and B is any collection of subsets ofX whose union equals X. Let T be the collection of all unions of finite intersections ofelements of B.

(1) T is a topology. (It is called the topology generated by B, and B is called asubbasis for T .)

(2) T is the “smallest” topology for which all the sets in B are open. More precisely,T is the intersection of all topologies containing B.

Proof. It is clear that ∅, X ∈ T , and that arbitrary unions of open sets are also opensets. If ⋃

α∈A

(Uα,1 ∩ · · · ∩ Uα,mα) and⋃β∈B

(Uβ,1 ∩ · · · ∩ Uβ,nβ)

are elements of T , then their intersection is⋃(α,β)∈A×B

Uα,1 ∩ · · · ∩ Uα,mα ∩ Uβ,1 ∩ · · · ∩ Uβ,nβ ,

which is also in T . This proves that T is a topology. To prove (2), we only need toshow that for every topology T ′ containing B, we have T ⊆ T ′. But this is clear fromthe fact that T ′ is closed under arbitrary unions and finite intersections. �

Theorem 32. [Problem 2-13] Let X be a totally ordered set with at least two elements.For any a ∈ X, define sets L(a), R(a) ⊆ X by

L(a) = {c ∈ X : c < a} ,R(a) = {c ∈ X : c > a} .

Give X the topology generated by the subbasis {L(a), R(a) : a ∈ X}, called the ordertopology.

(1) Each set of the form (a, b) is open in X and each set of the form [a, b] is closed.(2) X is Hausdorff.

(3) For any a, b ∈ X, (a, b) = [a, b].

Proof. We can write (a, b) = R(a) ∩ L(b) and [a, b] = (X \ L(a)) ∩ (X \ R(b)), whichshows that (a, b) is open and [a, b] is closed. For (2), let a and b be distinct points inX. If there exists some c with a < c < b then we have the neighborhoods L(c) of a andR(c) of b with L(c) ∩ R(c) = ∅. Otherwise, the neighborhoods L(b) of a and R(a) of bsatisfy L(b) ∩ R(a) = ∅. This shows that X is Hausdorff. For (3), if E is a closed setcontaining (a, b) then [a, b] ⊆ E since a and b are boundary points of (a, b). But [a, b]

is closed, so (a, b) = [a, b]. �

Theorem 33. [Problem 2-15] Let X be a first countable space.

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(1) For any set A ⊆ X and any point p ∈ X, we have p ∈ A if and only if there isa sequence {pn}∞n=1 in A such that pn → p.

(2) For any space Y , a map f : X → Y is continuous if and only if f takesconvergent sequences in X to convergent sequences in Y .

Proof. For (1), if p ∈ A then the constant sequence pn = p converges to p. Otherwise,p /∈ A and p is a limit point of A. Let B = {B1, B2, . . . } be a countable neighborhoodbasis for p and let En = B1 ∩ · · · ∩ Bn. For each n the set En is a neighborhood of p,so we can choose a point pn ∈ En ∩ A since p is a limit point of A. We want to showthat the sequence {pn} converges to p. Let U be a neighborhood of p. There exists aBN ∈ B with BN ⊆ U , and for all n ≥ N we have pn ∈ U since

pn ∈ En ⊆ B1 ∩ · · · ∩Bn ⊆ B1 ∩ · · · ∩BN .

Conversely, suppose that there is a sequence {pn} in A with pn → p. We may assumethat pn 6= p for all n, for otherwise p ∈ A and we are done. If U is a neighborhoodof p then there exists an integer N such that pn ∈ U for all n ≥ N . In particular,pN ∈ U ∩ A and pN 6= p by our previous assumption, which proves that p is a limitpoint of A. �

Theorem 34. [Problem 2-16] If X is a second countable topological space, then everycollection of disjoint open subsets of X is countable.

Proof. Let B be a countable basis for X and let {Uα}α∈A be a collection of disjointnonempty open subsets. For each α, choose a point xα ∈ Uα; there exists an elementBα ∈ B with xα ∈ Bα and Bα ⊆ Uα. Since Bα 6= Bβ whenever α 6= β, we have abijection between A and the countable set {Bα}. �

Lemma 35. Every metric space M is first countable.

Proof. Let p ∈M and consider the countable collection of open balls{B1/n(p) : n ∈ Z+

}.

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Theorem 36. Let X and Y be topological spaces with the property that every point inX has a neighborhood homeomorphic to an open set in Y . Then X is first countable ifY is first countable.

Proof. Let x ∈ X. There exists a neighborhood N of x with a homeomorphism f :N → f(N), and there exists a countable neighborhood basis B of f(x). If U is aneighborhood of x then f(U ∩ N) is a neighborhood of f(x), so there exists a B ∈ Bwith B ⊆ f(U ∩N). Then f−1(B) ⊆ U ∩N , which shows that {f−1(B) : B ∈ B} is acountable neighborhood basis of x. �

Corollary 37. [Problem 2-21] All locally Euclidean spaces and metric spaces are firstcountable.

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Example 38. [Problem 2-22] Let X = R2 as a set, but with the topology determinedby the following basis:

B = {sets of the form {(c, y) : a < y < b} , for fixed a, b, c ∈ R} .Determine which (if either) of the identity maps X → R2, R2 → X is continuous.

The map X → R2 is not continuous because the unit open ball B is not open in X.In particular, (1, 0) ∈ B but there is no element of B that both contains (1, 0) and is asubset of B. For if U = {(1, y) : a < y < b} ∈ B contains (1, 0) then there are no valuesof a, b for which U is a subset of B. The map R2 → X is not continuous either becauseno set of the form {(c, y) : a < y < b} is open in the usual metric topology on R2.

Theorem 39. [Problem 2-23] Any manifold has a basis of coordinate balls.

Proof. Suppose that (M, T ) is an n-manifold where T = {Uα}α∈A. For every p ∈ M ,choose a neighborhood Np of p that admits a homeomorphism ϕp : Np → Vp where Vpis an open subset of Rn. Then for every Uα ∈ T that contains p, the set ϕp(Uα ∩ Np)is open in Rn, so there exists an open ball B ⊆ ϕp(Uα ∩ Np) around ϕp(p); let Bp,α =ϕ−1p (B) ⊆ Uα ∩Np. Since ϕp|Bp,α : Bp,α → B is a homeomorphism, Bp,α is a coordinateball. By our construction, the set B = {Bp,α : (p, α) ∈M × A and p ∈ Uα} is a basisfor M . �

Theorem 40. [Problem 2-24] Suppose X is locally Euclidean of dimension n, andf : X → Y is a surjective local homeomorphism. Then Y is also locally Euclidean ofdimension n.

Proof. Let y ∈ Y so that y = f(x) for some x ∈ X. There exists a neighborhood N of xand homeomorphism ϕ : N → U where U is open in Rn, and there exists a neighborhoodN ′ of x such that f(N ′) is open and f |N ′ : N ′ → f(N ′) is a homeomorphism. Thenf(N ∩N ′) is a neighborhood of y and

ϕ|N∩N ′ ◦ (f |N∩N ′)−1 : f(N ∩N ′)→ ϕ(N ∩N ′)is a homeomorphism from a neighborhood of y to an open subset of Rn. �

Theorem 41. [Problem 2-25] Suppose M is an n-dimensional manifold with boundary.Then IntM is an n-manifold and ∂M is an (n− 1)-manifold (without boundary).

Proof. If p ∈ IntM then for some homeomorphism ϕ : N → U where p ∈ N and U isan open subset of Hn we have ϕ(p) ∈ IntHn. There exists an open ball B ⊆ IntHn ∩Uaround ϕ(p), so ϕ|ϕ−1(B) is a homeomorphism from the neighborhood ϕ−1(B) of p tothe open subset B of Rn. This shows that IntM is an n-manifold. Similarly, if p ∈ ∂Mthen for some homeomorphism ϕ : N → U where p ∈ N and U is an open subset of Hn

we have ϕ(p) ∈ ∂Hn. Let ψ : ∂Hn → Rn−1 be the projection onto Rn−1 that discardsthe last coordinate. There exists an open ball B in Rn−1 around ψ(ϕ(p)) such that

12

B ⊆ ψ(U ∩∂Hn), so some restriction of ψ◦ϕ is a homeomorphism from a neighborhoodof p to B. This shows that ∂M is an (n− 1)-manifold. �

Chapter 3. New Spaces from Old

Theorem 42. [Exercise 3.1] Let X be a space and let A ⊆ X be any subset. Define

TA = {U ⊆ A : U = A ∩ V for some open set V ⊆ X} .

Then TA is a topology on A.

Proof. It is clear that ∅, A ∈ TA. Let {Uα}α∈A be a subset of TA; for each α we haveUα = A ∩ Vα for set Vα open in X. Then⋃

α∈A

Uα =⋃α∈A

A ∩ Vα = A ∩⋃α∈A

Vα

which is an element of TA since⋃α∈A Vα is open in X. Finally, if U1, . . . , Un ∈ TA then

for each i we have Ui = A ∩ Vi for some Vi open in X, so

U1 ∩ · · · ∩ Un = A ∩ (V1 ∩ · · · ∩ Vn)

which is an element of TA since V1 ∩ · · · ∩ Vn is open in X. �

Theorem 43. [Exercise 3.3] Let M be a metric space, and let A ⊆ M be any sub-set. Then the subspace topology on A is the same as the metric topology obtained byrestricting the metric of M to points in A.

Proof. Denote the open ball in a metric space X of radius r around a point x by B(X)r (x).

To prove the result, it suffices to show that the two topologies have a common basis.Let U be a set in TA, so U = A ∩ V for some V open in M . If p ∈ U then there exists

some open ball B(X)r (p) contained in V , and by definition A ∩ B(X)

r (p) = B(A)r (p) is an

open ball around p contained in A. This shows that the set of all open balls{B

(A)r (p)

}is a basis for both topologies. �

Theorem 44. [Exercise 3.12] Let A be a subspace of a topological space X.

(1) The closed subsets of A are precisely the intersections of A with closed subsetsof X.

(2) If B ⊆ A ⊆ X, B is open in A, and A is open in X, then B is open in X.(3) If X is Hausdorff then A is Hausdorff.(4) If X is second countable then A is second countable.

13

Proof. If E is closed in A then A\E = A∩U for some U open in X, so E = A\(A∩U) =A∩ (X \U), where X \U is closed in X. The converse is similar. This proves (1). Forpart (2), we have B = A∩U for some U open in X, so B is open in X since A is openin X. Part (3) is obvious, and part (4) follows from part (b) of Proposition 3.11. �

Theorem 45. [Exercise 3.25] Let X1, . . . , Xn be topological spaces. Define the set

B = {U1 × · · · × Un : Ui is open in Xi, i = 1, . . . , n} .Then B is a basis for X1 × · · · ×Xn.

Proof. It is clear that⋃B = X1×· · ·×Xn. Let U = U1×· · ·×Un and U ′ = U ′1×· · ·×U ′n

be elements of B. Since

U ∩ U ′ = (U1 ∩ U ′1)× · · · × (Un ∩ U ′n)

and each Ui ∩ U ′i is open in Xi, we have U ∩ U ′ ∈ B. This proves that B is a basis. �

Theorem 46. [Exercise 3.26] The product topology on Rn = R × · · · × R is the sameas the metric topology induced by the Euclidean distance function.

Proof. It suffices to show that the set of all open balls forms a basis for the producttopology on Rn. Note first that every open ball is open in the product topology. IfU ⊆ Rn is open in the product topology and p = (p1, . . . , pn) ∈ U , then there exist setsU1, . . . , Un open in R such that p ∈ U1 × · · · × Un and U1 × · · · × Un ⊆ U . For eachi, choose an open ball (p − ri, p + ri) contained in Ui. Let r = min {r1, . . . , rn}; thenBr(p) is an open ball around p that is contained in U . This proves that the set of allopen balls is a basis for the product topology. �

Theorem 47. [Exercise 3.32] Let X1, . . . , Xn be topological spaces.

(1) The projection maps πi : X1×· · ·×Xn → Xi are all continuous and open maps.In particular, a set U1 × · · · × Un is open in X1 × · · · ×Xn if and only if everyUi is open in Xi.

(2) The product topology is “associative” in the sense that the three product topolo-gies X1×X2×X3, (X1×X2)×X3, and X1× (X2×X3) on the set X1×X2×X3

are all equal.(3) For any i and any points xj ∈ Xj, j 6= i, the map fi : Xi → X1×· · ·×Xn given

byfi(x) = (x1, . . . , xi−1, x, xi+1, . . . , xn)

is a topological embedding of Xi into the product space.(4) If for each i, Bi is a basis for the topology of Xi, then the set

B = {B1 × · · · ×Bn : Bi ∈ Bi}is a basis for the product topology on X1 × · · · ×Xn.

14

(5) If Ai is a subspace of Xi for i = 1, . . . , n, the product topology and the subspacetopology on A1 × · · · × An ⊆ X1 × · · · ×Xn are equal.

(6) If each Xi is Hausdorff, so is X1 × · · · ×Xn.(7) If each Xi is second countable, so is X1 × · · · ×Xn.

Proof. The first part of (1) follows by taking B = X1×· · ·×Xn and f = IdB in Theorem3.27. To show that πi is an open map, let U1 × · · · × Un be an element of the standardbasis of X1×· · ·×Xn. Since πi(U1×· · ·×Un) = Ui which is open, it follows that πi(U)is open for any open set U . For (3), it suffices to check that f−1i (U1× · · · ×Un) is openwhenever each Ui is open in Xi. If xj ∈ Uj for all j 6= i then f−1i (U1 × · · · × Un) = Ui,which is open. Otherwise, f−1i (U1 × · · · × Un) = ∅, which is also open. For (4), itis clear that every set in B is open. Let U be an open set in X1 × · · · × Xn and letx = (x1, . . . , xn) ∈ U . There exists some U1× · · ·×Un ⊆ U such that x ∈ U1× · · ·×Unand each Ui is open in Xi. Since Bi is a basis for Xi, there exists a Bi ∈ Bi with xi ∈ Bi

and Bi ⊆ Ui. Therefore x ∈ B1 × · · · × Bn and B1 × · · · × Bn ⊆ U1 × · · · × Un, whichproves that B is a basis for X1 × · · · ×Xn.

For (5), it suffices to show that the basis

B = {U1 × · · · × Un : Ui is open in Ai, i = 1, . . . , n}for the product topology is also a basis for the subspace topology on A1 × · · · × An.First, note that if U1 × · · · × Un ∈ B then for each i we have Ui = Ai ∩ Vi for some Viopen in Xi, so

U1 × · · · × Un = (A1 × · · · × An) ∩ (V1 × · · · × Vn)

which is open in the subspace topology. Now let U ⊆ A1 × · · · × An be an open set inthe subspace topology and let x ∈ U . We have U = (A1 × · · · × An) ∩ V for some Vopen in X1×· · ·×Xn. There exists some V1×· · ·×Vn ⊆ V such that x ∈ V1×· · ·×Vnand each Vi is open in Xi, so

(A1 ∩ V1)× · · · × (An ∩ Vn) = (A1 × · · · × An) ∩ (V1 × · · · × Vn) ⊆ U

is an element of B that contains x. This proves that B is a basis for the subspacetopology.

For (6), let x = (x1, . . . , xn) and y = (y1, . . . , yn) be distinct points in X1 × · · · × Xn.We have xi 6= yi for some i, so there exist neighborhoods U1 of xi and U2 of yi suchthat U1 ∩U2 = ∅. Then the sets X1× · · · ×Xi−1×Uj ×Xi+1× · · · ×Xn for j = 1, 2 aredisjoint neighborhoods of x and y respectively. Part (7) follows from (4). �

Theorem 48. Let X1, . . . , Xn be topological spaces and let Ai be a subset of Xi for eachi.

(1) If each Ai is closed in Xi then A1 × · · · × An is closed in X1 × · · · ×Xn.(2) A1 × · · · × An = A1 × · · · × An.

15

Proof. The projection maps πi : X1 × · · · × Xn → Xi are continuous, so π−1i (Ai) =X1 × · · · × Ai × · · · ×Xn is closed in X1 × · · · ×Xn and

A1 × · · · × An =n⋂i=1

π−1i (Ai)

is closed. For (2), it is clear that A1× · · · ×An is a closed set containing A1× · · · ×An,so A1 × · · · × An ⊆ A1 × · · · × An. Now let x = (x1, . . . , xn) ∈ A1 × · · · × An, let Ube a neighborhood of x, and let U1 × · · · × Un ⊆ U be a neighborhood of x with eachUi open in Xi. Since Ui is a neighborhood of xi ∈ Ai, it contains a point x′i 6= xi inAi. Thus x′ = (x′1, . . . , x

′n) is a point of A1 × · · · × An in U not equal to x, and x

is a limit point of A1 × · · · × An. Since the closure of a set contains its limit points,A1 × · · · × An ⊆ A1 × · · · × An. �

Theorem 49. [Exercise 3.43] Suppose we are given an indexed collection of nonemptytopological spaces {Xα}α∈A. Declare a subset of the disjoint union X =

∐α∈AXα to be

open if and only if its intersection with each Xα is open.

(1) This is a topology on X, called the disjoint union topology.(2) A subset of the disjoint union is closed if and only if its intersection with each

Xα is closed.(3) If each Xα is an n-manifold, then the disjoint union X is an n-manifold if and

only if the index set A is countable.

Proof. Let E be a closed subset of X. Since X \E is open, each (X \E)∩Xα = Xα \Eis open, so the intersection of E with Xα is closed. The converse is similar. For(3), suppose that X is an n-manifold. Since X has a countable base B and each Xα isnonempty and open, there is a surjection from B to A, which shows that A is countable.

Conversely, suppose that A is countable. Let x1, x2 ∈ X be distinct points. If x1, x2 ∈Xα for some α then there exists neighborhoods U1 of x1 and U2 of x2 such that U1∩U2 =∅ since Xα is Hausdorff, and if x1 ∈ Xα, x2 ∈ Xβ for α 6= β then Xα and Xβ areneighborhoods of x1 and x2 respectively with Xα ∩Xβ = ∅. Therefore X is Hausdorff.Next, we prove that X is second countable. For each Xα, let Bα be a countable basefor Xα, and let B =

⋃α∈A Bα, which is countable. If U is an open set in X and p ∈ U

then p ∈ Xα ∩U for some α, so there exists some B ∈ Bα with p ∈ B and B ⊆ Xα ∩U .Since B ∈ B and B ⊆ U , this shows that B is a countable base for X. Finally, it is easyto see that X is locally Euclidean of dimension n since each Xα is locally Euclidean ofdimension n. Therefore X is an n-manifold. �

Theorem 50. [Exercise 3.46] Let (X, T ) be a topological space, Y be any set, andπ : X → Y be a surjective map. Then

TY ={U ⊆ Y : π−1(U) ∈ T

}

16

is a topology on Y .

Proof. Since π−1(∅) = ∅ and π−1(Y ) = X, we have ∅, X ∈ TY . If {Uα}α∈A is a collectionof sets in TY then

π−1

(⋃α∈A

Uα

)=⋃α∈A

π−1(Uα) ∈ T

since each π−1(Uα) is in T , so⋃α∈A Uα ∈ TY . Similarly, if U1, . . . , Un ∈ TY then

π−1(U1 ∩ · · · ∩ Un) = π−1(U1) ∩ · · · ∩ π−1(Un) ∈ Tsince each π−1(Ui) is in T , so U1 ∩ · · · ∩ Un ∈ TY . �

Theorem 51. [Exercise 3.61] A continuous surjective map π : X → Y is a quotientmap if and only if it takes saturated open sets to open sets, or saturated closed sets toclosed sets.

Proof. Suppose that π takes saturated open sets to open sets. If U is open in Y thenπ−1(U) is open in X since π is continuous. Also, if U ⊆ Y and π−1(U) is open in Xthen U = π(π−1(U)) is open since π−1(U) is saturated. Conversely, if π is a quotientmap and U = π−1(V ) is a saturated open set then π(U) = π(π−1(V )) = V , which isopen. Finally, π takes saturated open sets to open sets if and only if π takes saturatedclosed sets to closed sets, since π−1(U) = X \ π−1(Y \ U) for any U open in Y . �


(1) Any composition of quotient maps is a quotient map.(2) An injective quotient map is a homeomorphism.(3) If q : X → Y is a quotient map, a subset K ⊆ Y is closed if and only if q−1(K)

is closed in X.(4) If q : X → Y is a quotient map and U ⊆ X is a saturated open or closed subset,

then the restriction q|U : U → q(U) is a quotient map.(5) If {qα : Xα → Yα}α∈A is an indexed family of quotient maps, then the map q :∐

αXα →∐

α Yα whose restriction to each Xα is equal to qα is a quotient map.

Proof. For (3), we have Y \K open if and only if q−1(Y \K) = X \ q−1(K) is open. For(4), let U = q−1(V ) be a saturated open set and consider q|U : U → q(U). If A ⊆ q(U)then

(q|U)−1(A) = q−1(A) ∩ q−1(V ) = q−1(A ∩ q(U)) = q−1(A),

so A is open if and only if (q|U)−1(A) = q−1(A) is open since q is a quotient map. Thecase for the restriction of q to a closed set is similar. For (5), let U ⊆

∐α Yα. We have

U ⊆open

∐α

Yα ⇔ U ∩ Yα ⊆open

Yα for every α ∈ A

17

⇔ q−1α (U) ⊆open

Xα for every α ∈ A

⇔ q−1(U) ∩Xα ⊆open

Xα for every α ∈ A

⇔ q−1(U) ⊆open

∐α

Xα.

�

Theorem 53. [Exercise 3.85] Any subgroup of a topological group is a topological groupwith the subspace topology. Any finite product of topological groups is a topological groupwith the direct product group structure and the product topology.

Proof. Let G be a topological group with operations µ : G×G→ G and ι : G→ G, andlet H be a subgroup of G. Since H×H is a subspace of G×G and H is a subspace of G,the restrictions µ|H×H and ι|H are continuous by Corollary 3.10. This shows that H isalso a topological group with the subspace topology. Now let G1, . . . , Gn be topologicalgroups with operations µi : Gi × Gi → Gi and ιi : Gi → Gi for i = 1, . . . , n. We wantto show that the maps

µ : (G1 × · · · ×Gn)× (G1 × · · · ×Gn)→ G1 × · · · ×Gn

((g1, . . . , gn), (g′1, . . . , g′n)) 7→ (g1g

′1, . . . , gng

′n)

and

ι : G1 × · · · ×Gn → G1 × · · · ×Gn

(g1, . . . , gn) 7→ (g−11 , . . . , g−1n )

are continuous. But

ϕ : (G1 × · · · ×Gn)× (G1 × · · · ×Gn)→ (G1 ×G1)× · · · × (Gn ×Gn)

((g1, . . . , gn), (g′1, . . . , g′n)) 7→ ((g1, g

′1), . . . , (gn, g

′n))

is continuous, being a simple rearrangement, and Proposition 3.33 shows that

µ1 × · · · × µn : (G1 ×G1)× · · · × (Gn ×Gn)→ G1 × · · · ×Gn

is continuous. Therefore µ is continuous since µ = (µ1 × · · · × µn) ◦ ϕ. Finally, ι =ι1 × · · · × ιn is continuous by Proposition 3.33. �

Example 54. [Problem 3-3] By considering the space X = [0, 1] ⊆ R and the setsA0 = {0}, Ai = [1/(i + 1), 1/i] for i = 1, 2, . . . , show that the gluing lemma (Lemma3.23) is false if {A1, . . . , Ak} is replaced by an infinite sequence of closed sets.

Define the sequence of maps fi : Ai → R by setting f0(0) = 1 and fi(x) = 0 for alli ≥ 1 and x ∈ Ai. By the gluing lemma, there is a unique continuous map f : X → Rsuch that f(0) = 1 and f(x) = 0 for all x ∈ (0, 1]. However, this map is clearly notcontinuous at 0.

18

Theorem 55. [Problem 3-4] Any closed ball in Rn is an n-dimensional manifold withboundary.

Proof. Let Bn = B1(0) be the closed unit ball in Rn, let N = (0, . . . , 0, 1) be the “northpole”, and let T = {0}n−1 × [0, 1] be the vertical line connecting 0 and N . Defineσ : Bn \ T → Hn given by

σ(x1, . . . , xn) =

(x1

‖x‖ − xn, . . . ,

xn−1‖x‖ − xn

, ‖x‖−1 − 1

),

with its inverse given by

σ−1(u1, . . . , un) =1

(un + 1)(‖u‖2 + 1)(2u1, . . . , 2un−1, ‖u‖2 − 1)

where u = (u1, . . . , un−1). Since both maps are continuous, this proves that Bn \ T ishomeomorphic to Hn. Furthermore, we can repeat the same argument by taking thevertical line T = {0}n−1 × [−1, 0] instead, giving us Euclidean neighborhoods of everypoint in Bn except for 0. But the identity map on the open unit ball Bn is a suitablecoordinate chart around 0, so this proves that Bn is an n-manifold with boundary. �

Theorem 56. [Problem 3-5] A finite product of open maps is open. A finite productof closed maps need not be closed.

