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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 discrete
topology 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 metric
topology on M .
We only check (4). It is clear that ∅ and M are in T . Let U1 , . . . , Un ∈ T and let
p ∈ U1 ∩ · · · ∩ Un . There exist open balls B1 ⊆ U1 , . . . , Bn ⊆ Un that contain p, so the
intersection B1 ∩ · · · ∩ Bn ⊆ U1 ∩ · · · ∩ Un is an open ball containing p. This shows that
U1 ∩ · · · ∩ Un ∈ T . Now let {UαS
}α∈A be some collection of elements in T which we can
assume to be nonempty. If p ∈ α∈A Uα weScan choose some α ∈ A, andSthere 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 contained
in A.
(2) A point q is in the exterior of A if and only if q has a neighborhood contained
in X \ A.
(3) A point q is in the boundary of A if and only if every neighborhood of q contains
both a point of A and a point of X \ A.
(4) Int A and Ext A are open in X, while ∂A is closed in X.
(5) A is open if and only if A = Int A, 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 = Int A ∪ ∂A.
(7) A = A ∪ ∂A = Int A ∪ ∂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 of
X \ A. Conversely, if every neighborhood of q contains both a point of A and a point
of X \ A then q ∈
/ (Int A ∪ Ext A) by (1) and (2). Parts (4) and (5) are obvious. If A
is closed and p ∈ ∂A then
p ∈ X \ (Int A ∪ Ext A) = X \ (Int A ∪ (X \ A)) ⊆ A.
If A contains all its boundary points then Int A ∪ ∂A ⊆ A, so A = Int A ∪ ∂A since
it is always true that A ⊆ Int A ∪ ∂A. If A = Int A ∪ ∂A then A is closed since
1
2
A = Int A ∪ (X \ (Int A ∪ Ext A)) = X \ Ext A and Ext A is open. This proves (6). For
(7),
Int A ∪ ∂A = Int A ∪ (X \ (Int A ∪ Ext A)) = X \ Ext A = A
and A ∪ ∂A = Int A ∪ ∂A since Int A ∪ ∂A ⊆ A ∪ ∂A and p ∈ A ∪ ∂A implies that
p ∈ ∂A or p ∈ A \ ∂A = Int A.
Theorem 3. [Exercise 2.10] A set A in a topological space X is closed if and only if it
contains all of its limit points.
Proof. Suppose A is closed and let q ∈ X \ A be a limit point of A. Since every
neighborhood 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 a
neighborhood 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 all
its boundary points, i.e. A is closed.
Theorem 4. [Exercise 2.11] A subset A ⊆ X is dense if and only if every nonempty
open 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 any
point in U . Since Ext A is empty, no neighborhood of p is contained in X \ A. But
U is a neighborhood of p, so U contains a point of A. Conversely, suppose that every
nonempty open set in X contains a point of A. Then every p ∈ X \ A is a limit point
of A, so p ∈ A and therefore X ⊆ A.
Theorem 5. [Exercise 2.12] Let (M, d) be a metric space with the usual topology. The
following 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 all
i ≥ 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 neighborhood
of q. Since U is open, there exists an open ball B ⊆ U of radius r around q, so there
exists an integer N such that d(qi , q) < ε for all i ≥ N . But then qi ∈ B ⊆ U for all
i ≥ N , which proves (1).
Theorem 6. [Exercise 2.13] Let X be a discrete topological space. Then the only
convergent 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 such
that 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 continuous
if 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 is
similar.
Theorem 8. [Exercise 2.18] Let X, Y , and Z be topological spaces.
(1)
(2)
(3)
(4)
Any constant map f : X → Y is continuous.
The identity map Id : X → X is continuous.
If f : X → Y is continuous, so is the restriction of f to any open subset of X.
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. If
y ∈ 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 under
finite intersections. For (4), let U ⊆ Z be an open set. Since g is continuous g −1 (U ) is
open, 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 between
X and itself. If ϕ : X → Y is a homeomorphism then ϕ−1 : Y → X is also a
homeomorphism. Finally, if ϕ : X → Y and ψ : Y → Z are homeomorphisms then
ψ ◦ 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 let
f : X1 → X2 be a bijective map. Then f is a homeomorphism if and only if f (T1 ) = T2
in 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
(x, y, z)
ϕ(x, y, z) = p
.
x2 + y 2 + z 2
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
(x, y, z)
ϕ(ϕ−1 (x, y, z)) = p
= (x, y, z)
x2 + y 2 + z 2
p
since (x, y, z) ∈ S2 implies that x2 + y 2 + z 2 = 1 and
ϕ−1 (ϕ(x, y, z)) =
(x, y, z)
= (x, y, z)
max(|x| , |y| , |z|)
since (x, y, z) ∈ C.