Proof. Let fi : Xi → Yi, i = 1, . . . , n, be a set of open maps and write f = f1×· · ·×fn.Let U ⊆ X1×· · ·×Xn be an open set; we want to show that f(U) is open. If y ∈ f(U)then y = f(x) for some x = (x1, . . . , xn) ∈ U . There exists some U1 × · · · × Un ⊆ Usuch that x ∈ U1× · · · ×Un and each Ui is open in Xi. For every i, fi(Ui) is open sincefi is an open map, so y ∈ f1(U1)× · · ·× f1(Un) ⊆ f(U). This proves that f(U) is open.However, a finite product of closed maps need not be closed. Let D = {0} be a discretespace, let f : R → R be the identity on R and let g : R → D be given by g(x) = 0.Then f and g are both closed, but

(f × g)({(x, y) : xy = 1}) = (R \ {0})× {0} ,which is not closed. �

Theorem 57. [Problem 3-6] Let X be a topological space. The diagonal of X ×X isthe subset 4 = {(x, x) : x ∈ X} ⊆ X × X. Then X is Hausdorff if and only if 4 isclosed in X ×X.

Proof. Suppose X is Hausdorff and let (x1, x2) ∈ (X × X) \ 4. Since x1 6= x2, thereexist neighborhoods U1 of x1 and U2 of x2 such that U1 ∩ U2 = ∅. This implies that(U1×U2)∩4 = ∅, so U1×U2 is a neighborhood of (x1, x2) with U1×U2 ⊆ (X×X)\4.This shows that (X ×X) \ 4 is open. Conversely, suppose that (X ×X) \ 4 is openand let x1, x2 be distinct points in X. Then (x1, x2) ∈ (X×X)\4, so there exist open

19

sets U1, U2 ⊆ X such that (x1, x2) ∈ U1 × U2 ⊆ (X ×X) \ 4. Therefore U1 ∩ U2 = ∅,which proves that X is Hausdorff. �

Theorem 58. [Problem 3-7] Let M = Rd × R where Rd is the set R with the discretetopology.

(1) M is homeomorphic to the space X of Example 38.(2) M is locally Euclidean and Hausdorff, but not second countable.

Proof. Let U × V ⊆M where V is open in R. For each u ∈ U we can write {u} × V =⋃α∈A(aα, bα), which is open in X. Therefore

U × V =⋃u∈U

{u} × V

is open in X. Also, every element {(c, y) : a < y < b} in the basis of X is clearly openin M . Part (1) then follows from Lemma 22. From Theorem 47 we know that Mis locally Euclidean (of dimension 2) and Hausdorff. To show that M is not secondcountable, suppose B is a countable basis for M . Consider the collection C of sets ofthe form {x} × R for x ∈ R, which are all disjoint and open in M . By Theorem 34,C is countable. But {x} × R 7→ x is a surjection from C to R, which contradicts theuncountability of R. �

Theorem 59. [Problem 3-10]

(1) (Characteristic Property of Disjoint Union Topologies) Let X =∐

α∈AXα bea disjoint union space. For any topological space B, a map f : X → B iscontinuous if and only if each fα = f ◦ iα is continuous, where iα : Xα → X isthe canonical injection:

X

Xα B

fiα

fα

(2) (Uniqueness of the Disjoint Union Topology) Let {Xα}α∈A be a collection oftopological spaces. The disjoint union topology on

∐α∈AXα is the unique topol-

ogy that satisfies the characteristic property.

Proof. We have

f is continuous⇔ f−1(U) ⊆open

X for all U ⊆open

B

⇔ f−1(U) ∩Xα ⊆open

Xα for all α ∈ A and U ⊆open

B

20

⇔ f−1α (U) ⊆open

Xα for all α ∈ A and U ⊆open

B

⇔ fα is continuous for all α ∈ A,

which proves (1). For (2), suppose that Xg =∐

α∈AXα has some other topology thatsatisfies the characteristic property, and write Xd for the usual disjoint union space on∐

α∈AXα. By applying the characteristic property to the identity map, we see thatevery injection iα into either Xg or Xd is continuous. Setting B = Xg and then B = Xd

shows that the identity map from the disjoint union topology to the given topology isa homeomorphism. Therefore the topologies on Xg and Xd are equal. �

Theorem 60. Let X1, . . . , Xn, Y be topological spaces. Then the identity map

ι : (X1 q · · · qXn)× Y → (X1 × Y )q · · · q (Xn × Y )

is a homeomorphism.

Proof. Let U be open in (X1 × Y ) q · · · q (Xn × Y ); then U ∩ (Xi × Y ) is open inXi × Y for every i. Let (x, y) ∈ U so that x ∈ Xi for some i. Since x ∈ U ∩ (Xi × Y ),we have x ∈ V × W ⊆ U ∩ (Xi × Y ) for some V open in Xi and some W open inY . But V ×W is also open in (X1 q · · · q Xn) × Y , which shows that U is open in(X1q· · ·qXn)×Y . Conversely, let U be open in (X1q· · ·qXn)×Y and let (x, y) ∈ U .We have (x, y) ∈ V ×W ⊆ U for some V open in X1 q · · · qXn and some W open inY . Since (V ×W )∩ (Xi×Y ) = (V ∩Xi)×W is open in Xi×Y for every i, V ×W ⊆ Uis a neighborhood of (x, y) in (X1× Y )q · · · q (Xn× Y ). This shows that U is open in(X1 × Y )q · · · q (Xn × Y ). �

Theorem 61. If X1, . . . , Xk are nonempty topological spaces then the projections πi :X1 × · · · ×Xk → Xi are quotient maps.

Proof. This follows from the fact that the projections are surjective, continuous andopen maps. See Theorem 47. �

Example 62. [Problem 3-11] Proposition 3.62(d) showed that the restriction of a quo-tient map to a saturated open set is still a quotient map. Show that the “saturated”hypothesis is necessary, by giving an example of a quotient map f : X → Y and anopen subset U ⊆ X such that f |U is surjective but not a quotient map.

The map f : [0, 1] → S1 given by f(s) = e2πis is a quotient map, but f |[0,1) is not aquotient map as Example 3.66 shows.

Theorem 63. [Problem 3-14] Real projective space Pn is an n-manifold.

Proof. Denote a line {λ(p1, . . . , pn+1) : λ ∈ R} ∈ Pn by [p1, . . . , pn+1]. Given some k, letVk = {[p1, . . . , pn+1] ∈ Pn : pk 6= 0} be the lines of Pn that are not parallel to the plane

21

xk = 1. Define π : Pn \ Vk → Rn by

π([p1, . . . , pn+1]) = p−1k (p1, . . . , pk, . . . , pn+1)

where pk is omitted. This map is well-defined since [p1, . . . , pn+1] = [q1, . . . , qn+1] impliesthat pi = λqi for every i. Furthermore, π is a homeomorphism. By choosing at leasttwo different values of k, we can find a coordinate ball around every element of Pn.This shows that Pn is an n-manifold. �

Theorem 64. [Problem 3-15] Let CPn denote the set of all 1-dimensional complexsubspaces of Cn+1, called n-dimensional complex projective space. TopologizeCPn as the quotient (Cn+1\{0})/C∗, where C∗ is the group of nonzero complex numbersacting by scalar multiplication. Then CPn is a 2n-manifold.

Proof. Proceed as in Theorem 63 to show that each Pn \ Vk is homeomorphic to Cn.Since Cn is a 2n-manifold, this proves that CPn is a 2n-manifold. �

Theorem 65. [Problem 3-16] Let X be the subset R × {0} ∪ R × {1} of R2. Definean equivalence relation on X by declaring (x, 0) ∼ (x, 1) if x 6= 0. Then the quotientspace X/ ∼ is locally Euclidean and second countable, but not Hausdorff. (This spaceis called the line with two origins.)

Proof. Let p ∈ X/ ∼. If p 6= [(0, 0)] and p 6= [(0, 1)] then there is clearly a neigh-borhood N of p homeomorphic to some open interval in R. If p = [(0, 0)] then letN = {[(x, 0)] : −1 < x < 1} and define ϕ : N → (−1, 1) by [(x, 0)] 7→ x. It is easy tocheck that this map is a well-defined homeomorphism. So X/ ∼ is locally Euclidean andtherefore second countable by Proposition 3.56. But X/ ∼ is not Hausdorff, since anytwo neighborhoods of [(0, 0)] and [(0, 1)] respectively must contain a common point. �

Theorem 66. [Problem 3-19] If G is a topological group and H ⊆ G is a subgroup,then H is also a subgroup.

Proof. Let µ : G×G→ G and ι : G→ G be the product and inverse maps respectively.Since ι(H) ⊆ H we have ι(H) ⊆ H by Lemma 21. Similarly, µ(H × H) ⊆ H impliesthat µ(H ×H) ⊆ H, and H ×H = H ×H by Theorem 48. �

Theorem 67. [Problem 3-20] If G is a group that is also a topological space, then Gis a topological group if and only if the map θ : G × G → G given by (x, y) 7→ xy−1 iscontinuous.

Proof. Since θ(x, y) = µ(x, ι(y)), θ is continuous if G is a topological group. Conversely,since ι(x) = θ(1, x) and µ(x, y) = θ(x, ι(y)), G is a topological group if θ is continuous.

�

Theorem 68. [Problem 3-21] Let G be a topological group and Γ ⊆ G be a subgroup.

22

(1) For any g ∈ G, the left translation Lg : G→ G passes to the quotient G/Γ anddefines a homeomorphism of G/Γ with itself.

(2) A topological space X is said to be homogeneous if for any x, y ∈ X, there isa homeomorphism ϕ : X → X taking x to y. Every coset space is homogeneous.

Proof. Let π : G→ G/Γ given by g 7→ gΓ be the quotient map. Since π◦Lg : G→ G/Γsatisfies (π ◦ Lg)(g′) = gg′Γ and is constant on the fibers of π, there exists a unique

continuous map Lg : G/Γ → G/Γ satisfying Lg ◦ π = π ◦ Lg. Furthermore, Lg−1 is a

continuous inverse of Lg since

Lg−1 ◦ Lg ◦ π = Lg−1 ◦ π ◦ Lg = π ◦ Lg−1 ◦ Lg = π

and similarly Lg ◦ Lg−1 ◦ π = π. This shows that Lg is a homeomorphism of G/Γ with

itself. For (2), if gΓ and g′Γ are cosets in G/Γ then Lg′g−1 is a homeomorphism thattakes gΓ to g′Γ. �

Theorem 69. [Problem 3-22] Let G be a topological group acting continuously on atopological space X.

(1) The quotient map π : X → X/G is open.(2) X/G is Hausdorff if and only if the orbit relation

D = {(x1, x2) ∈ X ×X : x2 = g · x1 for some g ∈ G}is closed in X ×X.

Proof. Let U ⊆ X be open; we want to show that π(U) is open, or equivalently, thatπ−1(π(U)) = {g · x : g ∈ G, x ∈ U} is open. Since the group action α : G×X → X iscontinuous, α−1(U) is open in G×X. Let π2 : G×X → X be the canonical projection.If g · x ∈ π−1(π(U)) then α(g−1, g · x) = x ∈ U , so g · x ∈ π2(α

−1(U)). Conversely,if x ∈ π2(α

−1(U)) then g · x ∈ U for some g ∈ G, so g−1 · (g · x) = x ∈ π−1(π(U)).This shows that π−1(π(U)) = π2(α

−1(U)). But this set is open by Theorem 47. For(2), notice that D being closed is equivalent to 4 = {(p, p) : p ∈ X/G} being closed inX/G, and apply Theorem 57. �

Theorem 70. [Problem 3-23] If Γ is a normal subgroup of the topological group G thenthe coset space G/Γ is a topological group.

Proof. Let π : G→ G/Γ be the quotient map (in the group sense). By Theorem 69, πis also a quotient map in the topology sense. Let µ : G×G→ G and ι : G→ G be theproduct and inverse maps respectively. By Theorem 56, π×π : G×G→ G/Γ×G/Γ isa continuous, surjective and open map, and therefore a quotient map. Since Γ is normalin G, π ◦ µ is constant on the fibers of π× π and there exists a unique continuous mapµ : G/Γ× G/Γ → G/Γ satisfying µ ◦ (π × π) = π ◦ µ. Similarly, there exists a unique

23

continuous map ι : G/Γ → G/Γ satisfying ι ◦ π = π ◦ ι. The group axioms are theneasily checked for µ and ι. �

Chapter 4. Connectedness and Compactness

Theorem 71. In a topological space X, the path connectivity relation ∼p is an equiv-alence relation.

Proof. For any point p ∈ X, the path f : [0, 1]→ X with f(t) = p is a path from p toitself. If f : [0, 1]→ X is a path from p to q, then g : [0, 1]→ X given by g(t) = f(1− t)is a path from q to p. Finally, if f : [0, 1]→ X is a path from p to q and g : [0, 1]→ Xis a path from q to r, then by the gluing lemma the map h : [0, 1]→ X given by

h(t) =

{f(2t) if 0 ≤ t ≤ 1

2,

g(2t− 1) if 12< t ≤ 1

is a path from p to r. �

Theorem 72. [Exercise 4.22] Let X be any space.

(1) Each path component is contained in a single component, and each componentis a disjoint union of path components.

(2) If A ⊆ X is path-connected, then A is a contained in a single path component.

Proof. (1) follows from the fact that every path component is also connected and Propo-sition 4.21. For (2), suppose B and C are path components both containing points ofA. It follows from Theorem 71 that B = C. �

Theorem 73. [Exercise 4.24] Every manifold is locally path-connected.

Proof. This follows from Theorem 39. �

Theorem 74. [Exercise 4.38] Let X be a compact space, and suppose {Fn} is a count-able collection of nonempty closed subsets of X that are nested, which means thatFn ⊇ Fn+1 for each n. Then

⋂n Fn is nonempty.

Proof. Suppose that⋂n Fn is empty. Then X =

⋃nX \Fn, so there is a finite subcover

{X \ Fn1 , . . . , X \ Fnk} where we take n1 < · · · < nk. But⋃ki=1X \Fni = X \Fnk since

Fn ⊇ Fn+1 for each n, and this is a contradiction since Fnk is nonempty. �

Theorem 75. [Exercise 4.49] Every compact metric space is complete.

Proof. Let {xn} be a Cauchy sequence in a metric space M . There is some subsequencethat converges to a point x, and it is well-known that a Cauchy sequence converges ifit has a convergent subsequence. Therefore xn → x. �

24

Theorem 76. [Exercise 4.67] Any finite product of locally compact spaces is locallycompact.

Proof. It suffices to show that if X, Y are locally compact then X×Y is locally compact.Let (x, y) ∈ X ×Y . By Proposition 4.63, there exist precompact neighborhoods U of xand V of y. Then U ×V is a precompact neighborhood of (x, y), since U × V = U ×Vis compact. �

Theorem 77. [Exercise 4.73] Suppose A is an open cover of X such that each elementof A intersects only finitely many others. Then A is locally finite. This need not betrue when the elements of A are not open.

Proof. Let x ∈ X and choose some A ∈ A such that x ∈ A. There are finitelymany elements of A that intersect A, so A is the required neighborhood of x. For acounterexample when the elements of A are not required to be open, take X = R andA = {{x} : x ∈ R}. �

Theorem 78. [Exercise 4.78,4.79]

(1) Every compact Hausdorff space is normal.(2) Every closed subspace of a normal space is normal.

Proof. Let X be a compact Hausdorff space and let A,B be closed subsets of X. ThenA,B are compact, so by Lemma 4.34 there exist disjoint open subsets U, V ⊆ X suchthat A ⊆ U and B ⊆ V . For (2), let X be a normal space and let E be closed in X.If A,B are closed subsets of E then there exist disjoint open subsets U, V ⊆ X suchthat A ⊆ U and B ⊆ V . So U ∩E and V ∩E are disjoint open subsets of E such thatA ⊆ U ∩ E and B ⊆ V ∩ E. �

Theorem 79. [Exercise 4.87] Every compact manifold with boundary is homeomorphicto a subset of some Euclidean space.

Proof. If M is a compact manifold with boundary then the double D(M) of M is a com-pact manifold (without boundary), and is homeomorphic to a subset of some Euclideanspace. The restriction of this homeomorphism to M is the desired homeomorphism. �


(1) If U is any open subset of R and x ∈ U , then U \ {x} is disconnected.(2) For n > 1, Rn is not homeomorphic to any open subset of R.

Proof. Let A = {t ∈ R : t < x} and B = {t ∈ R : t > x}; then {A ∩ U,B ∩ U} is aseparation of U \ {x}. This proves (1). Part (2) follows immediately. �

25

Theorem 81. [Problem 4-2] A nonempty topological space cannot be both a 1-manifoldand an n-manifold for any n > 1.

Proof. Let M be a nonempty topological space that is both a 1-manifold and an n-manifold for some n > 1. Choose some p ∈ M and let ϕ1 : U1 → V1 and ϕ2 : U2 → V2be homeomorphisms where U1 and U2 are neighborhoods of p, V1 is open in R, andV2 is open in Rn. Let B be an open ball around ϕ2(p) contained in ϕ2(U1 ∩ U2).Then W1 = B \ {ϕ2(p)} is homeomorphic to W2 = (ϕ1 ◦ ϕ−12 )(B) \ {ϕ1(p)}, but W2 isdisconnected by Theorem 80 while W1 is (path) connected. This is a contradiction. �

Theorem 82. [Problem 4-3] Suppose M is a 1-dimensional manifold with boundary.Then the interior and boundary of M are disjoint.

Proof. Suppose p ∈ M is both an interior and boundary point. Choose coordinatecharts (U,ϕ) and (V, ψ) such that U, V are neighborhoods of p, ϕ(U) is open in IntH1,ψ(V ) is open in H1, ϕ(p) > 0 and ψ(p) = 0. Let W = U ∩ V ; then ϕ(W ) is home-omorphic to ψ(W ). But this is impossible, for ϕ(W ) \ {ϕ(p)} is disconnected whileψ(W ) \ {ψ(p)} is connected. �

Theorem 83. [Problem 4-4] The following topological spaces are not manifolds:

(1) The union of the x-axis and the y-axis in R2.(2) The conical surface C ⊆ R3 defined by

C ={

(x, y, z) : z2 = x2 + y2}

Proof. Let M be the union of the x-axis and the y-axis, and suppose that M is a man-ifold. By Theorem 20, M cannot be a 0-manifold. Now suppose M is a 1-manifoldand let B be a coordinate ball around (0, 0) with a homeomorphism ϕ : B → (a, b).Removing the point ϕ(0, 0) from (a, b) produces two connected components, but remov-ing (0, 0) from B produces four connected components (the left, top, right and bottomparts of the cross shape). Therefore M must be a n-manifold for n > 1. But thisimplies that the positive x-axis is both a 1-manifold and an n-manifold, contradictingTheorem 81. Part (2) is similar, for there is no 2 dimensional coordinate ball around(0, 0). �

Theorem 84. [Problem 4-7] Suppose f : X → Y is a surjective local homeomorphism.If X is locally connected, locally path-connected, or locally compact, then Y has thesame property.

Proof. The result for the first two properties follow from Theorem 30. Suppose that Xis locally compact and let B be a basis of precompact open sets in X. Let y ∈ Y sothat y = f(x) for some x ∈ X. Let U be a neighborhood of x such that f(U) is openand f |U : U → f(U) is a homeomorphism. There exists some neighborhood E ⊆ U of

26

x that is precompact, so f(E) is a precompact neighborhood of y. This proves that Yis locally compact. �

Theorem 85. [Problem 4-9] Any n-manifold is a disjoint union of countably manyconnected n-manifolds.

Proof. Any n-manifold M is the disjoint union of its connected components, which areopen by Proposition 4.25 and therefore are n-manifolds by Proposition 2.53. Further-more, the collection of connected components of M is countable by Theorem 34. �

Example 86. [Problem 4-13] Let X be the topologist’s sine curve (Example 4.17).

(1) Show that X is connected but not path-connected or locally connected.(2) Determine the components and path components of X.

The topologist’s sine curve is the union of the two sets

A = {(x, y) : x = 0 and y ∈ [−1, 1]} ;

B = {(x, y) : y = sin(1/x) and x ∈ (0, 1]} .

As subsets of R2, since B = A ∪ B = X and B is connected, Proposition 4.9 showsthat X is connected. Suppose that X is path-connected and let γ : [0, 1] → X bepath connecting the points (0, 0) and (2/π, 1). Choose a δ > 0 such that ‖γ(t)‖ < 1/2whenever 0 ≤ t < δ. This is impossible, for (0, 1) is a limit point of B. Also, X isnot locally connected since any neighborhood of (0, 0) is disconnected. The two pathcomponents of X are exactly A and B.

Theorem 87. [Problem 4-15] Suppose G is a topological group.

(1) Every open subgroup of G is also closed.(2) For any neighborhood U of 1, the subgroup 〈U〉 generated by U is open and closed

in G.(3) For any connected subset U ⊆ G containing 1, 〈U〉 is connected.(4) If G is connected, then every neighborhood of 1 generates G.

Proof. Let µ : G×G→ G and ι : G→ G be the group operations. For (1), let H be anopen subgroup of G. Then every coset gH is open in G, and G\H =

⋃g∈G\H gH is open.

For (2), let x ∈ 〈U〉. Since µ(x, ·) is a homeomorphism, µ(x, U) ⊆ 〈U〉 is a neighborhoodof x. Then 〈U〉 =

⋃x∈〈U〉 µ(x, U) is open. For (3), let x ∈ 〈U〉. Then µ(x, U) ⊆ 〈U〉

is a connected set containing x, and µ(µ(x, U), U) ⊆ 〈U〉 is a connected set containingboth x and 1. By Proposition 4.9, 〈U〉 =

⋃x∈〈U〉 µ(µ(x, U), U) is connected. For (4),

if U is a neighborhood of 1 then 〈U〉 is both open and closed in G by (2). Therefore〈U〉 = G. �

Theorem 88. Every σ-compact space is Lindelof.

27

Proof. Let X be a σ-compact space and let A be a countable collection of compactsubsets that cover X. Let U be an open cover of X. Each A ∈ A is covered by finitelymany sets from U , so there is a countable subcover. �

Theorem 89. [Problem 4-16] A locally Euclidean Hausdorff space is a topological man-ifold if and only if it is σ-compact.

Proof. Let M be a locally Euclidean Hausdorff space. If M is a topological manifoldthen Proposition 4.60 shows that there is a countable collection B of regular coordinateballs that cover M , i.e. M =

⋃B∈B B. Conversely, suppose that M is σ-compact. Then

M is Lindelof by Theorem 88, so the argument of Proposition 4.60 shows that M issecond countable. �

Theorem 90. [Problem 4-17] Suppose M is a manifold of dimension n ≥ 1, andB ⊆ M is a regular coordinate ball. Then M \ B is an n-manifold with boundary,whose boundary is homeomorphic to Sn−1.

Proof. It suffices to show that there is a coordinate ball around every point of ∂B.There exists a neighborhood B′ of B and a homeomorphism ϕ : B′ → Br′(0) that takesB to Br(0) for some r′ > r > 0. In particular, B′ \B is homeomorphic to Br′(0)\Br(0).For any x ∈ ∂B there is a homeomorphism ψ : U → V where U is a neighborhood ofϕ(x) in Br′(0) \ Br(0), V is open in Hn, and ψ(ϕ(x)) = 0. Then ψ ◦ ϕ is the requiredcoordinate map from a neighborhood of x ∈ ∂B in B′ \ B. The boundary of M \ B is∂B ≈ ∂Br(0) ≈ Sn−1. �

Theorem 91. [Problem 4-18] Let M1 and M2 be n-manifolds. For i = 1, 2, let Bi ⊆Mi

be regular coordinate balls, and let M ′i = Mi \Bi. Choose a homeomorphism f : ∂M ′

2 →∂M ′

1 (such a homeomorphism exists by Problem 4-17). Let M1#M2 (called a connectedsum of M1 and M2) be the adjunction space M ′

1 ∪f M ′2.

(1) M1#M2 is an n-manifold (without boundary).(2) If M1 and M2 are connected and n > 1, then M1#M2 is connected.(3) If M1 and M2 are compact, then M1#M2 is compact.

Proof. Part (1) follows from Theorem 3.79. If M1 and M2 are connected and n > 1 thenM ′

1 and M ′2 are still connected. Let e : M ′

1 → M ′1 ∪f M ′

2 and f : M ′2 → M ′

1 ∪f M ′2 be

the canonical embeddings. By Theorem 3.79 we have e(M ′1) ∩ f(M ′

2) 6= ∅, so M1#M2

is connected. If M1 and M2 are compact then M ′1 and M ′

2 are closed and thereforecompact. This implies that M1#M2 = e(M ′

1) ∪ f(M ′2) is compact. �

Theorem 92. [Problem 4-19] Let M1#M2 be a connected sum of n-manifolds M1

and M2. There are open subsets U1, U2 ⊆ M1#M2 and points pi ∈ Mi such thatUi ≈Mi \ {pi}, U1 ∩ U2 ≈ Rn \ {0}, and U1 ∪ U2 = M1#M2.

28

Proof. For i = 1, 2, let Bi ⊆ Mi be the regular coordinate ball around pi ∈ Mi and letCi ⊇ Bi be the larger coordinate balls around pi. Let ji : Mi \ Bi → M1#M2 be theinjections. Take U1 = j1(M1 \ B1) ∪ j2(C2 \ B2) and U2 = j1(C1 \ B1) ∪ j2(M2 \ B2),noting that

U1 ∩ U2∼= j1(C1 \B1) ∪ j2(C2 \B2)

∼= Sn−1 × (0, 1)∼= Rn \ {0} .

�

Theorem 93. [Problem 4-20] Define a topology on Z by declaring a set A to be open ifand only if n ∈ A implies −n ∈ A. Then Z with this topology is second countable andlimit point compact but not compact.

Proof. Let Bi = {−i, i}. Then B = {B0, B1, . . . } is a countable basis for Z. Now let Ube a subset of Z with at least two elements, and choose some nonzero n ∈ U . Then −nis a limit point of U , since any neighborhood of −n must contain n. In particular, Zis limit point compact. However, the open cover B of Z has no finite subcover, so Z isnot compact. �

Theorem 94. [Problem 4-21] Let V be a finite-dimensional real vector space. A normon V is a real-valued function on V , written v 7→ |v|, satisfying

(1) Positivity: |v| ≥ 0, and |v| = 0 if and only if v = 0.(2) Homogeneity: |cv| = |c| |v| for any c ∈ R and v ∈ V .(3) Triangle inequality: |v + w| ≤ |v|+ |w|.

A norm determines a metric by d(v, w) = |v − w|. In fact, all norms determine thesame topology on V .

Proof. Since V is finite-dimensional, it is isomorphic to Rn. For all v, w ∈ V we have|v| ≤ |w|+ |v − w| and |w| ≤ |v|+ |v − w|, so

||v| − |w|| ≤ |v − w| .

This immediately shows that any norm on V is continuous. Let |·|1 and |·|2 be twonorms on V , and write Vi for V equipped with the metric induced by |·|i. Consider theunit sphere S1 = {v ∈ V : |v|1 = 1}, which is compact. The image of S1 under |·|2 istherefore compact, and there exists some x ∈ S1 such that |v|2 ≤ |x|2 for all v ∈ S1.

Let B2 be an open ball in V2 of radius r around a point x ∈ V and let B1 be an openball in V1 of radius r/ |x|2 around x. We want to show that B1 ⊆ B2. Let v ∈ B1 with

29

v 6= x so that 0 < |v − x|1 < r/ |x|2. Then

|v − x|2 = |v − x|1

∣∣∣∣ v − x|v − x|1

∣∣∣∣2

<r

|x|2

∣∣∣∣ v − x|v − x|1

∣∣∣∣2

≤ r

since (v − x)/ |v − x|1 ∈ S1. This shows that v ∈ B2. A similar argument shows thatevery open ball in V1 around a point x ∈ V contains an open ball in V2 around x. ByTheorem 23, V1 and V2 have the same topology. �

Theorem 95. [Problem 4-23] Let X be a locally compact Hausdorff space. The one-point compactification of X is the topological space X∗ defined as follows. Let ∞be some object not in X, and let X∗ = X q {∞} with the following topology:

T = {open subsets of X}∪ {U ⊆ X∗ : X∗ \ U is a compact subset of X} .

(1) T is a topology.(2) X∗ is a compact Hausdorff space.(3) X is dense in X∗if and only if X∗ is noncompact.(4) X is open and and has the subspace topology.(5) A sequence of points in X diverges to infinity if and only if it converges to ∞

in X∗.

Proof. Clearly ∅, X∗ ∈ T since X∗ \X∗ = ∅ is compact. Let {Uα}α∈A be a subset of T .Write A = A1 ∪A2 where for each α ∈ A1 the set Uα is open in X, and for each α ∈ A2

the set X∗ \ Uα is a compact subset of X (and therefore ∞ ∈ Uα). If A2 is empty then⋃α∈A Uα =

⋃α∈A1

Uα is open in X since each Uα is open in X. If A2 is not empty then

X∗ \⋃α∈A

Uα =

( ⋂α∈A1

X∗ \ Uα

)∩

( ⋂α∈A2

X∗ \ Uα

)

=

( ⋂α∈A1

X \ Uα

)∩

( ⋂α∈A2

X∗ \ Uα

)is a compact subset of X since each X \ Uα is closed for α ∈ A1 and each X∗ \ Uα iscompact for α ∈ A2. Therefore

⋃α∈A Uα ∈ T . Now let {Uα}α∈A be a finite subset of T

and partition A into the subsets A1 and A2 as above. If A1 is empty then

X∗ \⋂α∈A

Uα =⋃α∈A

X∗ \ Uα

is compact since A is finite, so⋂α∈A Uα ∈ T . If A1 is not empty then⋂

α∈A

Uα =

( ⋂α∈A1

Uα

)∩

( ⋂α∈A2

X∗ \ (X∗ \ Uα)

)

30

=

( ⋂α∈A1

Uα

)∩

( ⋂α∈A2

X \ (X∗ \ Uα)

)is open since for each α ∈ A2 the set X∗ \ Uα is compact and therefore closed since Xis Hausdorff. This proves that T is a topology.

We now prove that X∗ is compact Hausdorff. Since X is already Hausdorff, it sufficesto check that ∞ and any x ∈ X can be separated by neighborhoods. Let E be aprecompact neighborhood of x and let F = X∗ \ E. Then F is a neighborhood of ∞and E ∩ F = ∅, which shows that X∗ is Hausdorff. Now let U be an open cover ofX∗. Choose some U ∈ U containing ∞ so that X∗ \U is a compact subset of X. Thenthere exists some finite subcover {U1, . . . , Uk} of X∗ \U , and {U1, . . . , Uk, U} is a finitesubcover of X∗.

Suppose that X is noncompact and let U be a nonempty open set in X∗. If U containsno points from X then U = {∞}. Then X∗ \U = X is compact, which contradicts ourassumption that X is noncompact. This proves that X is dense in X∗. Conversely, ifX is compact then {∞} is open and therefore X cannot be dense in X∗.

Let {qn} be a sequence of points in X. Suppose that {qn} diverges to infinity and letU be a neighborhood of ∞. Then X∗ \U is compact and contains finitely many valuesof qn, so U contains infinitely many values of qn. Therefore qn → ∞. Conversely, ifqn →∞ and K ⊆ X is compact then X∗\K is a neighborhood of∞, so X∗\K containsall but finitely many values of qn and K contains finitely many values of qn. �

Theorem 96. [Problem 4-25] Let σ : Sn \ {N} → Rn be stereographic projection, asdefined in Example 3.6. Then σ extends to a homeomorphism of Sn with the one-pointcompactification of Rn.

Proof. Let S∗ = (Sn\{N})∗. By Theorem 98, σ extends to a continuous map σ∗ : S∗ →(Rn)∗ with σ∗(∞) =∞. Furthermore, (σ∗)−1 is also continuous since σ−1 extends to acontinuous map taking∞ to∞. Therefore σ∗ is a homeomorphism. From Theorem 97we have a homeomorphism ϕ : Sn → S∗, and σ∗◦ϕ is the required homeomorphism. �

Theorem 97. [Problem 4-26] Let M be a compact manifold of positive dimension, andlet p ∈M . Then M is homeomorphic to the one-point compactification of M \ {p}.

Proof. Let M = (M \ {p})∗ and define ϕ : M → M by

ϕ(x) =

{x if x 6= p,

∞ if x = p.

31

Let U be open in M . If∞ /∈ U then ϕ−1(U) is open. If∞ ∈ U then M \U is compact,so

ϕ−1(U) = M \ ϕ−1(M \ U) = M \ (M \ U)

is open since M \ U is closed. Therefore ϕ is continuous. By Lemma 4.50, ϕ is ahomeomorphism. �

Theorem 98. [Problem 4-27] If X and Y are noncompact, locally compact Hausdorffspaces, then a continuous map f : X → Y extends to a continuous map f ∗ : X∗ → Y ∗

taking ∞ to ∞ if and only if it is proper.

Proof. Suppose that f is proper. Set f ∗(x) = f(x) for all x ∈ X and f ∗(∞) =∞. LetU be open in Y ∗. If ∞ /∈ U then (f ∗)−1(U) is open since f is continuous. If ∞ ∈ Uthen Y ∗ \ U = Y \ U is compact, and (f ∗)−1(Y ∗ \ U) = f−1(Y \ U) where f−1(Y \ U)is compact since f is proper. Then

(f ∗)−1(U) = X∗ \ [(f ∗)−1(Y ∗ \ U)]

= X \ f−1(Y \ U)

is open since f−1(Y \U) is closed. Conversely, suppose that f extends to a continuousmap f ∗. Let E ⊆ Y be compact. Then Y ∗ \ E is open, so

(f ∗)−1(Y ∗ \ E) = X∗ \ (f ∗)−1(E)

is open. Since ∞ ∈ X∗ \ (f ∗)−1(E) we have that (f ∗)−1(E) = f−1(E) is compact. �

Theorem 99. [Problem 4-30] Suppose X is a topological space and {Aα} is a locallyfinite closed cover of X. If for each α ∈ A we are given a continuous map fα : Xα → Ysuch that fα|Xα∩Xβ = fβ|Xα∩Xβ for all α and β, then there exists a unique continuousmap f : X → Y whose restriction to each Xα is fα.

Proof. For each x ∈ X we have x ∈ Xα for some α, so we can set f(x) = fα(x).This makes f a well-defined map since fα|Xα∩Xβ = fβ|Xα∩Xβ for all α and β. Letx ∈ X and choose a neighborhood U of x that intersects with finitely many elementsAα1 , . . . , Aαk ∈ {Aα}. Let E be a closed subset of f(U) so that E = F ∩ f(U) for someF closed in Y . Then

(f |U)−1(E) = f−1(E) ∩ U= f−1(F ) ∩ U

=k⋃i=1

f−1αi (F ) ∩ U

which is closed in U since each f−1αi (F ) is closed. This shows that every point of X has aneighborhood U on which f |U is continuous. By Proposition 2.19, f is continuous. �

32

Theorem 100. [Problem 4-33] Suppose X is a topological space with the property thatfor every open cover of X, there exists a partition of unity subordinate to it. Then Xis paracompact.

Proof. Let U = {Uα}α∈A be an open cover X. By our hypothesis, there exists a partitionof unity {ψα}α∈A subordinate to U . For each α, let Vα = ψ−1α ((0, 1]). We want to showthat {Vα} is a locally finite open refinement of U . Each Vα is open in X since (0, 1] isalways open in ψα(X). Furthermore, Vα ⊆ suppψα ⊆ Uα, so {Vα} is a refinement of U .If x ∈ X then there is a neighborhood N of x that intersects with a finite number ofsets in {suppψα}. In particular, N intersects with a finite number of sets in {Vα} sinceVα ⊆ suppψα for every α. �

Chapter 5. Cell Complexes

Theorem 101. [Exericse 5.3] Suppose X is a topological space whose topology is co-herent with a family B of subspaces.

(1) If Y is another topological space, then a map f : X → Y is continuous if andonly if f |B is continuous for every B ∈ B.

(2) The map∐

B∈B B → X induced by inclusion of each set B ↪→ X is a quotientmap.

Proof. If f is continuous then it is clear that every f |B is continuous. Suppose thatevery f |B is continuous and let U be open in Y . Then for every B ∈ B the set(f |B)−1(U) = f−1(U) ∩ B is open in B, so f−1(U) is open since X is coherent with B.Part (2) follows directly from the definition of coherent. �

Theorem 102. [Exercise 5.19] Suppose X is an n-dimensional CW complex with n ≥ 1,and e0 is any n-cell of X. Then X \ e0 is a subcomplex, and X is homeomorphic to anadjunction space obtained from X \ e0 by attaching a single n-cell.

Proof. If e is a cell of X \e0 then e\e is contained in Xn−1, and in particular e∩e0 = ∅.Therefore X \ e0 is a subcomplex. Let φ : D0 → X be a characteristic map for e0 andform the adjunction space (X \ e0)∪φD0. Define ψ : (X \ e0)qD0 → X as being equalto inclusion on X \ e0 and to φ on D0. Then ψ makes the same identifications as thequotient map defining the adjunction space, so it remains to show that ψ is a quotientmap. The argument is identical to that of Proposition 5.18. �

Theorem 103. [Exercise 5.34] If K is a Euclidean simplicial complex, then the collec-

tion K consisting of the interiors of the simplices of K is a regular CW decompositionof |K|.

33

Proof. First note that the sets in K are disjoint, for the intersection of two simplices iseither empty or a face of each, and if a point is in the interior of a simplex σ, it cannot

be in a face of σ. This shows that K is a partition of |K|. Proposition 5.32 showsthat for each σ ∈ K we have a homeomorphism φ : 4k → σ where 4k is the standardk-simplex. Since φ restricts to a homeomorphism from Int4k to the interior of σ andφ(∂4k) maps into the boundary of σ, we can take φ as a characteristic map for the

interior of σ. Furthermore, since K is locally finite, K must also be locally finite. By

Proposition 5.4, K is a regular CW decomposition of |K|. �

Theorem 104. [Exercise 5.40] Let K and L be simplicial complexes. Suppose f0 :K0 → L0 is any map with the property that whenever {v0, . . . , vk} are the vertices ofa simplex of K, {f0(v0), . . . , f0(vk)} are the vertices of a simplex of L (possibly withrepetitions). Then there is a unique simplicial map f : |K| → |L| whose vertex map isf0. It is a simplicial isomorphism if and only if f0 is a bijection satisfying the followingadditional condition: {v0, . . . , vk} are the vertices of a simplex of K if and only if{f0(v0), . . . , f0(vk)} are the vertices of a simplex of L.

Proof. Suppose K is in Rn and L is in Rm. Let {Vα}α∈A be the collection of all subsetsK0 that define a vertex of a simplex of K. For each Vα = {v0, . . . , vk} we know that{f0(v0), . . . , f0(vk)} are the vertices of a simplex of L, so by Proposition 5.38 thereexists a unique map fα : σ → Rm that is the restriction of an affine map, takes vi tof0(vi) for each i, and takes σ onto a simplex of L. Since the maps fα agree on theintersection of their domains and K is locally finite, Theorem 99 shows that there isa unique (simplicial) map f : |K| → |L|. If in addition f0 is a bijection satisfying thegiven condition then a similar process gives an inverse to f . �

Theorem 105. [Problem 5-1] Suppose D and D′ are closed cells (not necessarily of thesame dimension).

(1) Every continuous map f : ∂D → ∂D′ extends to a continuous map f : D → D′,with F (IntD) ⊆ IntD′.

(2) Given points p ∈ IntD and p′ ∈ IntD′, F can be chosen to take p to p′.(3) If f is a homeomorphism, then F can also be chosen to be a homeomorphism.

Proof. We have homeomorphisms ϕ : Bn → D and ψ : Bm → D′ such that ϕ(Sn−1) =

∂D and ψ(Sm−1) = ∂D′. Define f : Sn−1 → Sm−1 by f = ψ−1 ◦ f ◦ ϕ. Define

F : Bn → Bm by F (0) = 0 and

(*) F (x) = ‖x‖ f(

x

‖x‖

)for x 6= 0. Then F is a continuous extension of f , and F = ψ ◦ F ◦ ϕ−1 is a continuous

extension of f . If f is a homeomorphism then F is easily seen to be a bijection, and by

34

Lemma 4.25, F is a homeomorphism. For part (2), Proposition 5.1 shows that thereexist homeomorphisms G : Bn → Bn and G′ : Bm → Bm such that G(0) = ϕ−1(p) and

G′(0) = ψ−1(p′). Modify the previous definitions by setting f = (G′)−1 ◦ψ−1 ◦ f ◦ϕ ◦Gand F = ψ ◦G′ ◦ F ◦G−1 ◦ ϕ−1. Then

F (p) = (ψ ◦G′ ◦ F ◦G−1)(ϕ−1(p))

= (ψ ◦G′ ◦ F )(0)

= (ψ ◦G′)(0)

= p′

as desired. �

Theorem 106. [Problem 5-2] Suppose D is a closed n-cell, n ≥ 1.

(1) Given any point p ∈ IntD, there is a continuous function F : D → [0, 1] suchthat F−1({1}) = ∂D and F−1({0}) = {p}.

(2) Any continuous function f : ∂D → [0, 1] extends to a continuous function F :D → [0, 1] that is strictly positive in IntD.

Proof. For (1), apply Theorem 105 to the map f : ∂D → [−1, 1] satisfying f(x) = 1 forall x ∈ D, obtaining a map F : D → [−1, 1] with F (p) = 0. From (*) it is easy to seethat F−1({1}) = ∂D and F−1({0}) = {p}. For (2), repeat the proof of Theorem 105

but set F (0) = 1 and

F (x) = 1− ‖x‖ f(

x

‖x‖

)for x 6= 0. �

Theorem 107. Let X be a connected topological space and let ∼ be an equivalencerelation on X. If every x ∈ X has a neighborhood U such that p ∼ q for every p, q ∈ U ,then p ∼ q for every p, q ∈ X.

Proof. Let p ∈ X and let S = {q ∈ X : p ∼ q}. If q ∈ S then there is a neighborhoodU of q such that q1 ∼ q2 for every q1, q2 ∈ U . In particular, for every r ∈ U we havep ∼ q and q ∼ r which implies that p ∼ r, and U ⊆ S. This shows that S is open. Ifq ∈ X \S then there is a neighborhood U of q such that q1 ∼ q2 for every q1, q2 ∈ U . Ifp ∼ r for some r ∈ U then p ∼ q since q ∼ r, which contradicts the fact that q ∈ X \S.Therefore U ⊆ X \ S, which shows that S is closed. Since X is connected, S = X. �


(1) Given any two points p, q ∈ Bn, there is a homeomorphism ϕ : Bn → Bn suchthat ϕ(p) = q and ϕ|∂Bn = Id∂Bn.

35

(2) For any topological manifold X, every point of X has a neighborhood U withthe property that for any p, q ∈ U , there is a homeomorphism from X to itselftaking p to q.

(3) Every connected topological manifold is topologically homogeneous.

Proof. Part (1) follows from applying Theorem 105 to the map Id∂Bn . Let x ∈ X andchoose a regular coordinate ball B around x. If p, q ∈ B then it follows from part (1)that there exists a homeomorphism ϕ : B → B such that ϕ(p) = q and ϕ|∂B = Id∂B.Since

{B,X \B

}is a closed cover of X, the gluing lemma shows that there is a (unique)

homeomorphism ψ : X → X taking p to q satisfying ψ|X\B = IdX\B. Part (3) followsby applying Theorem 107 to the equivalence relation defined by p ∼ q if and only ifthere exists a homeomorphism from X to itself taking p to q. �

Lemma 109. Let X be a Hausdorff space. If p1, . . . , pn are distinct points in X, thenthere exist neighborhoods U1, . . . , Un with pi ∈ Ui and Ui ∩ Uj = ∅ for every i 6= j.

Proof. We use induction on n. If n = 2 then the statement is clearly true. Assume thatthe statement is true for n distinct points and let p1, . . . , pn+1 be distinct points in X.There exist neighborhoods U1, . . . , Un with pi ∈ Ui and Ui ∩Uj = ∅ for every i 6= j. Foreach 1 ≤ i ≤ n, choose a neighborhoods Ei of pi and Fi of pn+1 such that Ei ∩ Fi = ∅.Then U1 ∩ E1, . . . , Un ∩ En, F1 ∩ · · · ∩ Fn are the desired neighborhoods of p1, . . . , pn+1

respectively. �

Theorem 110. [Problem 5-4] If M is a connected topological manifold with dimM > 1and (p1, . . . , pk) and (q1, . . . , qk) are two ordered k-tuples of distinct points in M , thenthere is a homeomorphism F : M →M such that F (pi) = qi for i = 1, . . . , k.

Proof. Choose neighborhoods U1, . . . , Uk with pi ∈ Ui and Ui ∩ Uj = ∅ for every i 6=j and similarly choose neighborhoods V1, . . . , Vk with qi ∈ Vi. Then choose regularcoordinate balls E1, . . . , Ek around p1, . . . , pk and regular coordinate balls F1, . . . , Fkaround q1, . . . , qk. For each 1 ≤ j ≤ k the manifold Mj = M \

⋃i 6=j Ei∪Fi is connected,

so Theorem 108 shows that there exists a homeomorphism ϕj : Mj → Mj that takespj to qj. By the gluing lemma, we can extend ϕj to a homeomorphism ψj : M → Msuch that ψj|Ei∪Fi is the identity for every i 6= j. Then ψ = ψ1 ◦ · · · ◦ ψk is the desiredhomeomorphism taking pj to qj for j = 1, . . . , k. �

Theorem 111. [Problem 5-5] Suppose X is a topological space and {Xα} is a familyof subspaces whose union is X. The topology of X is coherent with the subspaces {Xα}if and only if it is the finest topology on X for which all of the inclusion maps Xα ↪→ Xare continuous.

Proof. Suppose that the topology T of X is coherent with {Xα} and let T ′ be sometopology for which the inclusion maps iα : Xα ↪→ X are continuous. If U ∈ T ′ then

36

every i−1α (U) = U ∩Xα is open, so U is open in T since T is coherent with {Xα}. Thisshows that T ′ ⊆ T . Conversely, suppose that T is the finest topology for which theinclusion maps iα are continuous. Let U be a subset of X such that U ∩Xα ∈ T for allα. If U is not open in T then by defining a new topology T ′ ⊇ T such that U ∈ T ′,we obtain a finer topology such that every iα is continuous. This is a contradiction, soU must be open in T . �

Theorem 112. [Problem 5-6] Suppose X is a topological space. The topology of X iscoherent with each of the following collections of subspaces of X:

(1) Any open cover of X.(2) Any locally finite closed cover of X.

Proof. Let {Uα}α∈A be an open cover of X and let U be a subset of X such that U ∩Uαis open for every α ∈ A. Then

U = U ∩X = U ∩⋃α∈A

Uα =⋃α∈A

U ∩ Uα

is open, which proves (1). Now let {Eα}α∈A be a locally finite closed cover of X andlet E be a subset of X such that E ∩ Eα is closed for every α ∈ A. Let x be a limitpoint of E; we want to show that x ∈ E. Since {Eα} is locally finite, there exists aneighborhood U of x that intersects with finitely many elements Eα1 , . . . , Eαn ∈ {Eα}.If V is a neighborhood of x then U ∩V is also a neighborhood of x, so there exists somey ∈ U ∩ V not equal to x such that y ∈ E. But then y ∈ Eαi for some i, which showsthat x is a limit point of E ∩

⋃ni=1Eαi . Since

E ∩n⋃i=1

Eαi =n⋃i=1

E ∩ Eαi

is closed, x ∈ E ∩⋃ni=1Eαi ⊆ E. This proves (2). �

Theorem 113. [Problem 5-7] Suppose X is a topological space whose topology is co-herent with a collection {Xα}α∈A of subspaces of X, and for each α ∈ A we are givena continuous map fα : Xα → Y such that fα|Xα∩Xβ = fβ|Xα∩Xβ for all α and β. Thenthere exists a unique continuous map f : X → Y whose restriction to each Xα is fα(cf. Theorem 99).

Proof. For each x ∈ X we have x ∈ Xα for some α, so we can set f(x) = fα(x). Thismakes f a well-defined map since fα|Xα∩Xβ = fβ|Xα∩Xβ for all α and β. Let U be openin Y . Then for every α ∈ A we have that f−1α (U) = f−1(U) ∩ Xα is open in Xα, sof−1(U) is open since X is coherent with {Xα}. �

Theorem 114. [Problem 5-8] If X is any CW complex, the topology of X is coherentwith the collection of subspaces {Xn : n ≥ 0}.

37

Proof. Let U be a subset of X and suppose that U ∩Xn is closed in Xn for every n ≥ 0.If e is an n-cell in X then e∩U ∩Xn is closed in e since e is closed in Xn. By condition(W), U is closed in X. �

Theorem 115. [Problem 5-10] Every CW complex is compactly generated.

Proof. Let X be a CW complex and let U be a subset of X such that U ∩K is closedfor every compact set K ⊆ X. Then every U ∩ e is closed, so U is closed by condition(W). �

Theorem 116. [Problem 5-11] A CW complex is locally compact if and only if it islocally finite.

Proof. Let X be a CW complex. Suppose that X is locally finite and let x ∈ X.Since the collection {e : e ∈ E} is locally finite, there exists a neighborhood U of x thatintersects with finitely many elements e1, . . . , en. Then U is a precompact neighborhoodof x, since U is a closed subset of the compact set e1 ∪ · · · ∪ en. Conversely, supposethat X is locally compact, let x ∈ X and let U be a precompact neighborhood of x.By Theorem 5.14, U is contained in a finite subcomplex, so U intersects finitely manycells of X. �

Theorem 117. [Problem 5-12] Let Pn be n-dimensional projective space. The usualinclusion Rk+1 ⊆ Rn+1 for k < n allows us to consider Pk as a subspace of Pn. ThenPn has a CW decomposition with one cell in each dimension 0, . . . , n such that thek-skeleton is Pk for 0 < k < n.

Proof. We use induction on n. The result is clearly true for n = 0, so assume that theresult is true for Pn−1 and consider Pn. Define the map

F : Bn → Rn+1 \ {0}

(x1, . . . , xn) 7→(x1, . . . , xn,

√1− |x1|2 − · · · − |xn|2

)and let q : Rn+1\{0} → Pn be the quotient map making two nonzero points x, y ∈ Rn+1

equivalent if x = λy for some λ ∈ R. We can write Pn as the disjoint union of Pn−1and the set Q = {[p1, . . . , pn] ∈ Pn : pn 6= 0}. It is clear that F (∂Bn) ⊆ Rn × {0}, so(q ◦ F )(∂Bn) ⊆ Pn−1; it is also clear that (q ◦ F )(Bn) = Q since F (Bn) is the upperhemisphere of Sn−1. Finally, (F |Bn)−1 is the map that discards the last coordinate,which is continuous. This shows that Q is an n-cell with characteristic map q ◦ F , andthat Pn is a CW complex. �

Theorem 118. [Problem 5-13] Let CPn be n-dimensional complex projective space.Then CPn has a CW decomposition with one cell in each even dimension 0, 2, . . . , 2nsuch that the 2k-skeleton is CPk for 0 < k < n.

38

Proof. Proceed as in Theorem 117, replacing Bn with the complex unit ball

Bn =

{(z1, . . . , zn) ∈ Cn :

√|z1|2 + · · ·+ |zn|2 = 1

}which has dimension 2n. �

Theorem 119. [Problem 5-14] Every nonempty compact convex subset D ⊆ Rn is aclosed cell of some dimension.

Proof. Let x + S be an affine subspace of minimal dimension k containing D and letA = [v0, . . . , v`] be a simplex of maximal dimension contained in D. Suppose that` < k. There exists a point y ∈ D affinely independent from v0, . . . , v`, for otherwiseD would be contained in the affine subspace spanned by A, which has dimension lessthan k. Since D is convex, it contains the simplex [v0, . . . , v`, y], which has dimension`+1. But this contradicts the fact that A is a simplex of maximal dimension containedin D. Therefore ` = k, and considering D as a subset of x + S, the interior of D isnonempty. Now let ϕ : S → Rk be an isomorphism, which is also a homeomorphism.Applying Proposition 5.1 to ϕ(D) shows that D is a closed k-cell. �

Chapter 6. Compact Surfaces

Theorem 120. [Exercise 6.11] Each elementary transformation of a polygonal presen-tation produces a topologically equivalent presentation.

Proof. We only prove the result for subdividing and reflecting. Let P = 〈S | W1, . . . ,Wk〉be a polygonal presentation and let P ′ = 〈S, e | W ′

1, . . . ,W′k〉 be the presentation formed

by replacing every occurrence of a by ae and every occurrence of a−1 by e−1a−1,taking each word Wi to W ′

i . First assume P has words of length 3 or more. LetP1, . . . , Pk be polygonal regions for P , let P ′1, . . . , P

′k be the polygonal regions for

P , and let π :∐k

i=1 Pi → M and π′ :∐k

i=1 P′i → M ′ be the quotient maps. Let

f :∐k

i=1 Pi →∐k

i=1 P′i be a map that takes edges labeled with a to the two correspond-

ing edges labeled a and e (such that the preimage of a has length 1/2), and similarlyfor a−1. This map can be chosen to be a homeomorphism by Theorem 105. Sinceπ′ ◦ f makes the same identifications as π, M and M ′ are homeomorphic. If P hasa single word of length 2, it is easy to check that the individual cases are homeomor-phic to the sphere or projective plane (as is noted in Example 6.9). For reflection,

define f :∐k

i=1 Pi →∐k

i=1 P′i by sending each edge to itself, but with the opposite

orientation. �

Example 121. [Problem 6-1] Show that a connected sum of one or more projectiveplanes contains a subspace that is homeomorphic to the Mobius band.

39

Since P2#P2 is homeomorphic to the Klein bottle and the Klein bottle contains a copyof the Mobius band.

Example 122. [Problem 6-2] Note that both a disk and a Mobius band are manifoldswith boundary, and both boundaries are homeomorphic to S1. Show that it is possibleto obtain a space homeomorphic to a projective plane by attaching a disk to a Mobiusband along their boundaries.

We have the presentation 〈a, b, c, d, e | abcdec〉 for a Mobius band and the presentation〈a, b, c, d | abe−1d−1〉 for a (closed) disk with its edges identified with the boundary ofthe Mobius band. Attaching the disk to the Mobius band gives the presentation⟨

a, b, c, d, e | abe−1d−1, abcdec⟩≈⟨a, b, c, d, e | abe−1d−1, decabc

⟩≈ 〈a, b, c | abcabc〉≈ 〈a | aa〉 ,

which is the projective plane.

Example 123. [Problem 6-3] Show that the Klein bottle is homeomorphic to a quotientobtained by attaching two Mobius bands together along their boundaries.

This corresponds to the presentation⟨a, b, c, d | abcb, a−1dc−1d

⟩≈⟨a, b, c, d | bcba, a−1dc−1d

⟩≈⟨b, c, d | bcbdc−1d

⟩≈⟨b, c, d, e | bce, e−1bdc−1d

⟩≈⟨b, c, d, e | ebc, c−1de−1bd

⟩≈⟨b, d, e | ebde−1bd

⟩≈⟨e, f | efe−1f

⟩≈⟨e, f | fefe−1

⟩,

which is the Klein bottle.

Theorem 124. [Problem 6-5] Every compact 2-manifold with boundary is homeomor-phic to a compact 2-manifold with finitely many open cells removed.

Proof. Let M be a compact 2-manifold with boundary. By Theorem 5.27, there is ahomeomorphism ϕ : ∂M →

∐ki=1 S1. Let M ′ be the compact 2-manifold formed by

attaching half of a sphere to each ϕ−1(S1); then M is homeomorphic to M ′ with theinteriors of the half-spheres removed. �

Example 125. [Problem 6-6] For each of the following surface presentations, computethe Euler characteristic and determine which of our standard surfaces it represents.

40

(1) 〈a, b, c | abacb−1c−1〉(2) 〈a, b, c | abca−1b−1c−1〉

We run the classification algorithm on each of the presentations:⟨a, b, c | abacb−1c−1

⟩≈⟨a, b, c | cb−1c−1aba

⟩≈⟨a, b, c, d | cb−1c−1ad, d−1ba

⟩≈⟨a, b, c, d | dcb−1c−1a, a−1b−1d

⟩≈⟨b, c, d | ddcb−1c−1b−1

⟩≈⟨b, c, d, e | ddcb−1e, e−1c−1b−1

⟩≈⟨b, c, d, e | eddcb−1, bce

⟩≈ 〈c, d, e | ddccee〉 ,

so (1) is homeomorphic to P2#P2#P2 with Euler characteristic −1;⟨a, b, c | abca−1b−1c−1

⟩≈⟨a, b, c | c−1abca−1b−1

⟩≈⟨a, b, c, d | c−1ad, d−1bca−1b−1

⟩≈⟨a, b, c, d | dc−1a, a−1b−1d−1bc

⟩≈⟨b, c, d | dc−1b−1d−1bc

⟩≈⟨b, c, d, e | dc−1b−1e, e−1d−1bc

⟩≈⟨b, c, d, e | edc−1b−1, bce−1d−1

⟩≈⟨d, e | ede−1d−1

⟩,

so (2) is homeomorphic to the torus with Euler characteristic 0.

Chapter 7. Homotopy and the Fundamental Group

Theorem 126. [Exercise 7.6] Let B ⊆ Rn be any convex set, X be any topologicalspace, and A be any subset of X. Then any two continuous maps f, g : X → B thatagree on A are homotopic relative to A.

Proof. The straight-line homotopy between f and g is in fact a homotopy relative toA. �

Theorem 127. [Exercise 7.8] Let X be a topological space. For any points p, q ∈ X,path homotopy is an equivalence relation on the set of all paths in X from p to q.

Proof. This follows from Proposition 7.1, since the combined homotopy is still stationaryon {0, 1}. �

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Theorem 128. [Exercise 7.14] Let X be a path-connected topological space.

(1) Let f, g : I → X be two paths from p to q. Then f ∼ g if and only if f · g ∼ cp.(2) X is simply connected if and only if any two paths in X with the same initial

and terminal points are path-homotopic.

Proof. For (1), we have f ∼ g ⇔ f ·g ∼ g ·g ∼ cp. It follows that X is simply connectedif and only if every element of π1(X) is the identity, i.e. f · g ∼ cp for all paths f, gfrom p to q. �

Theorem 129. [Exercise 7.15] Every convex subset of Rn is simply connected, and Rn

itself is simply connected.


Theorem 130. [Exercise 7.23] The path homotopy relation is preserved by compositionwith continuous maps. That is, if f0, f1 : I → X are path-homotopic and ϕ : X → Y iscontinuous, then ϕ ◦ f0 and ϕ ◦ f1 are path-homotopic.

Proof. Let H : I × I → X be a path homotopy from f0 to f1. Then ϕ ◦H is easily seento be a path homotopy from ϕ ◦ f0 to ϕ ◦ f1. �


(1) A retract of a connected space is connected.(2) A retract of a compact space is compact.(3) A retract of a retract is a retract; that is, if A ⊆ B ⊆ X, A is a retract of B,

and B is a retract of X, then A is a retract of X.

Proof. Let r : X → A be a retraction; (1) and (2) follow from the fact that r iscontinuous. Let r1 : B → A and r2 : X → B be retractions. Then r1 ◦ r2 : X → A isalso a retraction, which proves (3). �

Theorem 132. [Exercise 7.33] The circle is not a retract of the closed disk.

Proof. The circle is not simply connected, but the closed disk is convex and thereforesimply connected. By Corollary 7.29, the circle cannot be a retract of the closed disk.

�

Theorem 133. [Exercise 7.36] Homotopy equivalence is an equivalence relation on theclass of all topological spaces.

42

Proof. It is clear that homotopy equivalence is reflexive and symmetric. Let ϕ : X → Yand ψ : Y → Z be homotopy equivalences with homotopy inverses ϕ : Y → X andψ : Z → Y . Then ϕ ◦ ψ : Z → X is a homotopy inverse for ψ ◦ ϕ : X → Z, since

ϕ ◦ ψ ◦ ψ ◦ ϕ ' ϕ ◦ IdY ◦ϕ ' ϕ ◦ ϕ ' IdX

and similarly ψ ◦ ϕ ◦ ϕ ◦ ψ ' IdZ by Theorem 130. �

Theorem 134. [Exercise 7.42] The following are equivalent:

(1) X is contractible.(2) X is homotopy equivalent to a one-point space.(3) Each point of X is a deformation retract of X.

Proof. (1)⇔ (2) and (3)⇒ (2) are obvious. If X is homotopy equivalent to a one-pointspace {p} ⊆ X then X is simply connected, so there is a path γ : I → X from p to anypoint q ∈ X. Let H : X × I → X be a deformation retraction to {p}; then

H ′(x, t) =

{H(x, 2t) if 0 ≤ t ≤ 1/2,

γ(2t− 1) if 1/2 ≤ t ≤ 1

is a deformation retraction from X to the one-point space {q}. �

Theorem 135. [Exercise 7.58] If a coproduct exists in a category, it is unique up toan isomorphism that respects the injections.

Proof. Let (S, (iα)) and (S ′, (i′α)) be two coproducts for a family of objects (Xα). Bythe universal property, there exist unique morphisms f : S → S ′ and f ′ : S ′ → S suchthat i′α = f ◦ iα and iα = f ′ ◦ i′α for every α. Then iα = f ′ ◦ f ◦ iα, so by uniqueness wehave f ′ ◦ f = IdS. Similarly, i′α = f ◦ f ′ ◦ i′α implies that f ◦ f ′ = IdS′ . Therefore f isan isomorphism (respecting the injections). �

Theorem 136. Let X1, . . . , Xn be topological spaces.

(1) Let Y be any topological space and let f, g : Y → X1 × · · · ×Xn be continuousmaps. Then f ' g if and only if fj ' gj for every j, where fj = πj ◦ f ,gj = πj ◦ g, and πj : X1 × · · · ×Xn → Xj is the canonical projection.

(2) Let Y be any topological space and let f, g : X1 q · · · qXn → Y be continuousmaps. Then f ' g if and only if fj ' gj for every j, where fj = f ◦ij, gj = g◦ij,and ij : Xj → X1 × · · · ×Xn is the canonical injection.

Proof. In both (1) and (2), if f ' g then fj ' gj for every j by Proposition 7.2. Nowsuppose that Hj : Y × I → Xj are homotopies from fj to gj in (1). Then the mapH : Y × I → X1 × · · · ×Xn given by

H(s, t) = (H1(s, t), . . . , Hn(s, t))

43

is a homotopy from f to g. Similarly, suppose that Hj : Xj×I → Y are homotopies fromfj to gj in (2). There exists a unique continuous map H : (X1×I)q· · ·q (Xn×I)→ Ysuch that H|Xj×I = Hj for every j. Let

ι : (X1 q · · · qXn)× I → (X1 × I)q · · · q (Xn × I)

be the identity map, which is continuous by Theorem 60. Then the map H ◦ ι is ahomotopy from f to g. �

Theorem 137. [Problem 7-1] Suppose f, g : X → Sn are continuous maps such thatf(x) 6= −g(x) for every x ∈ X. Then f and g are homotopic.

Proof. Define H : X × I → Sn by

H(x, t) =(1− t)f(x) + tg(x)

‖(1− t)f(x) + tg(x)‖;

we must check that the denominator is never zero. If (1 − t)f(x) + tg(x) = 0 and0 < t < 1 then

f(x) =(1− t)f(x)

‖(1− t)f(x)‖=−tg(x)

‖−tg(x)‖= −g(x)

since ‖f(x)‖ = ‖g(x)‖ = 1. We are given that this cannot happen, so H is a homotopyfrom f to g. �

Theorem 138. [Problem 7-2] Suppose X is a topological space, and g is any path inX from p to q. Let φg : π1(X, p) → π1(X, q) denote the group isomorphism defined inTheorem 7.13.

(1) If h is another path in X starting at q, then φg·h = φh ◦ φg.(2) For any continuous map ψ : X → Y the following diagram commutes:

π1(X, p) π1(Y, ψ(p))

π1(X, q) π1(Y, ψ(q)).

ψ∗

φψ◦gφg

ψ∗

Proof. (1) follows from the computation

φg·h[f ] = [g · h] · [f ] · [g · h]

= [h · g] · [f ] · [g · h]

= [h] · ([g] · [f ] · [g]) · [h]

= φh ◦ φg.

44

For (2), we need to show that φψ◦g ◦ ψ∗ = ψ∗ ◦ φg. Since ψ∗ is a homomorphism,

(ψ∗ ◦ φg)[f ] = ψ∗([g] · [f ] · [g])

= [ψ ◦ g] · [ψ ◦ f ] · [ψ ◦ g]

= [ψ ◦ g] · [ψ ◦ f ] · [ψ ◦ g]

= (φψ◦g ◦ ψ∗)[f ].

�

Theorem 139. [Problem 7-3] Let X be a path-connected topological space, and letp, q ∈ X. Then π1(X, p) is abelian if and only if all paths from p to q give the sameisomorphism of π1(X, p) with π1(X, q).

Proof. Note that π1(X, p) is abelian if and only if all of its inner automorphisms aretrivial. If g1 and g2 are paths from p to q then by 138,

φg1 = φg2 ⇔ φ−1g2 ◦ φg1 = Idπ1(X,p)

⇔ φg2 ◦ φg1 = Idπ1(X,p)

⇔ φg1·g2 = Idπ1(X,p) .

Therefore it suffices to show that φg1·g2 = Idπ1(X,p) for all paths g1, g2 from p to q if andonly if every inner automorphism of π1(X, p) is trivial. One direction is immediate, forφg1·g2 is always an inner automorphism. Conversely, if φg is an inner automorphism(where g is a loop based at p) and h is a path from p to q then φg = φ(g·h)·h =Idπ1(X,p). �

Theorem 140. [Problem 7-4] Let F : I × I → X be a continuous map, and let f , g,h, and k be the paths in X defined by

f(s) = F (s, 0);

g(s) = F (1, s);

h(s) = F (0, s);

k(s) = F (s, 1).

Then f · g ∼ h · k.

Proof. Define G : I × I → I × I by

F (s, t) =

{2s(1− t, t) if s ∈ [0, 1/2],

(1− t, t) + (2s− 1)(t, 1− t) if s ∈ (1/2, 1].

Then F ◦G is a path homotopy from f · g to h · k. �

Theorem 141. [Problem 7-5] Let G be a topological group.

45

(1) Up to isomorphism, π1(G, g) is independent of the choice of the base point g ∈ G.(2) π1(G, g) is abelian.

Proof. Let g1, g2 ∈ G. The map x 7→ g2g−11 x is a homeomorphism of G with itself, so the

induced map from π1(X, g1)→ π1(X, g2) is an isomorphism. Therefore we can assumethat g = 1 for part (2). Let f and g be loops based at 1 ∈ G and define F : I × I → Gby (s, t) 7→ f(s)g(t). Then f · g ∼ g · f by Theorem 140, so [f ] · [g] = [g] · [f ]. Thisshows that π1(G, 1) is abelian. �

Lemma 142. Let f1, . . . , fn be paths in a topological space X such that fk(1) = fk+1(0)for every k = 1, . . . , n−1, and fn(1) = f1(0). Let f = f1 ·· · ··fn and f ′ = fn ·f1 ·· · ··fn−1.Let f and f ′ be the circle representatives of f and f ′ respectively. Then f is (freely)

homotopic to f ′.

Proof. Let µk : I → S1 be given by s 7→ exp((k + s)2πi/n). By reparametrizing f

and f ′, we may assume that fk = f ◦ µk−1 = f ′ ◦ µk for each k = 1, . . . , n. Define

H : S1 × I → X by H(z, t) = f(e−2πit/nz) where z is taken to be a complex number.

Then H is a homotopy from f to f ′. �

Theorem 143. [Problem 7-6] For any path-connected space X and any base pointp ∈ X, the map sending a loop to its circle representative induces a bijection betweenthe set of conjugacy classes of elements of π1(X, p) and [S1, X] (the set of free homotopyclasses of continuous maps from S1 to X).

Proof. Let C be the set of conjugacy classes of elements of π1(X, p) and denote theconjugacy class of an element [f ] ∈ π1(X, p) by [[f ]]. Let ϕ : C → [S1, X] be the mapthat sends an element [[f ]] ∈ C to the free homotopy class of the circle representative

f of f . We first check that ϕ is well-defined. Let [g] ∈ [[f ]] so that [g] = [h] · [f ] · [h] forsome [h] ∈ π1(X, p). Then Lemma 142 shows that the circle representative of h · f ·h is

(freely) homotopic to the circle representative of h ·h ·f ∼ f , so g ' f . Let α : S1 → Xbe a continuous map, let ω : I → X be given by ω(s) = α(e2πis), and let γ be a pathfrom p to ω(0). We have ω ' γ · γ · ω and by Lemma 142 the circle representativeof γ · γ · ω is homotopic to the circle representative of γ · ω · γ, so α ∈ ϕ([[γ · ω · γ]]).

This shows that ϕ is surjective. Finally, suppose that ϕ([[f ]]) = ϕ([[g]]) so that f ishomotopic to g. Then [f ] = [g], and in particular we have [[f ]] = [[g]]. This shows thatϕ is bijective. �

Theorem 144. [Problem 7-7] Suppose (M1, d1) and (M2, d2) are metric spaces. A mapf : M1 → M2 is said to be uniformly continuous if for every ε > 0 there exists aδ > 0 such that for all x, y ∈ M1, d1(x, y) < δ implies d2(f(x), f(y)) < ε. If M1 iscompact, then every continuous map f : M1 →M2 is uniformly continuous.

46

Proof. Let ε > 0 be given. For each t ∈M1, choose a number δ(t) such that d2(f(x), f(t)) <ε/2 whenever x ∈ M1 and d1(x, t) < δ(t). Since M1 is compact, the open coverU =

{Bδ(t)(t) : t ∈M1

}of M1 has a Lebesgue number δ. If x ∈M1 then the set Bδ(x)

is contained in some Bδ(t)(t) ∈ U , so for all y with d1(x, y) < δ we have

d2(f(x), f(y)) ≤ d2(f(x), f(t)) + d2(f(t), f(y)) < ε.

�

Theorem 145. [Problem 7-8] A retract of a Hausdorff space is a closed subset.

Proof. Let X be a Hausdorff space and let r : X → A be a retraction. Let x ∈X \A. Since X is Hausdorff, there exist neighborhoods U of x and V of r(x) such thatU ∩ V = ∅. Now U ∩ r−1(V ∩ A) is a neighborhood of x contained in X \ A, sincer(A ∩ U ∩ r−1(V ∩ A)) ⊆ A ∩ U ∩ V ∩ A = ∅. �

Theorem 146. [Problem 7-9] Suppose X and Y are connected topological spaces, andthe fundamental group of Y is abelian. If F,G : X → Y are homotopic maps such thatF (x) = G(x) for some x ∈ X, then F∗ = G∗ : π1(X, x)→ π1(Y, F (x)).

Proof. Let H : X × I → Y be a homotopy from F to G and let α : I → X be aloop based at x. Consider the map H ◦ (α × IdI) : I × I → Y . By Lemma 7.17 wehave f · g ∼ h · k where f = F ◦ α, g(s) = h(s) = H(x, s), and k(s) = G ◦ α. But[f ] · [g] = [g] · [f ] = [g] · [k] implies that [F ◦ α] = [f ] = [k] = [G ◦ α] since π1(Y, F (x))is abelian, which shows that F∗ = G∗. �

Theorem 147. Let X and Y be topological spaces. If F : X → Y is null-homotopicthen F∗ : π1(X, x)→ π1(Y, F (x)) is the trivial map for all x ∈ X.

Proof. Let H : X×I → Y be a homotopy from F to a constant map and let α : I → Xbe a loop based at x. Consider the map H ◦ (α× IdI) : I × I → Y . By Lemma 7.17 wehave f · g ∼ h · k where f = F ◦α, g(s) = h(s) = H(x, s), and k is a constant path. Wehave f ∼ f · g · g ∼ g · k · g ∼ cF (x) from Theorem 7.11, and therefore F∗ is trivial. �

Theorem 148. [Problem 7-10] Let X and Y be topological spaces. If either X or Y iscontractible, then every continuous map from X to Y is homotopic to a constant map.

Proof. Let f : X → Y be a continuous map. If IdX is homotopic to a constant map cthen f = f ◦ IdX ' f ◦ c, and f ◦ c is a constant map. Similarly, if IdY is homotopic toa constant map c then f = IdY ◦f ' c ◦ f , and c ◦ f is a constant map. �

Theorem 149. [Problem 7-11] The Mobius band is homotopy equivalent to S1.

47

Proof. We define the Mobius band B to be the geometric realization of the presentation〈a, b, c | abcb〉. In other words, it is the quotient space formed from I × I by identifyingthe edge {0} × I with {1} × I. Let q : I × I → B be the associated quotient map anddefine H : (I × I)× I → B by

H((x, y), t) = q

(x,

1

2+ t

(y − 1

2

)).

Then H descends to a continuous map H : B × I → B and in fact H is a (strong)deformation retraction from B to q(I × {1/2}) ≈ S1. �

Example 150. [Problem 7-12] Let X be the space of Example 5.9.

(1) {(0, 0)} is a strong deformation retract of X.(2) {(1, 0)} is a deformation retract of X, but not a strong deformation retract.

The map H : X × I → X given by

H(x, t) =

{x/(1− t) if 0 ≤ t < 1,

(0, 0) if t = 1

is a strong deformation retraction from X to {(0, 0)}. Furthermore,

H ′(x, t) =

{H(x, 2t) if 0 ≤ t ≤ 1/2,

(2t− 1, 0) if 1/2 < t ≤ 1.

is a deformation retraction from X to {(1, 0)}. However, {(1, 0)} cannot be a strongdeformation retract of X since it is a limit point of X but no retraction is possibledirectly towards (1, 0).

Theorem 151. [Problem 7-14] Let M be a compact connected surface that is not home-omorphic to S2. Then there is a point p ∈M such that M \ {p} is homotopy equivalentto a bouquet of circles.

Proof. This follows from Theorem 6.15. �

Theorem 152. [Problem 7-16] Given any family (Xα)α∈A of topological spaces, thedisjoint union space

∐αXα is their coproduct in the category Top.


Theorem 153. [Problem 7-17] The wedge sum is the coproduct in the category Top∗.

Proof. Let ((Xα, pα))α∈A be a family of pointed spaces, let X =∨α∈AXα and define

jα : Xα → X by jα = q ◦ iα where iα : Xα →∐

α∈AXα and q :∐

α∈AXα →∨α∈AXα is

the quotient map. Note that jα(pα) = jβ(pβ) for all α, β; denote this common value byp. Let (W, r) be a pointed space and let fα : Xα → W be a pointed continuous map.

48

There exists a unique continuous map f :∐

α∈AXα → W such that fα = f ◦ iα for allα. Since f(pα) = r for all α, there exists a unique continuous map g :

∨α∈AXα → W

such that f = g ◦ q. Then fα = f ◦ iα = g ◦ q ◦ iα = g ◦ jα for all α. This shows that(X, p) is the coproduct of ((Xα, pα))α∈A. �

Theorem 154. [Problem 7-18] Let (Gα)α∈A be a family of abelian groups. The directsum, together with the obvious injections iα : Gα ↪→

⊕αGα, is the coproduct of the

Gα’s in the category Ab.

Proof. Let X be an abelian group and let fα : Gα → X be homomorphisms. Letf :⊕

αGα → X be given by (gα) 7→∑

α∈A fα(gα), which is well-defined since gα = 0for all but finitely many α. It is clear that fα = f ◦ iα for all α. Suppose that f ′ isanother homomorphism such that fα = f ′ ◦ iα for all α. Then for all (gα) ∈

⊕αGα we

have

f ′((gα)) = f ′

(∑α∈A

iα(gα)

)=∑α∈A

(f ′ ◦ iα)(gα) =∑α∈A

fα(gα),

which shows that f ′ = f . �

Remark 155. [Problem 7-19] The direct sum does not yield the coproduct in the categoryGrp. Take G1 = G2 = Z, H = GL(2,R) and the maps fk : Gk → H defined by

f1(n) =

[1 n0 1

]and f2(n) =

[1 0n 1

].

There is no homomorphism f : Z⊕ Z→ H such that fk = f ◦ ik for k = 1, 2 since

f((1, 1)) = f((1, 0) + (0, 1)) =

[1 10 1

] [1 01 1

]=

[2 11 1

]but

f((1, 1)) = f((0, 1) + (1, 0)) =

[1 01 1

] [1 10 1

]=

[1 11 2

].

Chapter 8. The Circle

Theorem 156. [Exercise 8.7] A rotation of S1 is a map ρ : S1 → S1 of the formρ(z) = eiθz for some fixed eiθ ∈ S1. If ρ is a rotation, then N(ρ ◦ f) = N(f) for everyloop f in S1.

Proof. Define g : S1 → R by z 7→ θ/2π + f(z). Then g is a lift of ρ ◦ f , so

N(ρ ◦ f) = g(1)− g(0)

= θ/2π + f(1)− θ/2π − f(0)

= f(1)− f(0)

49

= N(f).

�

Theorem 157. [Problem 8-1] (cf. Theorem 80)

(1) If U ⊆ R2 is an open subset and x ∈ U , then U \ {x} is not simply connected.(2) If n > 2 then Rn is not homeomorphic to any open subset of R2.

Proof. Let B ⊆ U be an open ball of radius r > 0 around x. A loop γ that traversesthe circle ∂B counterclockwise once has a winding number of 1. If γ is null-homotopicin U \ {x} then it is also null-homotopic in R2 \ {x}, which contradicts Corollary 8.11.This shows that U \ {x} is not simply connected. Part (2) follows immediately fromCorollary 7.38. �

Theorem 158. [Problem 8-2] A nonempty topological space cannot be both a 2-manifoldand an n-manifold for any n > 2 (cf. Theorem 81).

Proof. Let M be a nonempty topological space that is both a 2-manifold and an n-manifold for some n > 2. Choose some p ∈ M and let ϕ1 : U1 → V1 and ϕ2 : U2 → V2be homeomorphisms where U1 and U2 are neighborhoods of p, V1 is open in R2, andV2 is open in Rn. Let B be an open ball around ϕ2(p) contained in ϕ2(U1 ∩ U2). ThenW1 = B\{ϕ2(p)} is homeomorphic toW2 = (ϕ1◦ϕ−12 )(B)\{ϕ1(p)}, butW2 is not simplyconnected by Theorem 157 while W1 is simply connected. This is a contradiction. �

Theorem 159. [Problem 8-3] Suppose M is a 2-dimensional manifold with boundary.Then the interior and boundary of M are disjoint (cf. Theorem 82).

Proof. Suppose p ∈ M is both an interior and boundary point. Choose coordinatecharts (U,ϕ) and (V, ψ) such that U, V are neighborhoods of p, ϕ(U) is open in IntH2,ψ(V ) is open in H2, and ψ(p) ∈ ∂H2. Let W = U ∩ V ; then ϕ(W ) is homeomorphic toψ(W ). But this is impossible, for ϕ(W ) \ {ϕ(p)} is not simply connected by Theorem157 while ψ(W ) \ {ψ(p)} is simply connected. �

Theorem 160. [Problem 8-4] A continuous map ϕ : S1 → S1 has an extension to a

continuous map φ : B2 → S1 if and only if it has degree zero.

Proof. Let ω : I → S1 be standard generator of π1(S1, 1). By Proposition 7.16, ϕ ◦ ωhas a winding number of zero if and only if ϕ extends to a continuous map from B2 toS1. �

Theorem 161. [Problem 8-5] Every nonconstant polynomial in one complex variablehas a zero.

50

Proof. Suppose that p(z) = zn+an−1zn−1 + · · ·+a1z+a0 is a polynomial with no zeros

and n > 0. We can assume that a0 6= 0, for otherwise p(0) = 0. Let

pε(z) = zn + an−1εzn−1 + · · ·+ a1ε

n−1z + a0εn

so that pε(z) = εnp(z/ε) when ε 6= 0. Since p has no zeros, the map H : S1×I → C\{0}given by H(z, t) = ptε(z) is a homotopy from z 7→ zn to pε|S1 . Therefore pε|S1 has awinding number of n, and the map φ : S1 → C \ {0} given by φ(z) = p(z/ε) = ε−npε(z)also has a winding number of n. But φ is homotopic to the constant loop cp(0) = ca0 bythe homotopy (z, t) 7→ p(tz/ε), which is a contradiction. �

Theorem 162. [Problem 8-6] Every continuous map f : B2 → B2 has a fixed point.

Proof. If f has no fixed point then we can define a continuous map ϕ : B2 → S1 by

ϕ(x) =x− f(x)

‖x− f(x)‖;

by Theorem 160, ϕ|S1 has degree zero. Define H : S1 × I → S1 by

H(x, t) =x− (1− t)f(x)

‖x− (1− t)f(x)‖.

If t = 0 then the denominator is never zero. Otherwise,

‖x− (1− t)f(x)‖ ≥ ‖‖x‖ − (1− t) ‖f(x)‖‖ ≥ t > 0

for t ∈ (0, 1] since ‖x‖ = 1 and ‖f(x)‖ ≤ 1. This shows that H is a well-definedhomotopy from ϕ|S1 to IdS1 , which contradicts the fact that ϕ|S1 has degree zero. �

Lemma 163. Let f : I → S1 be a loop with winding number n. Then there exists a lift

f : I → R of f such that f(0) = 0 and f(1) = n.

Proof. We can assume that f is based at 1 ∈ S1. Let α : I → S1 be the map s 7→ e2πins

and let α : I → R be the lift of α given by s 7→ ns. By Corollary 8.5, there exists

a lift f of f such that f(0) = 0 = α(0), and since α ∼ f , Corollary 8.6 shows that

f(1) = α(1) = n. �

Theorem 164. [Problem 8-7] If ϕ : S1 → S1 is continuous and degϕ 6= ±1, then ϕ isnot injective.

Proof. The result is clear when degϕ = 0, so assume otherwise; since degϕ 6= ±1, wehave |degϕ| > 1. Let ε : I → S1 be the map s 7→ e2πis. By Lemma 163, there existsa lift ϕ : I → R of ϕ ◦ ε such that ϕ(0) = 0 and |ϕ(1)| = |degϕ| > 1. Choose a point0 < s < 1 such that ϕ(s) = ±1. Noting that ε ◦ ϕ = ϕ ◦ ε, we have ϕ(ε(0)) = ϕ(ε(s))while ε(0) 6= ε(s), which shows that ϕ is not injective. �

51

Theorem 165. [Problem 8-8] Suppose ϕ, ψ : S1 → S1 are continuous maps of differentdegrees. Then there is a point z ∈ S1 where ϕ(z) = −ψ(z).


Theorem 166. [Problem 8-9] Suppose f : I → C \ {0} is a continuously differentiableloop. Then its winding number is given by

N(f) =1

2πi

ˆ 1

0

f ′(s)

f(s)ds.

Proof. Let f : I → R be a lift of f/ |f |; then exp(2πif(s)) = f(s)/r(s) for all s ∈ I,where r = |f |. Now

1

2πi

ˆ 1

0

f ′(s)

f(s)ds =

1

2πi

ˆ 1

0

2πir(s)f ′(s) exp(2πif(s)) + r′(s) exp(2πif(s))

r(s) exp(2πif(s))ds

=1

2πi

ˆ 1

0

(2πif ′(s) +

r′(s)

r(s)

)ds

=1

2πi

[2πif(s) + log r(s)

]10

=1

2πi[2πif(1)− 2πif(0) + log r(1)− log r(0)]

= f(1)− f(0)

= N(f).

�

Theorem 167. [Problem 8-10] A vector field on Rn is a continuous map V : Rn →Rn. If V is a vector field, a point p ∈ Rn is called a singular point of V if V (p) = 0,and a regular point if V (p) 6= 0. A singular point is isolated if it has a neighborhoodcontaining no other singular points. Suppose V is a vector field on R2, and let RV ⊆ R2

denote the set of regular points of V . For any loop f : I → RV , define the windingnumber of V around f , denoted by N(V, f), to be the winding number of the loopV ◦ f : I → R2 \ {0}.

(1) N(V, f) depends only on the path class of f .(2) Suppose p is an isolated singular point of V . Then N(V, fε) is independent

of ε for ε sufficiently small, where fε(s) = p + εω(s), and ω is the standardcounterclockwise loop around the unit circle. This integer is called the index ofV at p, and is denoted by Ind(V, p).

(3) Now assume V has finitely many singular points in the open unit disk. Thenthe index of V around the loop ω is equal to the sum of the indices of V at theinterior singular points.

52

pnC

γ

pnβ α

ω

Figure 0.1. A keyhole path.

Proof. Part (1) follows from Proposition 7.2. For (2), choose r > 0 so that Br(p)has no other singular points. If 0 < ε1, ε2 < r then H : I × I → RV given byH(s, t) = p+[(1−t)ε1+tε2]ω(s) is a homotopy from fε1 to fε2 , and N(V, fε1) = N(V, fε2)by part (1). For part (3), we use induction on the number n of singular points. Theresult is clear when n = 1. Assuming the result for n− 1, suppose we have n singularpoints p1, . . . , pn. Let γ be a keyhole path in B2\{p1, . . . , pn} that traverses the unit diskcounterclockwise and traverses a small circle C around pn clockwise, such that γ enclosesp1, . . . , pn−1 (see Figure 0.1). Then N(V, γ) encloses p1, . . . , pn−1, and ignoring pn, γ ishomotopic to ω. Therefore N(V, γ) =

∑n−1i=1 Ind(V, pi) by the induction hypothesis. Let

γ be the path γ with the gap closed, so that γ is homotopic to both γ and ω·α·β ·α whereα is a path from 1 ∈ S1 to a point x on C and β is a clockwise loop around C based atx. By applying Theorem 166 we have N(V, γ) = N(V, ω ·α ·β ·α) = N(V, ω) +N(V, β)since α cancels α in the integral. But N(V, β) = − Ind(V, pn), so

N(V, ω) = N(V, γ)−N(V, β)

= N(V, γ)−N(V, β)

=n∑i=1

Ind(V, pi).

�

Theorem 168. [Problem 8-11] For each continuous map ϕ : T2 → T2 there is a 2× 2integer matrix D(ϕ) with the following properties:

(1) D(ψ ◦ ϕ) is equal to the matrix product D(ψ)D(ϕ).(2) Two continuous maps ϕ and ψ are homotopic if and only if D(ϕ) = D(ψ).

53

(3) For every 2× 2 integer matrix E, there is a continuous map ϕ : T2 → T2 suchthat D(ϕ) = E. If E is invertible then ϕ is a homeomorphism.

(4) ϕ is homotopic to a homeomorphism only if and only if D(ϕ) is invertible overthe integers.

Proof. We can consider π1 (T2, (1, 1)) as a free Z-module with basis {[ω1], [ω2]} whereωj is the standard loop in the jth copy of S1. For a continuous map ϕ : T2 → T2 suchthat ϕ(1, 1) = (1, 1), define D(ϕ) to be the unique 2× 2 integer matrix that representsthe Z-map

ϕ∗ : π1(T2, (1, 1)

)→ π1

(T2, (1, 1)

).

Note that if ij : S1 → T2 and pj : T2 → S1 are the canonical injections and projectionsthen

(*) D(ϕ) =

[ν11(1) ν12(1)ν21(1) ν22(1)

]where ϕjk = pj ◦ ϕ ◦ ik and νjk : Z → Z is the unique homomorphism such that(ϕjk)∗([ω]) = [ω]νjk(1). In other words, the entries of the matrix D(ϕ) are the degreesof the component maps ϕjk. Part (1) follows immediately since (ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗. Ifϕ and ψ are homotopic then ϕjk ' ψjk for 1 ≤ j, k ≤ 2, and D(ϕ) = D(ψ) from (*).Conversely, if D(ϕ) = D(ψ) then ϕjk ' ψjk for 1 ≤ j, k ≤ 2, so ϕ ' ψ by Theorem 136.

For part (3), let Ejk be the entries of E. If we set ϕjk(s) = sEjk , then there exist uniquecontinuous maps ϕk : S1 → T2 such that ϕjk = pj ◦ϕk for each j. Define ϕ : T2 → T2 byϕ(s1, s2) = ϕ1(s1)ϕ2(s2) where the multiplication is the component-wise multiplicationof complex numbers; then D(ϕ) = E. More explicitly, we have

ϕ

[s1s2

]=

[ϕ11(s1)ϕ12(s2)ϕ21(s1)ϕ22(s2)

]=

[sE111 sE12

2

sE211 sE22

2

].

If ϕE : T2 → T2 denotes the above map for the matrix E then ϕEF = ϕE ◦ϕF by directcomputation. It follows that ϕ is a homeomorphism if E is invertible.

Finally, if ϕ is homotopic to a homeomorphism ψ then ψ−1 ◦ ϕ ' ψ−1 ◦ ψ = IdT2 , soD(ψ−1 ◦ϕ) = D(ψ−1)D(ϕ) = I and D(ϕ) is invertible. Conversely, if D(ϕ) is invertiblethen there is a homeomorphism ψ such that D(ψ) = D(ϕ) by part (3), so ϕ ' ψ.

For the general case when ϕ(1, 1) is not necessarily equal to (1, 1), define D(ϕ) =D(ρϕ ◦ ϕ) where ρϕ is the rotation (z1, z2) 7→ ϕ(1, 1)−1(z1, z2) that takes ϕ(1, 1) to(1, 1); the preceding proofs can then be easily modified. �

54

Chapter 9. Some Group Theory

Theorem 169. [Exercise 9.10] Let S be a set. For any group H and any map ϕ : S →H, there exists a unique homomorphism φ : F (S)→ H extending ϕ:

F (S)

S H

φi

ϕ

Proof. For each σ ∈ S there is a unique homomorphism ϕσ : F (σ) → H such thatϕσ(σ) = ϕ(σ). If iσ : F (σ)→ F (S) is the canonical injection, by Theorem 9.5 there isa unique homomorphism φ : F (S) → H such that ϕσ = φ ◦ iσ for every σ ∈ S. Sinceφ(σ) = (φ ◦ iσ)(σ) = ϕ(σ), the map φ extends ϕ. Furthermore, if φ′ : F (S) → H isanother homomorphism that extends ϕ then

ϕσ(σn) = ϕ(σ)n = [(φ′ ◦ iσ)(σ)]n = (φ′ ◦ iσ)(σn)

for every σ ∈ S, so φ′ = φ. �

Theorem 170. [Exercise 9.11] The free group on S is the unique group (up to isomor-phism) satisfying the characteristic property of Theorem 169.

Proof. Let G and G′ be two groups satisfying the characteristic property, and let i :S → G and i′ : S → G′ be the inclusion maps. There exist unique homomorphismsϕ : G′ → G and ϕ′ : G→ G′ such that i = ϕ◦i′ and i′ = ϕ′◦i. Then ϕ◦ϕ′◦i = ϕ◦i′ = i,so ϕ ◦ ϕ′ = IdG since both ϕ ◦ ϕ′ and IdG satisfy the diagram with G in place of F (S)and H. Similarly, ϕ′ ◦ ϕ = IdG′ . This shows that G ∼= G′. �

Theorem 171. [Exercise 9.15] Let S be a nonempty set.

(1) Given any abelian group H and any map ϕ : S → H, there exists a uniquehomomorphism φ : ZS → H extending ϕ.

(2) The free abelian group Z {σ1, . . . , σn} on a finite set is isomorphic to Zn via themap (k1, . . . , kn) 7→ k1σ1 + · · ·+ knσn.

Proof. If φ : ZS → H is a homomorphism extending ϕ then

φ(k1σ1 + · · ·+ knσn) = k1φ(σ1) + · · ·+ knφ(σn)

= k1ϕ(σ1) + · · ·+ knϕ(σn)

for all k1, . . . , kn ∈ Z and σ1, . . . , σn ∈ S, which shows that φ is unique. But the aboveequation clearly defines a homomorphism, so φ exists. Part (2) is obvious. �

55

Theorem 172. [Exercise 9.16] For any set S, the identity map of S induces an isomor-phism between the free abelian group on S and the direct sum of infinite cyclic groupsgenerated by elements of S:

ZS ∼=⊕σ∈S

Z {σ} .

Proof. This is clear from the definition of the direct sum. �


(1) An abelian group is free abelian if and only if it has a basis.(2) Any two free abelian groups whose bases have the same cardinality are isomor-

phic.

Proof. If G is a free abelian group then there is a subset S ⊆ G such that the inclusionS ↪→ G induces an isomorphism ϕ : ZS → G. Then {ϕ(σ) : σ ∈ S} is clearly abasis for G. Conversely, if an abelian group G has a basis S then the induced mapϕ : ZS → G has trivial kernel and is surjective. For (2), let G and G′ be two free abeliangroups with bases B and B′ respectively; assume that there is a bijection ϕ : B → B′.Since G ∼= ZB and G′ ∼= ZB′, it suffices to show that ZB ∼= ZB′. If i : B → ZBand i′ : B′ → ZB′ are the canonical injections, there exist unique homomorphismsφ : ZB′ → ZB and φ′ : ZB → ZB′ such that i′ ◦ ϕ = φ′ ◦ i and i ◦ ϕ−1 = φ ◦ i′. Thenφ ◦ φ′ ◦ i = φ ◦ i′ ◦ ϕ = i ◦ ϕ−1 ◦ ϕ = i, so φ ◦ φ′ = IdZB by uniqueness in Theorem 171.Similarly, φ′ ◦ φ = IdZB′ . This shows that ZB ∼= ZB′. �

Theorem 174. [Problem 9-1] The free product of two or more nontrivial groups isinfinite and nonabelian.

Proof. Let G1 and G2 be nontrivial groups. Choose non-identity elements g1 ∈ G1 andg2 ∈ G2; then wn =

∏ni=1 g1g2 produces an infinite sequence of distinct elements in

G1 ∗G2. If w′n = g1∏n

i=1 g−12 g−11 then w′nwn = g1 but wnw

′n 6= g1. �

Theorem 175. [Problem 9-2] A free group on two or more generators has center con-sisting of the identity alone.

Proof. We can assume that there are exactly two generators x and y. Let xmwyn ∈F (x, y) where m 6= 0 or n 6= 0 and w is some word. Suppose m 6= 0. Then

(xmwyn)(y−nw−1x−my) = y

but

(y−nw−1x−my)(xmwyn) 6= y.

Similarly, if n 6= 0 then xmwyn and xy−nw−1x−m do not commute. �

56

Theorem 176. [Problem 9-3] A group G is free if and only if it has a generating setS ⊆ G such that every element g ∈ G other than the identity has a unique expressionas a product of the form

g = σn11 · · · σ

nkk

where σi ∈ S, ni are nonzero integers, and σi 6= σi+1 for each i = 1, . . . , k − 1.

Proof. If G is free with basis S ⊆ G then the result follows from Proposition 9.2, sinceany expression of the indicated form is a reduced word. Conversely, let S ⊆ G. If Sgenerates G then the induced homomorphism φ : F (S) → G is surjective; if S alsosatisfies the unique expression property then φ is injective. �

Theorem 177. [Problem 9-4] Let G1, G2, H1, H2 be groups, and let fj : Gj → Hj begroup homomorphisms for j = 1, 2.

(1) There exists a unique homomorphism f1 ∗ f2 : G1 ∗G2 → H1 ∗H2 such that thefollowing diagram commutes for j = 1, 2:

G1 ∗G2 H1 ∗H2

Gj Hj

f1 ∗ f2

ij i′j

fj

where ij : Gj → G1 ∗G2 and i′j : Hj → H1 ∗H2 are the canonical injections.(2) Let S1 and S2 be disjoint sets, and let Ri be a subset of the free group F (Si) for

i = 1, 2. Then 〈S1 ∪ S2 | R1 ∪R2〉 is a presentation of the free product group〈S1 | R1〉 ∗ 〈S2 | R2〉.

Proof. Consider the homomorphisms i′j ◦ fj : Gj → H1 ∗ H2 for j = 1, 2. By thecharacteristic property there exists a unique homomorphism f1 ∗f2 : G1 ∗G2 → H1 ∗H2

such that i′j ◦ fj = (f1 ∗ f2) ◦ ij for j = 1, 2, which proves (1).

For (2), we first show that F (S1 ∪ S2) ∼= F (S1) ∗ F (S2). For i = 1, 2, let

ji : Si → S1 ∪ S2,

ki : Si → F (Si),

ì : F (Si)→ F (S1) ∗ F (S2),

j : S1 ∪ S2 → F (S1 ∪ S2)

be the canonical injections. There exist unique homomorphisms mi : F (Si) → F (S1 ∪S2) such that mi ◦ ki = j ◦ ji, and these maps induce a unique homomorphism ϕ :F (S1) ∗ F (S2)→ F (S1 ∪ S2) satisfying mi = ϕ ◦ ì. Since S1 and S2 are disjoint, there

57

is a unique map k : S1 ∪ S2 → F (S1) ∗ F (S2) satisfying k ◦ ji = ì ◦ ki. This induces aunique homomorphism ψ : F (S1 ∪ S2)→ F (S1) ∗ F (S2) satisfying k = ψ ◦ j. Now

ϕ ◦ ψ ◦ j ◦ ji = ϕ ◦ k ◦ ji = ϕ ◦ ì ◦ ki = mi ◦ ki = j ◦ ji,so ϕ ◦ ψ ◦ j = j and ϕ ◦ ψ = IdF (S1∪S2) by uniqueness. Similarly,

ψ ◦ ϕ ◦ ì ◦ ki = ψ ◦mi ◦ ki = ψ ◦ j ◦ ji = k ◦ ji = ì ◦ ki,so ψ ◦ ϕ ◦ ì = ì and ψ ◦ ϕ = IdF (S1)∗F (S2) by uniqueness. This proves that ϕ is anisomorphism. We now show that

F (S1 ∪ S2)/R1 ∪R2∼= (F (S1)/R1) ∗ (F (S2)/R2).

Let

π : F (S1 ∪ S2)→ F (S1 ∪ S2)/R1 ∪R2,

πi : F (Si)→ F (Si)/Ri

be the quotient maps and let

ni : F (Si)/Ri → (F (S1)/R1) ∗ (F (S2)/R2)

be the canonical injections. Since Ri ⊆ ker(π ◦mi), there are unique homomorphismsfi : F (Si)/Ri → F (S1 ∪ S2)/R1 ∪R2 such that fi ◦ πi = π ◦mi. Then there is a uniquehomomorphism f : (F (S1)/R1) ∗ (F (S2)/R2) → F (S1 ∪ S2)/R1 ∪R2 such that fi =f ◦ni. There is a unique homomorphism p : F (S1) ∗F (S2)→ (F (S1)/R1) ∗ (F (S2)/R2)satisfying ni ◦ πi = p ◦ ì. Since R1 ∪R2 ⊆ ker(p ◦ψ), there is a unique homomorphismg : F (S1 ∪ S2)/R1 ∪R2 → (F (S1)/R1) ∗ (F (S2)/R2) satisfying g ◦ π = p ◦ ψ. Now

g ◦ f ◦ ni ◦ πi = g ◦ fi ◦ πi = g ◦ π ◦mi

= p ◦ ψ ◦mi = p ◦ ψ ◦ ϕ ◦ ì = p ◦ ì = ni ◦ πi,so g ◦ f ◦ ni = ni and g ◦ f = Id(F (S1)/R1)∗(F (S2)/R2)

by uniqueness. Similarly,

f ◦ g ◦ π ◦mi = f ◦ p ◦ ψ ◦mi = f ◦ p ◦ ψ ◦ ϕ ◦ ì= f ◦ p ◦ ì = f ◦ ni ◦ πi = fi ◦ πi = π ◦mi

implies that

f ◦ g ◦ π ◦ j ◦ ji = f ◦ g ◦ π ◦mi ◦ ki = π ◦mi ◦ ki = π ◦ j ◦ ji,so f ◦ g ◦ π = π and f ◦ g = IdF (S1∪S2)/R1∪R2

by uniqueness. This proves that

〈S1 ∪ S2 | R1 ∪R2〉 ∼= 〈S1 | R1〉 ∗ 〈S2 | R2〉 .�

Theorem 178. [Problem 9-5] Let S be a set, let R and R′ be subsets of the free groupF (S), and let π : F (S) → 〈S | R〉 be the projection onto the quotient group. Then

〈S | R ∪R′〉 is a presentation of the quotient group 〈S | R〉 /π(R′).

58

Proof. We want to show that

F (S)/R ∪R′ ∼= (F (S)/R)/π(R′).

Let

π′ : F (S)→ F (S)/R ∪R′,

π′′ : F (S)/R→ (F (S)/R)/π(R′)

be the quotient maps. Since R ⊆ ker(π′), there is a unique homomorphism f1 :

F (S)/R→ F (S)/R ∪R′ such that π′ = f1 ◦ π. But π(R′) ⊆ ker(f1) since R′ ⊆ ker(π′),

so there is a unique homomorphism f : (F (S)/R)/π(R′) → F (S)/R ∪R′ such thatf1 = f ◦ π′′. Since R ∪R′ ⊆ ker(π′′ ◦ π), there exists a unique homomorphism

g : F (S)/R ∪R′ → (F (S)/R)/π(R′) such that π′′ ◦ π = g ◦ π′. Now

f ◦ g ◦ π′ = f ◦ π′′ ◦ π = f1 ◦ π = π′

and

g ◦ f ◦ π′′ ◦ π = g ◦ f1 ◦ π = g ◦ π′ = π′′ ◦ π,so f ◦ g = IdF (S)/R∪R′ and g ◦ f = Id(F (S)/R)/π(R′) by uniqueness. �


(1) The free group on generators α1, . . . , αn has the presentation

F (α1, . . . , αn) ∼= 〈α1, . . . , αn | ∅〉 .In particular, Z has the presentation 〈α | ∅〉.

(2) The group Z× Z has the presentation 〈β, γ | βγ = γβ〉.(3) The cyclic group Zn has the presentation

Zn ∼= 〈α | αn = 1〉 .(4) The group Zm × Zn has the presentation

Zm × Zn ∼= 〈β, γ | βm = 1, γn = 1, βγ = γβ〉 .

Proof. Part (3) follows from Theorem 178. Let ϕ : Z × Z → 〈β, γ | βγ = γβ〉 be theisomorphism in (2). We have

Zm × Zn ∼= (Z× Z)/(mZ× nZ)∼= 〈β, γ | βγ = γβ〉 /ϕ(mZ× nZ).

∼= 〈β, γ | βγ = γβ〉 /{βm, γn}∼= 〈β, γ | βm = 1, γn = 1, βγ = γβ〉

by Theorem 178. �

59

Theorem 180. [Problem 9-7] The free abelian group on a set S is uniquely determinedup to isomorphism by the characteristic property (Proposition 9.14).

Proof. The proof is identical to that of Theorem 170. �

Theorem 181. [Problem 9-8] Suppose G is a free abelian group of finite rank. Thenevery basis of G is finite.

Proof. Let n be the rank of G. If G has an infinite basis then it has a linearly indepen-dent set S with n+ 1 elements, and ZS is a free abelian group of rank n+ 1 containedin G. This contradicts Proposition 9.19. �

Theorem 182. [Problem 9-9] Suppose C is a concrete category with faithful functorF : C → Set. If S is a set, a free object on S in C is an object F ∈ Ob(C) togetherwith a map i : S → F(F ), such that for any object Y ∈ Ob(C) and any map ϕ : S →F(Y ) there exists a unique morphism φ ∈ HomC(F, Y ) such that the following diagramcommutes:

F(F )

S F(Y )

F(φ)i

ϕ

(1) Any two free objects on the same set are isomorphic in C.(2) A free group is a free object in Grp, and a free abelian group is a free object in

Ab.(3) The free objects in Top are discrete topological spaces.

Proof. Let F1 and F2 be free objects on a set S with injections i1 : S → F(F1) andi2 : S → F(F2). By definition, there exist unique morphisms φ1 : F2 → F1 andφ2 : F1 → F2 such that i1 = F(φ1) ◦ i2 and i2 = F(φ2) ◦ i1. Now

F(φ1) ◦ F(φ2) ◦ i1 = F(φ1) ◦ i2 = i1,

so F(φ1 ◦φ2) = F(φ1) ◦F(φ2) = IdF(F1) by uniqueness. Since F is faithful, this impliesthat φ1 ◦ φ2 = IdF1 . Similarly, φ2 ◦ φ1 = IdF2 . This proves (1). Part (2) follows fromTheorem 169 and Theorem 171. Part (3) is obvious. �

60

Chapter 10. The Seifert-Van Kampen Theorem

Theorem 183. [Exercise 10.9] A graph Γ is connected if and only if given any twovertices v, v′ ∈ Γ, there is an edge path from v to v′. In a connected graph, any twovertices can be connected by a simple edge path.

Proof. One direction is evident. Suppose that Γ is connected. Apply Theorem 107 withthe equivalence relation on Γ defined by x ∼ y if and only if there is an edge pathcontaining x and y; this proves the converse.

In a connected graph, we can connect any two vertices by an edge path. If the edgepath has a vertex v appearing twice then it is of the form (v0, e1, . . . , v, . . . , v, . . . , ek, vk).Deleting all edges and vertices between the two appearances of v creates a shorter edgepath between v0 and vk. By repeating the process we eventually create a simple edgepath between v0 and vk. Note that if v0 = vk then we can let the edge path be thetrivial edge path (v0). �

Theorem 184. [Exercise 10.18] Let G be a group.

(1) [G,G] is a normal subgroup of G.(2) [G,G] is trivial if and only if G is abelian.(3) The quotient group G/[G,G] is always abelian.

Proof. First note that

gaba−1b−1g−1 = (gag−1)(gbg−1)(gag−1)−1(gbg−1)−1.

If x ∈ [G,G] then x = x1 · · ·xn for elements xi = aibia−1i b−1i , so the above computation

shows that gxg−1 ∈ [G,G] and g[G,G]g−1 ⊆ [G,G]. Similarly we have g−1[G,G]g ⊆[G,G], so g[G,G]g−1 = [G,G]. This proves (1). For (2), if G is abelian then [G,G] isclearly trivial since aba−1b−1 = e for all a, b ∈ G. If [G,G] is trivial and a, b ∈ G thenaba−1b−1 ∈ [G,G], so aba−1b−1 = e and therefore ab = ba. For (3), let a[G,G] andb[G,G] be elements of G/[G,G]. We have

ab[G,G] = baa−1b−1ab[G,G] = ba[G,G]

since a−1b−1ab ∈ [G,G], which shows that G/[G,G] is abelian. �

Theorem 185. [Exercise 10.20] Let G be a group. For any abelian group H and anyhomomorphism ϕ : G→ H, there exists a unique homomorphism ϕ : Ab(G)→ H suchthat the following diagram commutes:

61

G H

Ab(G)

ϕ

ϕ

Proof. We haveϕ(aba−1b−1) = ϕ(a)ϕ(b)ϕ(a)−1ϕ(b)−1 = e

for all a, b ∈ G since H is abelian. Therefore [G,G] ⊆ ker(ϕ) and there exists a uniquehomomorphism ϕ : Ab(G) → H such that ϕ = ϕ ◦ π where π : G → Ab(G) is thequotient map. �

Theorem 186. [Problem 10-1] Sn is simply connected when n ≥ 2.

Proof. Let N be the north pole and let S be the south pole. Let U = Sn \ {N} andV = Sn \{S}. Then U ∪V = Sn, U and V are open in Sn, and the sets U , V and U ∩Vare all path-connected (since n ≥ 2). Therefore

π1(S1) ∼= (π1(U) ∗ π1(V ))/C

by the Seifert-van Kampen theorem. But U, V ≈ Rn, so π1(U) and π1(V ) are bothtrivial. This implies that π1(S1) is also trivial. �

Example 187. [Problem 10-2] Let X ⊆ R3 be the union of the unit 2-sphere with theline segment {0} × {0} × [−1, 1]. Compute π1(X,N), where N = (0, 0, 1) is the northpole, giving explicit generator(s).

Let P = (1, 0, 0) and let B be a small coordinate ball around P in X. Let U = X \{P}so that U ∪B = X and U ∩B is homeomorphic to S1. By Corollary 10.5,

π1(X,P ) ∼= π1(U, P )/i∗π1(U ∩B)

where i : U ∩ B → U is the canonical injection. But for any [f ] ∈ π1(U ∩ B) the loopi ◦ f is a loop in S2 \P , which is simply connected. Therefore i∗π1(U ∩B) is trivial andπ1(X,P ) ∼= π1(U, P ). But U is homotopy equivalent to S1, so π1(X,N) ∼= π1(X,P ) ∼= Z.A generator for π1(X,N) is the loop that traverses the line segment from the north poledown to the south pole and then returns to the north pole by a path on S2.

Theorem 188. [Problem 10-3] Any two vertices in a tree are joined by a unique simpleedge path.

Proof. Suppose that v and w are two vertices in a tree that are joined by two differentsimple edge paths P and Q. Let Q be the edge path from w to v obtained by reversingQ, and let P ·Q be the edge path formed by concatenating P and Q and deleting the

62

extra occurrence of w. By deleting duplicates from both ends, we can see that P · Qcontains a cycle. �

Theorem 189. [Problem 10-4] Every vertex u in a finite tree T is a strong deformationretract of the tree.

Proof. If the tree has no edges, there is nothing to prove. Suppose that u is incidentwith exactly one edge. The proof follows as in Theorem 10.10, but we use the fact thatevery tree with at least two vertices contains two vertices each incident with exactly oneedge. By always choosing v 6= u in the proof, we obtain a strong deformation retractiononto u. Now suppose that u is incident with two edges e, e′ which have endpoints v, v′

(not equal to u). We have T \ (e ∪ u ∪ e′) = U ∪ U ′ where U and U ′ are disjoint treescontaining v and v′ respectively, and where v and v′ are both incident with at most oneedge. Choose strong deformation retractions from U to v and U ′ to v′. By combiningthese with the obvious strong deformation retraction from e ∪ u ∪ e′, we have a strongdeformation retraction from T onto u. �

Theorem 190. The fundamental groups of the following spaces are isomorphic to Z∗n =Z ∗ · · · ∗ Z:

(1) S1 ∨ · · · ∨ S1, the bouquet of n circles.(2) R2 \ {p1, . . . , pn}, the plane with n isolated points removed.(3) S2 \ {p1, . . . , pn+1}, the sphere with n+ 1 isolated points removed.

Proof. (1) follows from Theorem 10.7. (2) follows from the fact that R2 \ {p1, . . . , pn}is homotopy equivalent to S1 ∨ · · · ∨ S1 (see Example 7.44). (3) follows from the factthat S2 \ {pn+1} is homeomorphic to R2. �

Example 191. [Problem 10-5] Compute the fundamental group of the complement ofthe three coordinate axes in R3, giving explicit generator(s).

Using the map x 7→ x/ ‖x‖ we see that the space X is homeomorphic to the sphere S2

with the six points (±1, 0, 0), (0,±1, 0) and (0, 0,±1) removed. By Theorem 190,

π1(X) ∼= Z ∗ Z ∗ Z ∗ Z ∗ Z.

We can view X as R3\{0} with six rays removed. Choose one of the six rays to exclude,and choose five loops that wind once around each of the remaining rays. Then theseloops generate π1(X).

Theorem 192. [Problem 10-6] Suppose M is a connected manifold of dimension atleast 3, and p ∈ M . Then the inclusion M \ {p} ↪→ M induces an isomorphismπ1(M \ {p}) ∼= π1(M).

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Proof. Let B be a coordinate ball around p and let U = B and V = M \{p} in Corollary10.5. Choose some base point q in B \ {p}. Then the inclusion M \ {p} ↪→M inducesan isomorphism

π1(M, q) ∼= π1(M \ {p} , q)/j∗π1(B \ {p} , q)where j : B \ {p} →M \ {p} is the inclusion. But π1(B \ {p} , q) is trivial by Corollary7.38, so π1(M, q) ∼= π1(M \ {p} , q). �

Theorem 193. [Problem 10-7] Suppose M1 and M2 are connected n-manifolds withn ≥ 3. Then the fundamental group of M1#M2 is isomorphic to π1(M1) ∗ π1(M2).

Proof. By Theorem 92, there are open subsets U1, U2 ⊆ M1#M2 and points pi ∈ Mi

such that Ui ≈ Mi \ {pi}, U1 ∩ U2 ≈ Rn \ {0}, and U1 ∪ U2 = M1#M2. Choose a basepoint q in U1 ∩U2. Since Rn \ {0} is simply connected for n ≥ 3, by Corollary 10.4 andTheorem 192 we have

π1(M1#M2, q) ∼= π1(U1, q) ∗ π1(U2, q)∼= π1(M1 \ {p1}) ∗ π1(M2 \ {p2})∼= π1(M1) ∗ π1(M2).

�

Theorem 194. [Problem 10-8] Suppose M1 and M2 are nonempty, compact, connected2-manifolds. Then any two connected sums of M1 and M2 are homeomorphic.

Proof. Since any connected sum M1#M2 is also a nonempty, compact, connected 2-manifold, it follows from Theorem 10.22 that it suffices to prove that any two connectedsums have isomorphic fundamental groups. This follows from 92. �

Example 195. [Problem 10-9] Let Xn be the union of the n circles of radius 1 thatare centered at the points {0, 2, 4, . . . , 2n− 2} in C, which are pairwise tangent to eachother along the x-axis. Prove that π1(Xn, 1) is a free group on n generators, and describeexplicit loops representing the generators.

This is identical to Example 10.8 - we have π1(Xn, 1) ∼= Z ∗ · · · ∗ Z. For each k =0, . . . , n−1, let ωk be a loop based at (2k−1, 0) that traverses the (k+1)th circle once,and let γk be a path from 1 to (2k − 1, 0). Then {γk · ωk · γk : k = 0, . . . , n− 1} are nloops representing the generators of π1(Xn, 1).

Theorem 196. [Problem 10-10] For any finitely presented group G, there is a finiteCW complex whose fundamental group is isomorphic to G.

Proof. Let 〈α1, . . . , αn | r1, . . . , rm〉 be a presentation of G. Let Γ be a graph with asingle vertex v and n loops at v; denote these loops by α1, . . . , αn. By Theorem 10.12,π1(Γ, v) ∼= 〈α1, . . . , αn | ∅〉. Let Γ0 = Γ. For each i = 1, . . . ,m, write ri = αp1k1 · · ·α

p`k`

,

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let β : I → Γ be a representative of [αp1k1 ] · · · [αp`k`

] ∈ π1(Γ, v) and let β : S1 → Γ be the

circle representative of β. Attach a 2-cell Di = B2 to Γi−1 along the attaching map βto obtain a new CW complex Γi; then

π1(Γi, v) ∼= 〈α1, . . . , αn | r1, . . . , ri〉

by Proposition 10.13. By repeating the process, we have π1(Γm, v) ∼= G. �

Example 197. [Problem 10-11] For each of the following spaces, give a presentationof the fundamental group together with a specific loop representing each generator.

(1) A closed disk with two interior points removed.(2) The projective plane with two points removed.(3) A connected sum of n tori with one point removed.(4) A connected sum of n tori with two points removed.

Denote these spaces by X1, . . . , X4.

(1) π1(X1) ∼= 〈α, β | ∅〉 where α is a loop around the first interior point and β is aloop around the second interior point.

Example 198. [Problem 10-12] Give a purely algebraic proof that the groups 〈α, β | αβαβ−1〉and 〈ρ, γ | ρ2γ2〉 are isomorphic.

This follows from Lemma 6.16.

Theorem 199. [Problem 10-16] Abelianization defines a functor from Grp to Ab.

Proof. For any two groups G,H and any homomorphism f : G → H, define Ab(f) :Ab(G)→ Ab(H) as the unique homomorphism satisfying Ab(f) ◦ πG = πH ◦ f , whereπG : G→ Ab(G) and πH : H → Ab(H) are the quotient maps. If G = H and f = IdGthen Ab(f) ◦π = π, so Ab(f) = IdAb(G) by uniqueness. If K is a group and g : H → Kis a homomorphism then

Ab(g ◦ f) ◦ πG = πK ◦ g ◦ f= Ab(g) ◦ πH ◦ f= Ab(g) ◦ Ab(f) ◦ πG,

so Ab(g ◦ f) = Ab(g) ◦ Ab(f) by uniqueness. This shows that Ab : Grp → Ab is afunctor. �

Theorem 200. [Problem 10-17] Ab(G) is the unique group (up to isomorphism) thatsatisfies the characteristic property expressed in Theorem 10.19, for any group G.

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Proof. Suppose Ab(G) and Ab(G)′ are two groups that satisfy the characteristic prop-erty. Let π : G → Ab(G) and π′ : G → Ab(G)′ be the canonical maps. There existunique homomorphisms ϕ : Ab(G)′ → Ab(G) and ϕ′ : Ab(G) → Ab(G)′ satisfyingϕ◦π′ = π and ϕ′◦π = π′. Since ϕ◦ϕ′◦π = ϕ◦π′ = π and ϕ′◦ϕ◦π′ = ϕ′◦π = π′, we haveϕ ◦ ϕ′ = IdAb(G) and ϕ′ ◦ ϕ = IdAb(G)′ by uniqueness. Therefore Ab(G) ∼= Ab(G)′. �

Theorem 201. [Problem 10-18] For any groups G1 and G2,

Ab(G1 ∗G2) ∼= Ab(G1)⊕ Ab(G2).

Proof. For i = 1, 2, let

αi : Gi → Ab(Gi),

α : G1 ∗G2 → Ab(G1 ∗G2),

ji : Ab(Gi)→ Ab(G1)⊕ Ab(G2),

ki : Gi → G1 ∗G2

be the canonical maps. There exists a unique homomorphism ` : G1 ∗G2 → Ab(G1)⊕Ab(G2) satisfying ` ◦ ki = ji ◦αi, and there exists a unique homomorphism ϕ : Ab(G1 ∗G2)→ Ab(G1)⊕Ab(G2) satisfying ϕ◦α = `. Also, there exist unique homomorphismsmi : Ab(Gi) → Ab(G1 ∗ G2) satisfying mi ◦ αi = α ◦ ki, so there exists a uniquehomomorphism ψ : Ab(G1)⊕ Ab(G2)→ Ab(G1 ∗G2) satisfying ψ ◦ ji = mi. Now

ϕ ◦ ψ ◦ ji ◦ αi = ϕ ◦mi ◦ αi = ϕ ◦ α ◦ ki = ` ◦ ki = ji ◦ αi,so ϕ ◦ ψ = IdAb(G1)⊕Ab(G2) by uniqueness. Similarly,

ψ ◦ ϕ ◦ α ◦ ki = ψ ◦ ` ◦ ki = ψ ◦ ji ◦ αi = mi ◦ αi = α ◦ ki,so ψ ◦ ϕ = IdAb(G1∗G2) by uniqueness. �

Corollary 202. The abelianization of a free group on n generators is free abelian of rankn, and isomorphic finitely generated free groups have the same number of generators.

Theorem 203. [Problem 10-19] For any set S, the abelianization of the free groupF (S) is isomorphic to the free abelian group ZS.

Proof. Let

j : S → ZS,k : S → F (S),

` : F (S)→ Ab(F (S))

be the canonical maps. There exist unique homomorphisms ϕ : ZS → Ab(F (S)) andψ : Ab(F (S))→ ZS such that ϕ ◦ j = ` ◦ k and ψ ◦ ` ◦ k = j. Then

ϕ ◦ ψ ◦ ` ◦ k = ϕ ◦ j = ` ◦ k,

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so ϕ ◦ ψ = IdAb(F (S)) by uniqueness. Similarly,

ψ ◦ ϕ ◦ j = ψ ◦ ` ◦ k = j,

so ψ ◦ ϕ = IdZS by uniqueness. �

Theorem 204. [Problem 10-20] Let Γ be a finite connected graph. The Euler charac-teristic of Γ is χ(Γ) = V −E, where V is the number of vertices and E is the number ofedges. The fundamental group of Γ is a free group on 1−χ(Γ) generators, and thereforeχ(Γ) is a homotopy invariant.

Proof. Let T be a spanning tree in Γ. Since the number of vertices in a tree is alwaysone more than the number of edges, we have χ(T ) = 1. By Theorem 10.12, there is onegenerator of π1(Γ) for each edge of Γ not in T . Therefore χ(Γ) = 1− n where n is thenumber of generators of π1(Γ), and the result follows. �


(1) If a pushout of a pair of morphisms exists, it is unique up to isomorphism inthe category C.

(2) The amalgamated free product is the pushout of two group homomorphisms withthe same domain.

(3) Let S1 and S2 be sets with nonempty intersection. In the category of sets, thepushout of the inclusions S1 ∩ S2 ↪→ S1 and S1 ∩ S2 ↪→ S2 is S1 ∪ S2 togetherwith appropriate inclusion maps.

(4) Suppose X and Y are topological spaces, A ⊆ Y is a closed subset, and f : A→X is a continuous map. The adjunction space X ∪f Y is the pushout of (ιA, f)in the category Top.

(5) In the category Top, given two continuous maps with the same domain, thepushout always exists.

Proof. Let fi : A0 → Ai for i = 1, 2 be a pair of morphisms. Suppose there are twopushouts, P and P ′, with morphisms gi : Ai → P and g′i : Ai → P ′. Then there existunique morphisms h : P ′ → P and h′ : P → P ′ such that h ◦ g′i = gi and h′ ◦ gi = g′i.Since

h ◦ h′ ◦ gi = h ◦ g′i = gi and h′ ◦ h ◦ g′i = h′ ◦ gi = g′i,

we have h ◦ h′ = IdP and h′ ◦ h = IdP ′ by uniqueness. This proves (1).

Let fi : A0 → Ai be a pair of group homomorphisms. We want to show that A1 ∗A0 A2

is the pushout of A. Let ji : Ai → A1 ∗ A2 and π : A1 ∗ A2 → A1 ∗A0 A2 be the usualinclusions, and choose gi = π ◦ ji. Let

C ={

(j1 ◦ f1)(a)(j2 ◦ f2)(a)−1 : a ∈ A0

}

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so that A1 ∗A0 A2 = (A1 ∗ A2)/C. If a ∈ A0 then

(g1 ◦ f1)(a) = (j1 ◦ f1)(a)C

= (j1 ◦ f1)(a)(j1 ◦ f1)(a)−1(j2 ◦ f2)(a)C

= (j2 ◦ f2)(a)C

= (g2 ◦ f2)(a),

which shows that g1 ◦ f1 = g2 ◦ f2. Now let B be a group and let hi : Ai → B be a pairof homomorphisms such that h1 ◦ f1 = h2 ◦ f2. There exists a unique homomorphism

h : A1 ∗ A2 → B such that h ◦ ji = hi. For a ∈ A0 we have

h((j1 ◦ f1)(a)(j2 ◦ f2)(a)−1) = (h ◦ j1 ◦ f1)(a)(h ◦ j2 ◦ f2)(a)−1

= (h1 ◦ f1)(a)(h2 ◦ f2)(a)−1

= e,

so C ⊆ ker(h) and there exists a unique homomorphism h : A1 ∗A0 A2 → B such that

h ◦ π = h. Sinceh ◦ gi = h ◦ π ◦ ji = h ◦ ji = hi,

this proves (2).

For (3), let fi : Si → T be a pair of functions such that f1|S1∩S2 = f2|S1∩S2 . Defineg : S1∪S2 → T by letting g|Si = fi; this is well-defined due to the conditions on f1 andf2. For (4), let g1 : X → Z and g2 : Y → Z be continuous maps such that g1 ◦f = g2|A.Let j1 : X → XqY and j2 : Y → XqY be the canonical injections. Let g : XqY → Zbe the unique continuous map such that g ◦j1 = g1 and g ◦j2 = g2; then g descends to aunique continuous map g : X ∪f Y → Z satisfying g ◦ q = g where q : X qY → X ∪f Yis the quotient map. Since g ◦ (q ◦ ji) = g ◦ ji = gi for i = 1, 2, this proves (4). For (5),let fi : A0 → Ai be a pair of continuous maps and define a relation ∼ on A1 q A2 asthe smallest equivalence relation such that f1(x) ∼ f2(x) for every x ∈ A0. Take thepushout P to be the quotient space (A1 qA2)/ ∼; the proof is then similar to (4). �

Chapter 11. Covering Maps


(1) Every covering map is a local homeomorphism, an open map, and a quotientmap.

(2) An injective covering map is a homeomorphism.(3) A finite product of covering maps is a covering map.(4) The restriction of a covering map to a saturated, connected, open subset is a

covering map onto its image.

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Proof. (1) and (2) follow from Theorem 13 and Proposition 3.69. Part (3) is obvious.For (4), let q : E → X be a covering map and let F be a saturated, connected, opensubset of E. If x ∈ q(F ) then x has some evenly covered neighborhood U ⊆ X. Then(q|F )−1(U ∩ q(F )) = q−1(U ∩ q(F )) since F is saturated, which shows that q|F : F →q(F ) is a covering map. �

Example 207. [Exercise 11.7] Let Xn be the union of n circles in C as described inExample 195. Define a map q : X3 → X2 by letting A, B, and C denote the unit circlescentered at 0, 2, and 4, respectively, and defining

q(z) =

z, z ∈ A;

2− (z − 2)2, z ∈ B;

4− z, z ∈ C.Show that q is a covering map.

Let δ > ε > 0 be small numbers; then X2 ∩ Bδ(1) and X2 \ Bε(1) are evenly coveredopen sets whose union is X2.

Example 208. [Exercise 11.9] Let E be the interval (0, 2) ⊆ R, and define f : E → S1

by f(x) = e2πix. Then f is a local homeomorphism and is clearly surjective, but f isnot a covering map since the point 1 ∈ S1 has no evenly covered neighborhood.

If a small neighborhood U around 1 is chosen then the connected components of f−1(U)do not map onto U ; if U is a large neighborhood around 1 then the connected compo-nents of f−1(U) do not map injectively into U .

Theorem 209. [Exercise 11.25] If S1 and S2 are right G-sets and ϕ : S1 → S2 is aG-isomorphism, then ϕ−1 is also a G-isomorphism.

Proof. Let s2 ∈ S2 and g ∈ G. Then ϕ−1(s2 · g) = ϕ−1(ϕ(ϕ−1(s2) · g)) = ϕ−1(s2) · g,which shows that ϕ−1 is G-equivariant. �

Theorem 210. Let q : E → X be a covering map and let f : X → Y be any function.Then f is continuous if and only if f ◦ q is continuous.

Proof. One direction is evident. Suppose that f ◦ q is continuous. Let x ∈ X, let U bean evenly covered neighborhood of x and let σ : U → E be a local section of q so thatq ◦ σ = IdU . Then f |U = (f |U ◦ q) ◦ σ which is continuous since f ◦ q is continuous. ByProposition 2.19, f is continuous. �

Theorem 211. [Problem 11-1] Suppose q : E → X is a covering map.

(1) If X is Hausdorff then E is too.(2) If X is an n-manifold then E is too.(3) If E is an n-manifold and X is Hausdorff then X is an n-manifold.

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Figure 0.2. A covering of the Klein bottle by the torus.

Proof. For (1), let x and y be distinct points in E. Let U be an evenly covered neigh-

borhood of q(x) and let U be the sheet of q−1(U) containing x so that q|U : U → U

is a homeomorphism. If y ∈ U then we are done since U is Hausdorff, and if y /∈ U

then U is a neighborhood of x and E \ U is a neighborhood of y. Part (2) follows fromProposition 4.40 of [1]. Part (3) follows from Proposition 3.56. �

Theorem 212. [Problem 11-2] For any n ≥ 1, the map q : Sn → Pn defined in Example11.6 is a covering map.

Proof. It is clear that q is continuous and surjective. For each i = 1, . . . , n, let πi :Sn → R be the projection onto the ith coordinate, let Sni+ = π−1i ((0,∞)) and letSni− = π−1i ((−∞, 0)). Let Pni = Pn \ q(π−1i ({0})); then q−1(Pni ) is the disjoint union ofSni+ and Sni− while q|Sni+ and q|Sni− are homeomorphisms. Since the sets Pn1 , . . . ,Pnn coverPn, this shows that q is a covering map. �

Example 213. [Problem 11-3] Let S be the following subset of C2:

S ={

(z, w) : w2 = z, w 6= 0}.

(It is the graph of the two-valued complex square root “function” described in Chapter1, with the origin removed.) Show that the projection π1 : C2 → C onto the firstcoordinate restricts to a two-sheeted covering map q : S → C \ {0}.

Let R be any ray extending from (and including) the origin; then q−1(C\R) consists oftwo sheets. The sets C \R cover C \ {0} as R varies, which shows that q is a coveringmap.

Example 214. [Problem 11-4] Show that there is a two-sheeted covering of the Kleinbottle by the torus.

See Figure 0.2.

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Theorem 215. [Problem 11-5] Let M and N be connected manifolds of dimension

n ≥ 2, and suppose q : M →M is a k-sheeted covering map. Then there is a connected

sum M#N that admits a k-sheeted covering by a manifold of the form M#N · · ·#N(connected sum of M with k disjoint copies of N).

Proof. Let U be some evenly covered neighborhood in M and let B be a regular co-ordinate ball that lies inside U . Then q−1(B) is the disjoint union of k coordinate

balls B1, . . . , Bk in M . Choose a regular coordinate ball C in N . Form the connected

sum M#N by identifying ∂B with ∂C, and form the connected sum M#N · · ·#N by

identifying Bi with Ci for i = 1, . . . , k, where Ci ⊆ Ni is C in the ith copy Ni of N .

Define a covering map q′ : M#N · · ·#N → M#N by setting q′|M\Bi = q|M\Bi and

q′|Ni\Ci = IdN\C for each i. �

Theorem 216. [Problem 11-6] Every nonorientable compact surface of genus n ≥ 1has a two-sheeted covering by an orientable one of genus n− 1.

Proof. If X is a surface, write X#k for X# · · ·#X, the connected sum of k copies of X.We use induction on n. The case n = 1 follows from Theorem 212, and the case n = 2follows from Example 214. Assume that the result holds for n = 2, . . . , k, and considera nonorientable compact surface (P2)#k+1. By the induction hypothesis, there is a two-sheeted covering map q : (T2)#k−2 → (P2)#k−1. By Theorem 215, there exists a two-sheeted covering map q′ : (T2)#k → (P2)#k−1#T2. But Lemma 6.17 shows that T2#P2

is homeomorphic to (P2)#3, so there is a covering map q′′ : (T2)#k → (P2)#k+1. �

Theorem 217. [Problem 11-9] Every proper local homeomorphism between connected,locally path-connected and compactly generated Hausdorff spaces is a covering map.

Proof. Let q : X → Y be such a map and let y ∈ Y . Since {y} is compact and q isproper, q−1({y}) is also compact. This implies that q−1({y}) is a finite discrete set sinceq is a local homeomorphism. Write q−1({y}) = {p1, . . . , pn}; by Lemma 109, we canchoose pairwise disjoint open sets U1, . . . , Un with pi ∈ Ui. By shrinking each set, we canassume that q|Ui : Ui → q(Ui) is a homeomorphism for i = 1, . . . , n. Let U =

⋃ni=1 Ui.

By Theorem 4.95, q(X \ U) is closed in Y , and V =⋂ni=1 q(Ui) ∩ (Y \ q(X \ U)) is a

neighborhood of y that is evenly covered: it is clear that V is open, and if q(x) ∈ Vthen q(x) /∈ q(X \U), so x ∈ U . Therefore q−1(V ) is the disjoint union of the open setsq−1(V ) ∩ Ui, each of which is mapped homeomorphically onto V . �

Theorem 218. [Problem 11-10] A covering map is proper if and only if it is finite-sheeted.

Proof. Every covering map is a local homeomorphism, so one direction follows from theargument used in Theorem 217. Conversely, suppose that the covering map q : E → X

71

is finite-sheeted, let B ⊆ X be a compact set, and let U be an open cover of q−1(B).For each x ∈ B the set q−1({x}) is finite. Write q−1({x}) = {x1, . . . , xk} and let W be

an evenly covered neighborhood of x so that q−1(W ) = W1 ∪ · · · ∪ Wk where each Wi

is connected and open, q|Wi: Wi → W is a homeomorphism, and xi ∈ Wi. For each i

choose a set Ui ∈ U containing xi. Let Vx =⋂ki=1 q(Ui∩ Wi); then Vx is a neighborhood

of x and q−1(Vx) is covered by U1, . . . , Un. Since B is compact, there are finitely manysuch sets Vx1 , . . . , Vxn that cover B. Since q−1(B) is covered by q−1(Vx1), . . . , q

−1(Vxn)and each q−1(Vxi) is covered by finitely many sets in U , this shows that q−1(B) iscompact. �

Theorem 219. [Problem 11-11] Let q : E → X be a covering map. Then E is compactif and only if X is compact and q is a finite-sheeted covering.

Proof. If E is compact then X is compact. If x ∈ X then q−1({x}) is closed andtherefore compact. Since q is a local homeomorphism, q−1({x}) must be a finite discreteset. The converse follows from Theorem 218. �

Theorem 220. [Problem 11-12] A continuous map f : S1 → S1 is said to be odd iff(−z) = −f(z) for all z ∈ S1, and even if f(z) = f(−z) for all z ∈ S1.

(1) Let p2 : S1 → S1 be the two-sheeted covering map of Example 11.4. If f is odd,there exists a continuous map g : S1 → S1 with deg f = deg g such that thefollowing diagram commutes:

G H

Ab(G)

ϕ

ϕ

(2) If deg f is also even, then g lifts to a map g : S1 → S1 such that p2 ◦ g = g.Furthermore, g ◦ p2 and f are both lifts of g ◦ p2 that agree at either 1 or −1, sothey are equal everywhere.

(3) Every odd map has odd degree.

Proof. Suppose that f is odd and define g : S1 → S1 by g(z) = f(√z)2. We have

g(z2) = f(|z|)2 = f(z)2 for all z ∈ S1 since f(−z)2 = (−f(z))2 = f(z)2, so g◦p2 = p2◦f .The continuity of g follows from Theorem 210, and deg f = deg g by Proposition 8.15.This proves (1). Choose a point e ∈ S1 such that p2(e) = g(1). The existence of g in (2)follows from Theorem 11.18, for if we identify π1(S1, g(1)) with Z then (p2)∗π1(S1, e) =2Z and g∗π1(S1, 1) = nZ ⊆ 2Z since n is even. We have p2 ◦ g ◦ p2 = g ◦ p2 = p2 ◦ f ,so g ◦ p2 and f are both lifts of g ◦ p2. Also, since g(1)2 = g(1) = f(1)2 we have

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(g ◦p2)(1) = (g ◦p2)(−1) = ±f(1), so g ◦p2 and f agree at either 1 or −1. By Theorem11.12, g ◦ p2 = f . This proves (2). But f is odd while g ◦ p2 is even, which is acontradiction. This proves (3). �

Theorem 221. [Problem 11-13] Every even map f : S1 → S1 has even degree.

Proof. Define g : S1 → S1 by g(z) = f(√z). Then g(z2) = f(|z|) = f(z) for all z ∈ S1

since f is even, and the continuity of g follows from Theorem 210. Now deg(f) =deg(g ◦ p2) = 2 deg(g), so deg(f) is even. �

Theorem 222. [Problem 11-14] For any continuous map F : S2 → R2, there is a pointx ∈ S2 such that F (x) = F (−x).

Proof. Suppose that F (x) 6= F (−x) for all x ∈ S2 and define a continuous map f :S2 → S1 by

f(x) =F (x)− F (−x)

‖F (x)− F (−x)‖.

Let g : S1 × I → S2 be given by

((x, y), t) 7→(√

1− t2x,√

1− t2y, t)

;

then f ◦ g is a homotopy from f |S1 to a constant map. But f |S1 is odd and has odddegree by Theorem 220, so f |S1 cannot be null-homotopic. �

Theorem 223. [Problem 11-15] Given three disjoint, bounded, connected open subsetsU1, U2, U3 ⊆ R3, there exists a plane that simultaneously bisects all three, in the sensethat the plane divides R3 into two half-spaces H+ and H− such that for each i, Ui∩H+

has the same volume as Ui ∩H−.

Proof. We first show that for any x ∈ S2 and any open set U ⊆ R3, there is a realnumber λ such that the plane through λx and orthogonal to x bisects U . Fix some i anddefine V : R → [0,∞) by λ 7→ Vol(U ∩ H+

λ ) where H+λ = {v ∈ R3 : 〈x, v〉 > 〈x, λx〉}.

Since V is monotonic and continuous and 0,Vol(U) ∈ V (R), there exists a uniqueλ ∈ R such that V (λ) = Vol(U)/2. For x ∈ S2 and an open set U ⊆ R3, denotethis value of λ by ΛU(x). Note that ΛU(−x) = −ΛU(x). Define F : S2 → R2 byx 7→ (ΛU1(x)−ΛU2(x),ΛU2(x)−ΛU3(x)); by Theorem 222 there exists some x ∈ S2 suchthat F (x) = F (−x), i.e. ΛU1(x) = ΛU2(x) = ΛU3(x). �

Example 224. [Problem 11-16] Let T be the topologist’s sine curve (Example 4.17),and let Y be the union of T with a semicircular arc that intersects T only at (0, 1) and(2/π, 1).

(1) Show that Y is simply connected.(2) Show that there is a continuous map f : Y → S1 that has no lift to R.

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From Example 86 we know that Y has exactly two path components, each of which issimply connected. Therefore Y is simply connected. Let A be the semicircular arc, letγ : I → A be a path that traces out A from (0, 1) to (2/π, 1) and let ω : [0, 1]→ S1 bethe loop s 7→ e2πis. Let g : T → S1 be the constant map x 7→ 1. By the gluing lemma,we can construct a continuous map f : Y → S1 such that f |T = g and f |A = ω ◦ γ−1. If

f : Y → R is a lift of f then f ((0, 1)) = f ((2/π, 1)), but this is impossible for f |A ◦ γwould be a lift of a loop with winding number 1 while (f |A ◦ γ)(0) = (f |A ◦ γ)(1).

Theorem 225. [Problem 11-18] If X is a topological space that has a universal coveringspace then X is semilocally simply connected.

Proof. Let q : E → X be a covering map with E simply connected. Let x ∈ X, let U

be an evenly covered neighborhood of x, choose some x ∈ q−1({x}) and let U be thecomponent of q−1(U) containing x. If γ : I → U is a loop based at x then by Corollary11.14 there is a path γ : I → E such that γ = q ◦ γ and γ(0) = x. But γ(I) ⊆ q−1(U)

is connected, so γ(I) ⊆ U ; the fact that γ(1) ∈ q−1({x}) implies that γ(1) = x. SinceE is simply connected, γ is null-homotopic in E and therefore γ is null-homotopic inX. This shows that U is relatively simply connected. �

Example 226. [Problem 11-19] For each n ∈ N, let Cn denote the circle in R2 withcenter (1/n, 0) and radius 1/n. The Hawaiian earring is the space H =

⋃n∈NCn,

with the subspace topology.

(1) H is not semilocally simply connected, and therefore has no universal coveringspace.

(2) The cone on H is simply connected and semilocally simply connected, but notlocally simply connected.

The point (0, 0) ∈ H does not have a relatively simply connected neighborhood, for anyneighborhood of (0, 0) must contain some circle Cn for n sufficiently large. The coneCH = (H × I)/(H × {0}) on H is contractible and is therefore simply connected andsemilocally simply connected. Consider the open set U = H × (1/2, 1] as a subset ofCH; the point (0, 0, 1) has no simply connected neighborhood in U . Therefore CH isnot locally simply connected.

Theorem 227. [Problem 11-20] Suppose X is a connected space that has a contractibleuniversal covering space. For any connected and locally path-connected space Y , acontinuous map f : Y → X is null-homotopic if and only if for each y ∈ Y , the inducedhomomorphism f∗ : π1(Y, y)→ π1(X, f(y)) is the trivial map. This result need not holdif the universal covering space is not contractible.

Proof. We can assume that Y is nonempty. Let q : E → X be a covering map with Econtractible. Suppose that the induced homomorphism f∗ : π1(Y, y) → π1(X, f(y)) is

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trivial for some y ∈ Y . Choose some e ∈ E such that q(e) = f(y). By Theorem 11.18,

there is a lift f : Y → E such that f = q ◦ f and f(y) = e. Let G : E × I → E be a

deformation retraction of E to a point; then q ◦G ◦ (f × IdI) is a homotopy from f toa constant map. The converse follows from Theorem 147. �

Example 228. [Problem 11-21] For which compact, connected surfaces M do thereexist continuous maps f : M → S1 that are not null-homotopic? Prove your answercorrect.

If M = P2# · · ·#P2 = (P2)#k (with k copies of the projective plane) then we have aninduced homomorphism f∗ : π1((P2)#k) → Z. If k = 1 then the only such homomor-phism is the trivial map, so f must be null-homotopic by Theorem 227. If M = S2

then all continuous maps f : M → S1 are null-homotopic since π1(S2) is trivial. IfM = (T2)#k then f might not be null-homotopic.

Chapter 12. Group Actions and Covering Maps

Theorem 229. [Exercise 12.12] For any covering map q : E → X, the action ofAutq(E) on E is a covering space action.

Proof. If e ∈ E then we can choose a neighborhood U of e such that q(U) is open andq|U is a homeomorphism. If ϕ ∈ Autq(E) and x ∈ U ∩ ϕ(U) then x = ϕ(y) for somey ∈ U . By Proposition 12.1 we have that y ∈ q−1({q(x)}), so y = x since q|U is ahomeomorphism. Applying Proposition 12.1 again shows that ϕ must be the identitymap. �

Theorem 230. [Exercise 12.13] Given a covering space action of a group Γ on atopological space E, the restriction of the action to any subgroup of Γ is a coveringspace action.

Proof. Obvious. �

Theorem 231. [Problem 12-1] Suppose q1 : E → X1 and q2 : E → X2 are normalcoverings. There exists a covering X1 → X2 making the obvious diagram commute ifand only if Autq1(E) ⊆ Autq2(E).

Proof. Choose any e ∈ E, let x1 = q1(e) and let x2 = q2(e). If there is a coveringq : X1 → X2 such that q ◦ q1 = q2 then for every ϕ ∈ Autq1(E) we have q2 ◦ ϕ =q ◦ q1 ◦ ϕ = q ◦ q1 = q2, i.e. ϕ ∈ Autq2(E). Conversely, if Autq1(E) ⊆ Autq2(E) then byTheorem 12.14 we have a chain of covering maps

Eπ1−→ E/Autq1(E)

q2−→ X2

such that q2 = q2 ◦ π1. Since there is a homeomorphism ψ : X1 → E/Autq1(E) suchthat ψ ◦ q1 = π1 by Theorem 236, the map q2 ◦ ψ is the desired covering map. �

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Example 232. [Problem 12-2] Let q : X3 → X2 be the covering map of Example 207.

(1) Determine the automorphism group Autq(X3).(2) Determine whether q is a normal covering.(3) For each of the following maps f : S1 → X2, determine whether f has a lift to

X3 taking 1 to 1.(a) f(z) = z.(b) f(z) = z2.(c) f(z) = 2− z.(d) f(z) = 2− z2.

It is easy to check manually that q is a normal covering. Example 11.17 shows thatif we write π1(X3, 1) = 〈ω1, ω2, ω3 | ∅〉 where ωi goes counterclockwise around the ithcircle and similarly π1(X2, 1) = 〈a, b | ∅〉 then q∗π1(X3, 1) = 〈a, b2, bab−1〉. ThereforeAutq(X3) ∼= 〈a, b | a, b2, bab−1〉. For (3), we have the following images of π1(S1, 1) underf∗: 〈a〉, 〈a2〉, 〈b〉, 〈b2〉. So all maps except for (c) have a lift to X3.

Example 233. [Problem 12-3] Let Xn be the union of n circles described in Problem 10-9, and let A, B, C, and D denote the unit circles centered at 0, 2, 4, and 6, respectively.Define a covering map q : X4 → X2 by

q(z) =

z, z ∈ A,2− (2− z)2, z ∈ B,(z − 4)2, z ∈ C,z − 4, z ∈ D.

(1) Identify the subgroup q∗π1(X4, 1) ⊆ π1(X2, 1) in terms of the generators de-scribed in Example 11.17.

(2) Prove that q is not a normal covering map.

Let ω1, . . . , ω4 be loops that go once counterclockwise around A, B, C and D, startingthat 1, 1, 3 and 5. Let c1 be the lower half of B and let c2 be the lower half ofC. Then π1(X4, 1) is the free group on {[ω1], [ω2], [c1 · ω3 · c1], [c1 · c2 · ω4 · c2 · c1]}, andq∗π1(X4, 1) = G = 〈a, b2, ba2b−1, bab−1a−1b−1〉. But G is not normal in π1(X2, 1) sinceb ∈ a−1b−1Gba but b /∈ G, so the covering map q is not normal.

Example 234. [Problem 12-4] Let E be the figure-eight space of Example 7.32, and letX be the union of the x-axis with infinitely many unit circles centered at {2πk + i : k ∈ Z}.Let q : X → E be the map that sends each circle in X onto the upper circle in E bytranslating in the x-direction and sends the x-axis onto the lower circle by x 7→ ieix− i.You may accept without proof that q is a covering map.

(1) Identify the subgroup q∗π1(X, 0) of π1(E , 0) in terms of the generators for π1(E , 0).(2) Determine the automorphism group Autq(X).

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(3) Determine whether q is a normal covering.

Write π1(E , 0) = 〈a, b | ∅〉, where a represents a loop that traverses the top circle coun-terclockwise and b traverses the bottom circle counterclockwise. Since X is homotopicto the wedge sum of infinitely many circles, π1(X, 0) is the free group on {ωk : k ∈ Z},where ωk is a loop that approaches (2πk, 0) along the x-axis, traverses the circle cen-tered at 2πk+i counterclockwise, and returns to (0, 0). It is clear that [q◦ωk] = bkab−k,so q∗π1(X, 0) is generated by

{bkab−k : k ∈ Z

}. This group is normal in π1(E , 0), so q

is a normal covering and Autq(X) ∼= Z.

Theorem 235. [Problem 12-5] Let q : E → X be a covering map. The discrete topologyis the only topology on Autq(E) for which its action on E is continuous.

Proof. Fix x ∈ E and define F : Autq(E)→ E by F (ϕ) = ϕ(x); if the action of Autq(E)on E is continuous then F must be continuous. By Proposition 12.1, F is injective.Let ϕ ∈ Autq(E), y = q(ϕ(x)) = q(x), let U be an evenly covered neighborhood of y

and let U be the component of q−1(U) containing ϕ(x). Then F−1(U) = {ϕ} is open,which shows that Autq(E) is discrete. �

Theorem 236. [Problem 12-7] Suppose q : E → X is a covering map (not necessarilynormal). Let E ′ = E/Autq(E) be the orbit space, and let π : E → E ′ be the quotientmap. Then there is a covering map q′ : E ′ → X such that q′ ◦ π = q.

Proof. Since q is constant on the fibers of π, it descends to a continuous map q′ : E ′ → Xsuch that q′ ◦ π = q, and it remains to show that q′ is a covering map. Let x ∈ X andlet U be an evenly covered neighborhood of x. We have (q′)−1(U) = π(q−1(U)), soit remains to show that π(q−1(U)) is the disjoint union of open sets that are mappedhomeomorphically onto U by q′. But Proposition 12.1 shows that each element ofAutq(E) permutes the components of q−1(U), so it is clear that π(q−1(U)) is the disjoint

union of open sets. If U ′ is one of these open sets and U is component of π−1(U ′) then π|Uand q|U are homeomorphisms, so (q′)|U ′ = q|U ◦ (π|U)−1 is also a homeomorphism. �

Example 237. [Problem 12-8] Consider the action of Z on Rm \ {0} defined by n ·x =2nx.

(1) Show that this is a covering space action.(2) Show that the orbit space (Rm \ {0})/Z is homeomorphic to Sm−1 × S1.(3) Show that if m ≥ 2, the universal covering space of Sm × S1 is homeomorphic

to Rm+1 \ {0}.

(1) is obvious. Let q : Rm \ {0} → (Rm \ {0})/Z be the quotient map. For (2), definef : Rm \ {0} → Sm−1 × S1 by

x 7→(

x

‖x‖, exp (2πi log2 ‖x‖)

).

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Since f is constant on each orbit, it descends to a continuous map f : (Rm \ {0})/Z→Sm−1 × S1 such that f = f ◦ q. Let θ : S1 → [0, 1] be an inverse to s 7→ exp(2πis) suchthat θ(1) = 0 and θ|S1\{1} is continuous. Define g : Sm−1 × S1 → (Rm \ {0})/Z by

(s, t) 7→ q(2θ(t)s);

then g is continuous. Since f ◦ g and g ◦ f are identity maps, this shows that (Rm \{0})/Z ≈ Sm−1 × S1. By Theorem 12.14, q is a (normal) covering map. If m ≥ 2then Rm+1 \ {0} is simply connected, so the universal covering space of Sm−1 × S1 ishomeomorphic to Rm+1 \ {0}.

Example 238. [Problem 12-11] Let M = T2#T2.

(1) Show that the fundamental group of M has a subgroup of index 2.

(2) Prove that there exists a manifold M and a two-sheeted covering map q : M →M .

We have π1(M) ∼= Z2∗Z2 which has the presentation 〈S | R〉, where S = {α, β, γ, δ} andR = {αβα−1β−1, γδγ−1δ−1}. But Z ∼= 〈S | R ∪ {β, γ, δ}〉, so Z2

∼= 〈S | R ∪ {α2, β, γ, δ}〉.It follows from Theorem 178 that the index of the normal closure G of {α2, β, γ, δ} in

π1(M) is 2. By Theorem 12.18, there is a covering map q : M → M such that

q∗π1(M) ∼= G, and by Corollary 12.8, Autq(M) has order 2. This implies that q istwo-sheeted.

Example 239. [Problem 12-14] Give an example to show that a subgroup of a finitelygenerated nonabelian group need not be finitely generated.

See Example 234.

Theorem 240. [Problem 12-15] Suppose X is a topological space that has a universalcovering space. Let x ∈ X, and write G = π1(X, x). Let CovX denote the categorywhose objects are coverings of X and whose morphisms are covering homomorphisms;and let SetG denote the category whose objects are transitive right G-sets and whosemorphisms are G-equivariant maps. Define a functor F : CovX → SetG as follows: forany covering q : E → X, F(q) is the set q−1({x}) with its monodromy action; and forany covering homomorphism ϕ : E1 → E2, F(ϕ) is the restriction of ϕ to q−1({x}).Then F is an equivalence of categories.

Proof. Let S ∈ SetG. By Theorem 12.18, there is a covering map q : E → X suchthat the conjugacy class of q∗π1(E) is the same as the isotropy type of S, and byTheorem 11.29, the isotropy type of F(q) is equal to the isotropy type of S. ThenF(q) is isomorphic to S by Proposition 11.26. Furthermore, Proposition 12.1 showsthat F : HomCovX (q1, q2) → HomSetG(F(q1),F(q2)) is injective. Suppose qi : Ei → Xare covering maps for i = 1, 2 and f : F(q1) → F(q2) is G-equivariant. Choose

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e1 ∈ F(q1) and let e2 = f(e1) ∈ F(q2); then Ge1 ⊆ Ge2 by Proposition 11.24, andq∗π1(E1, e1) ⊆ q∗π1(E2, e2) by Theorem 11.29. Applying Theorem 11.37 shows thatthere is a covering homomorphism q : E1 → E2 from q1 to q2 taking e1 to e2. SinceF(q) agrees with f at e1, we must have F(q) = f by Proposition 11.24. �

Theorem 241. [Problem 12-16] Suppose G is a topological group acting continuouslyon a Hausdorff space E. If the map α : G × E → E defining the action is a propermap, then the action is a proper action.

Proof. Define Θ : G×E → E ×E by (g, e) 7→ (α(g, e), e); we want to show that Θ is aproper map. Let π : E × E → E be the projection onto the first coordinate. SupposeL ⊆ E × E is compact; then π(L) is also compact, and α−1(π(L)) is compact sinceα is proper. Since E is Hausdorff, L is closed and Θ−1(L) is closed. But Θ−1(L) ⊆α−1(π(L)), so Θ−1(L) must be compact. �

Chapter 13. Homology

Definition 242. Given a continuous map f : X → Y , let f# : Cp(X) → Cp(Y ) bethe homomorphism defined by setting f#σ = f ◦ σ for each singular p-simplex σ. Ifc ∈ Zp(X) then ∂(f#c) = f#(∂c) = 0, so f#Zp(X) ⊆ Zp(Y ). Similarly, if c ∈ Bp(X)then c = ∂b for some b ∈ Cp+1(X), so f#c = f#(∂b) = ∂(f#b) which shows thatf#Bp(X) ⊆ Bp(Y ). Let πX : Zp(X) → Hp(X) and πY : Zp(Y ) → Hp(Y ) be thequotient maps. Since πY ◦ f# : Zp(X) → Hp(Y ) satisfies Bp(X) ⊆ ker(πY ◦ f#), thereis a unique homomorphism f∗ : Hp(X) → Hp(Y ) such that the following diagramcommutes:

Zp(X) Zp(Y )

Hp(X) Hp(Y )

f#

πX πY

f∗

This map is called the homomorphism induced by f .

Theorem 243. Let X, Y , and Z be topological spaces.

(1) The homomorphism (IdX)∗ : Hp(X) → Hp(X) induced by the identity map ofX is the identity of Hp(X).

(2) If f : X → Y and g : Y → Z are continuous maps, then

(g ◦ f)∗ = g∗ ◦ f∗ : Hp(X)→ Hp(Z).

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Thus the pth singular homology group defines a covariant functor from the category oftopological spaces to the category of abelian groups.

Proof. Part (1) follows from the fact that (IdX)# = IdCp(X) and that IdHp(X) satisfiesthe diagram in Definition 242. For (2), since (g ◦ f)# = g# ◦ f# we have

g∗ ◦ f∗ ◦ πX = g∗ ◦ πY ◦ f# = πZ ◦ g# ◦ f# = πZ ◦ (g ◦ f)#.

Therefore (g ◦ f)# = g∗ ◦ f∗ by uniqueness. �

Theorem 244. [Exercise 13.10] Suppose f : X → Y is a homotopy equivalence. Thenfor each p ≥ 0, f∗ : Hp(X)→ Hp(Y ) is an isomorphism.

Proof. Let g : Y → X be a continuous map such that g ◦ f ' IdX and f ◦ g ' IdY . ByTheorem 13.8 we have

g∗ ◦ f∗ = (g ◦ f)∗ = IdHp(X) and f∗ ◦ g∗ = (f ◦ g)∗ = IdHp(Y ),

so f∗ is an isomorphism. �

Theorem 245. [Exercise 13.12] If F,G : C∗ → D∗ are chain homotopic maps, thenF∗ = G∗ : Hp(C∗)→ Hp(D∗) for all p.

Proof. Let h : Cp → Dp+1 be a chain homotopy from F to G. Fix some p and letπC , πD : ker ∂p → ker ∂p/ im ∂p+1 be the quotient maps. For all c ∈ ker ∂p we have

(G∗ ◦ πC)(c) = (πD ◦G)(c)

= πD(F (c) + (G− F )(c))

= πD(F (c) + (h ◦ ∂ + ∂ ◦ h)(c))

= πD(F (c) + (∂ ◦ h)(c))

= (πD ◦ F )(c)

since (∂ ◦ h)(c) ∈ im ∂p+1. By uniqueness, F∗ = G∗. �

Theorem 246. Let

0→ C∗F−→ D∗

G−→ E∗ → 0

be a short exact sequence of chain maps. Then for each p there is a connecting homo-morphism ∂∗ : Hp(E∗)→ Hp−1(C∗) such that the following sequence is exact:

(*) · · · ∂∗−→ Hp(C∗)F∗−→ Hp(D∗)

G∗−→ Hp(E∗)∂∗−→ Hp−1(C∗)

F∗−→ · · · .

Proof. Consider the diagram

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0 Cp+1 Dp+1 Ep+1 0

0 Cp Dp Ep 0

0 Cp−1 Dp−1 Ep−1 0

0 Cp−2 Dp−2 Ep−2 0.

F G

F G

F G

F G

∂

∂

∂

∂

∂

∂

∂

∂

∂

Every square in this diagram commutes, and the horizontal rows are exact. Supposee ∈ Ep with ∂pe = 0. Since G is surjective, there is some d ∈ Dp such that e = Gd.Then G∂pd = ∂pGd = ∂pe = 0, so ∂pd ∈ kerG and ∂pd = Fc for some c ∈ Cp−1 byexactness at Dp−1. We have F∂p−1c = ∂p−1Fc = ∂p−1∂pd = 0, so ∂p−1c = 0 since F isinjective. Therefore we can define a map ∂# : ker ∂p → Hp−1(C∗) by setting ∂#e = πCcwhere πC : ker ∂p → Hp−1(C∗) is the quotient map. We first ensure that ∂# is well-defined. Suppose we have d′ ∈ Dp and c′ ∈ Cp−1 with e = Gd′ and ∂pd

′ = Fc′. Thend− d′ ∈ kerG, so d− d′ = Fa for some a ∈ Cp. Now F (c− c′) = ∂p(d− d′) = ∂pFa =F∂pa, so c− c′ = ∂pa since F is injective. We have πCc = πCc

′, which shows that ∂# iswell-defined. Next, we show that ∂# is a homomorphism. If e′ ∈ Ep then e′ = Gd′ and∂pd

′ = Fc′ for some d′ ∈ Dp and c′ ∈ Cp−1. So e+e′ = G(d+d′) and ∂p(d+d′) = F (c+c′),which implies that ∂#(e + e′) = ∂#e + ∂#e

′. Finally, we show that im ∂p+1 ⊆ ker ∂#.Suppose that e = ∂p+1e

′ for some e′ ∈ Ep+1; then e′ = Gd′ for some d′ ∈ Dp+1. Wehave G∂p+1d

′ = ∂p+1Gd′ = ∂p+1e

′ = e = Gd, so d− ∂p+1d′ ∈ kerG and d− ∂p+1d

′ = Fafor some a ∈ Cp. Now Fc = ∂pd = ∂p(d− ∂p+1d

′ + ∂p+1d′) = ∂pFa = F∂pa, so c = ∂pa

since F is injective. Then πCc = 0, which shows that im ∂p+1 ⊆ ker ∂#. Therefore ∂#descends to a homomorphism ∂∗ : Hp(E∗) → Hp−1(C∗) such that ∂∗πE = ∂#, whereπE : ker ∂p → Hp(E∗) is the quotient map.

Now we prove exactness of the sequence (*). Given a cycle c, we write [c] for thehomology class of c. Suppose [c] ∈ Hp(C∗) with [c] = ∂∗[e] for some [e] ∈ Hp(E∗). Fromthe definition of ∂∗ there is some d ∈ Dp+1 such that ∂p+1d = Fc, so F∗[c] = [Fc] = 0.Therefore im ∂∗ ⊆ kerF∗. Conversely, if F∗[c] = [Fc] = 0 then Fc = ∂p+1d for somed ∈ Dp+1. We have ∂p+1Gd = G∂p+1d = GFc = 0, and ∂∗[Gd] = [c] from the definitionof ∂∗. This shows that kerF∗ ⊆ im ∂∗, and proves exactness at Hp(C∗). Next, we proveexactness at Hp(D∗). Since GF = 0 and G∗F∗ = 0, it is immediate that imF∗ ⊆ kerG∗.Suppose G∗[d] = 0, i.e. Gd = ∂p+1e for some e ∈ Ep+1. Since G is surjective, thereis some d′ ∈ Dp+1 such that e = Gd′. Then G∂p+1d

′ = ∂p+1Gd′ = ∂p+1e = Gd, so

d− ∂p+1d′ ∈ kerG and d− ∂p+1d

′ = Fc for some c ∈ Cp. Since F∗[c] = [Fc] = [d], this

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shows that kerG∗ ⊆ imF∗ and proves exactness at Hp(D∗). Finally, we prove exactnessat Hp(E∗). Suppose [e] ∈ Hp(E∗) such that [e] = G∗[d] = [Gd] for some d ∈ Dp with∂pd = 0. Let [c] = ∂∗[e] = ∂∗[Gd]. From the definition of ∂∗ we have Fc = ∂pd = 0,so c = 0 since F is injective. This shows that imG∗ ⊆ ker ∂∗. Conversely, suppose[e] ∈ Hp(E∗) with ∂∗[e] = 0. This means that e = Gd and ∂pd = Fc for somed ∈ Dp and some boundary c ∈ Cp−1. Choose c′ ∈ Cp such that c = ∂pc

′. Then∂pd = F∂pc

′ = ∂pFc′, so ∂p(d − Fc′) = 0 and G(d − Fc′) = Gd = e. Therefore

G∗[d− Fc′] = [Gd] = [e], so ker ∂∗ ⊆ imG∗. This proves exactness at Hp(E∗). �

Theorem 247. [Exercise 13.39] The induced cohomology homomorphism satisfies thefollowing properties.

(1) If f : X → Y and g : Y → Z are continuous, then (g ◦ f)∗ = f ∗ ◦ g∗.(2) The homomorphism induced by the identity map is the identity.

Therefore, the assignments X 7→ Hp(X;G), f 7→ f ∗ define a contravariant functorfrom the category of topological spaces to the category of abelian groups. Furthermore,if f : X → Y is a homeomorphism then for every abelian group G and every integerp ≥ 0 the map f ∗ : Hp(Y ;G)→ Hp(X;G) is an isomorphism.

Proof. We have

((g ◦ f)#ϕ)(c) = ϕ((g ◦ f)#c) = ϕ((g# ◦ f#)(c)) = (g#ϕ)(f#c) = ((f# ◦ g#)ϕ)(c),

so (g ◦ f)# = f# ◦ g# and (1) follows. For (2) we have ((IdX)#ϕ)(c) = ϕ((IdX)#c) =ϕ(c), so (IdX)# = IdCp(X;G). �

Theorem 248. [Problem 13-1] Let X1, . . . , Xk be spaces with nondegenerate base points.Then for every p > 0,

Hp(X1 ∨ · · · ∨Xk) ∼= Hp(X1)⊕ · · · ⊕Hp(Xk).

Proof. By Lemma 10.6, it suffices to prove the theorem for the case k = 2. Let p1, p2be nondegenerate base points. For i = 1, 2, let Wi be a neighborhood of pi that admitsa strong deformation retraction onto {pi}. Let q : X1 qX2 → X1 ∨X2 be the quotientmap, let U = q(X1 q W2) and let V = q(W1 q X2). Since X1 q W2 and W1 q X2

are saturated open sets in X1 qX2, the sets U and V are open in X1 ∨X2. ApplyingTheorem 13.16 and noting that U ∩ V is contractible, we have an exact sequence

0→ Hp(U)⊕Hp(V )k∗−l∗−−−→ Hp(X1 ∨X2)→ 0

where k : U → X1∨X2 and l : V → X1∨X2 are the inclusions. This implies that k∗− l∗is an isomorphism. Since X1 is a deformation retract of U and X2 is a deformationretract of V , we have Hp(X1)⊕Hp(X2) ∼= Hp(U)⊕Hp(V ) ∼= Hp(X1∨X2) as desired. �

Theorem 249. [Problem 13-2] (cf. Theorem 157)

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(1) If U ⊆ Rn is an open subset with n ≥ 2 and x ∈ U , then Hn−1(U \ {x}) 6= 0.(2) If m > n then Rm is not homeomorphic to any open subset of Rn.

Proof. For (1), we can assume n > 2 for the case n = 2 is covered by Theorem 157. LetBr(x) ⊆ U be an open ball around x. Applying Theorem 13.16 with the open subsets

Br(x) \ {x} and U \Br/2(x) gives an exact sequence

Hn−1(Br(x) \ {x})⊕Hn−1(U \Br/2(x))→ Hn−1(U \ {x})→ Hn−2(Br(x) \Br/2(x)).

Since Br(x) \ Br/2(x) is homotopic to Sn−1 and Br(x) \ {x} ∼= Rn \ {0}, the sequencereduces to

Z⊕Hn−1(U \Br/2(x))→ Hn−1(U \ {x})→ 0.

This shows that Hn−1(U \{x}) 6= 0. For (2), the cases n = 1, 2 are covered by Theorem80 and Theorem 157. Suppose U ⊆ Rn is open with n > 2 and ϕ : U → Rm is ahomeomorphism. Choose any x ∈ U ; then U \ {x} ≈ Rm \ {ϕ(x)}, so Hn−1(U \ {x}) ∼=Hn−1(Rm \ {ϕ(x)}) = 0. This contradicts part (1). �

Theorem 250. [Problem 13-3] A nonempty topological space cannot be both an m-manifold and an n-manifold for any m > n (cf. Theorem 158).

Proof. The cases n = 1, 2 are covered by Theorem 81 and Theorem 158, so we canassume that n > 2. Let M be a nonempty topological space that is both a m-manifoldand an n-manifold for m > n > 2. Choose some p ∈ M and let ϕ1 : U1 → V1 andϕ2 : U2 → V2 be homeomorphisms where U1 and U2 are neighborhoods of p, V1 isopen in Rm, and V2 is open in Rn. Let B be an open ball around ϕ1(p) contained inϕ1(U1 ∩ U2). Then B ≈ (ϕ2 ◦ ϕ−11 )(B), but this contradicts Theorem 249. �

Theorem 251. [Problem 13-4] Suppose M is an n-dimensional manifold with bound-ary. Then the interior and boundary of M are disjoint (cf. Theorem 159).

Proof. Suppose p ∈ M is both an interior and boundary point. Choose coordinatecharts (U,ϕ) and (V, ψ) such that U, V are neighborhoods of p, ϕ(U) is open in IntHn,ψ(V ) is open in Hn, and ψ(p) ∈ ∂Hn. Let W = U ∩ V ; then ϕ(W ) is homeomorphicto ψ(W ). But this is impossible, for Hn−1(ϕ(W ) \ {ϕ(p)}) 6= 0 by Theorem 249 whileHn−1(ψ(W ) \ {ψ(p)}) = 0. �

Theorem 252. [Problem 13-5] Let n ≥ 1. If f : Sn → Sn is a continuous map that has

a continuous extension to a map F : Bn+1 → Sn, then f has degree zero (cf. Theorem160).

Proof. Define a homotopyH : Sn×I → Sn from a constant map to f byH(x, t) = F (tx);then deg f = 0 by Proposition 13.25 and Proposition 13.27. �

Theorem 253. [Problem 13-6] Sn is not a retract of Bn+1 for any n.

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Proof. The case n = 0 is clear, so we can assume n ≥ 1. Since Bn+1 is contractible,Hn(Bn+1) = 0. But Hn(Sn) = Z, so Sn cannot be a retract of Bn+1 by Corollary13.4. �

Theorem 254. [Problem 13-7] For every integer n ≥ 0, every continuous map f :Bn → Bn has a fixed point (cf. Theorem 162).

Proof. We can assume that n ≥ 1. If f has no fixed point then we can define acontinuous map ϕ : Bn → Sn−1 by

ϕ(x) =x− f(x)

‖x− f(x)‖.

This contradicts Theorem 252, since we can define a homotopy H : Sn−1 × I → Sn−1from ϕ|Sn−1 to IdSn−1 by

H(x, t) =x− (1− t)f(x)

‖x− (1− t)f(x)‖using an argument identical to Theorem 162. �

Theorem 255. [Problem 13-8] If n is even then Z2 is the only nontrivial group thatcan act freely on Sn by homeomorphisms.

Proof. Suppose G acts freely on Sn by homeomorphisms. For any g ∈ G, write ϕg forthe homeomorphism x 7→ g · x. Then degϕg = ±1 by Proposition 13.25, so deg definesa homomorphism from G to {±1}. If degϕg = 1 and ϕg 6= IdSn then ϕg has no fixedpoint since G acts freely. But degϕ = −1 by Theorem 13.29, which is a contradiction.This shows that deg is an injective homomorphism, so G is either the trivial group oris isomorphic to Z2. �

Example 256. [Problem 13-9] Use the CW decomposition of Theorem 117 and theresults of this chapter to compute the singular homology groups of the 3-dimensionalreal projective space P3.

We first compute H2(P2). By Proposition 13.33 there is a short exact sequence

0→ H2(P1)→ H2(P2)→ K → 0,

where K is the kernel of ((q ◦ F )|∂B2)∗ : H1(∂B2) → H1(P1). But H2(P1) = K = 0, soH2(P2) = 0.

Let ϕ = (q◦F )|∂B3 as defined in Theorem 117, and consider P2 as a subspace of P3. SinceH2(P2) = 0, the kernel K of ϕ∗ : H2(∂B3)→ H2(P2) is H2(∂B3) = H2(S2) = Z and theimage L is 0. It is clear that H0(P3) = Z. By Proposition 13.33, if p = 1 or p > 3 thenthe inclusion P2 ↪→ P3 induces an isomorphism. Therefore H1(P3) = H1(P2) = Z2 andHp(P3) = 0 for p > 3. If p = 2 then we have a short exact sequence

0→ L ↪→ H2(P2)→ H2(P3)→ 0.

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Since H2(P2) = L = 0, we have H2(P3) = 0. If p = 3, we have a short exact sequence

0→ H3(P2)→ H3(P3)→ K → 0

that reduces to an exact sequence

0→ H3(P3)→ Z→ 0,

so H3(P3) = Z. Therefore

Hp(P3) =

Z, p = 0,

Z2, p = 1,

0, p = 2,

Z, p = 3,

0, p > 3.

Theorem 257. [Problem 13-11] For any field F of characteristic zero, the functorG 7→ Hom(G,F), f 7→ f ∗ is exact.

Proof. Suppose that the sequence

· · · → Af−→ B

g−→ C → · · ·

is exact at B; we want to show that

· · · ← Hom(A,F)f∗←− Hom(B,F)

g∗←− Hom(C,F)← · · ·

is exact at Hom(B,F). If β = g∗γ for some γ ∈ Hom(C,F) then for all a ∈ A we have

(f ∗g∗γ)(a) = (γgf)(a) = 0

since im f ⊆ ker g. Therefore f ∗β = f ∗g∗γ = 0, which shows that im g∗ ⊆ ker f ∗.Conversely, suppose that f ∗β = 0; then βfa = 0 for all a ∈ A. We define a mapγ ∈ Hom(im g,F) by setting γ(c) = βb for any c = gb ∈ im g. To show that γ is well-defined, suppose b, b′ ∈ B with gb = gb′. Then b − b′ ∈ ker g, so b − b′ = fa for somea ∈ A, and βb− βb′ = β(b− b′) = βfa = 0. It is clear that γ is a homomorphism sinceg and β are homomorphisms. By Lemma 13.42 there is an extension γ′ ∈ Hom(C,F)of γ, so β = g∗γ′ and ker f ∗ ⊆ im g∗ as desired. �

Theorem 258. [Problem 13-12] Let X be a topological space and let U, V ⊆ X beopen subsets whose union is X. Then there is an exact Mayer-Vietoris sequence forcohomology with coefficients in a field F of characteristic zero:

· · · → Hp−1(U ∩ V ;F)→ Hp(X;F)→ Hp(U ;F)⊕Hp(V ;F)→ Hp(U ∩ V ;F)→ · · · .


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References

[1] John M. Lee. Introduction to Smooth Manifolds. Springer, 2nd edition, 2013.

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