Theorem 12. [Exercise 2.28] Let X denote the half-open interval [0, 1) ⊆ R, and let
S 1 denote the unit circle in R2 . Define a map a : X → S 1 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 S 1 , the inverse
of a cannot be continuous.
Theorem 13. [Exercise 2.31]
(1)
(2)
(3)
(4)
Every
Every
Every
Every
local homeomorphism is an open map.
homeomorphism is a local homeomorphism.
bijective continuous open map is a homeomorphism.
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 of
x such that f (V ) is open and f |V : V → f (V ) is aShomeomorphism; 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 with
the 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 neighborhood
of 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 ∈ X
there exists a continuous function f : X → R such that f −1 ({0}) = {p}. Then X is
Hausdorff.
Proof. Let p, q be distinct points of X and let f : X → R be a continuous function such
that f −1 ({0}) = {p}. Then f (q) 6= 0, and we can choose disjoint neighborhoods U of 0
and V of f (q) in R. We have f −1 (U ) ∩ f −1 (V ) = ∅, while f −1 (U ) is a neighborhood of
p 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 discrete
topology.
Proof. Let X be a finite set with a Hausdorff topology T . By Proposition 2.37, every
one-point set in X is closed. But every subset of X can be written as a finite union of
one-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 given
topology.
(1) M is a metric space with the metric topology, and B is the collection of all open
balls in M .
(2) X is a set with the discrete topology, and B is the collection of all one-point
subsets 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 open
set in X. If x ∈ U then x ∈ {x} ⊆ U , so B is a basis. For (3), if U is an open set then
it 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 Euclidean
topology on Rn :
(1) B1 = {Cs (x) : x ∈ Rn and s > 0}, where Cs (x) is the open cube of side s around
x:
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 of
radius 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 r0 with 0 < r0 ≤ r/2 and any point x0 ∈ Br0 (x) with
rational coordinates; then Br0 (x0 ) ⊆ B is an element of B2 that contains x.
Theorem 19. [Exercise 2.51]
(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. Let
E be the countable set formed by choosing a point from each B ∈ B. Let U be a
nonempty 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 show
that 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 ball
of radius r containing x. Since E is dense in M , there exists a point p ∈ E ∩ Br/2 (x).
Let r0 be a rational number with d(p, x) < r0 < r/2. Then Br0 (p) is an element of B
with x ∈ Br0 (p) and Br0 (p) ⊆ B, since
d(x, y) ≤ d(x, p) + d(p, y) < r
for every y ∈ Br0 (p).
Theorem 20. [Exercise 2.54] A topological space X is a 0-manifold if and only if it is
a countable discrete space.
Proof. Let x ∈ X and let N be a neighborhood of x homeomorphic to an open subset
of R0 , i.e. a point. Then N must contain exactly one element, so N = {x}. This shows
that every one-point subset of X is open, and therefore X is discrete. Also, any second
countable 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 continuous
map. If A ⊆ X and B ⊆ Y are subsets with f (A) ⊆ B, then f (A) ⊆ B.
Proof. Suppose that f (x) ∈ Ext B 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 ⊆ T1
and 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 union
of elements from B1 , which are open in T2 by assumption. Therefore every open set in
T1 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 (but
different 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 0 ⊆ B in M2 around
x. 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 0 ⊆ B in M2 around x. This shows that U is open
in M2 . Similarly, every set that is open in M2 is also open in M1 . This shows that the
two spaces have the same topology.
Example 24. [Problem 2-1] Let X be an infinite set. Consider the following collections
of 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 \ Int B.
(2) Int(X \ B) = X \ B.
Proof. Suppose that x ∈ X \ B. We can assume that x ∈ B, for otherwise x ∈ X \ B ⊆
X \ Int B. Then x is a limit point of X \ B, so any neighborhood of x contains a point
of X \ B. This shows that x ∈
/ Int B, i.e. x ∈ X \ Int B. 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 ∈ Int B. This proves (1). For (2), apply (1) with
X \ 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 that
any 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 is
a 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 point
q1 ∈ A not equal to p. Suppose we have chosen the distinct points q1 , . . . , qn , none of
which are equal to p. For each i, there exist neighborhoods Ui of qi and Vi of p such
that Ui ∩ Vi = ∅. Then V = V1 ∩ · · · ∩ Vn is a neighborhood of p, so V contains some
point qn+1 ∈ A not equal to p. Furthermore, qn+1 is not equal to any qi for i ≤ n since
V ∩ Ui = ∅ for every i.
Theorem 28. [Problem 2-8] Let X be a Hausdorff space, let A ⊆ X, and let A0 denote
the set of limit points of A. Then A0 is closed in X.
Proof. Let x be a limit point of A0 and let E be a neighborhood of x. Then there exists
some y ∈ E ∩ A0 not equal to x. Choose neighborhoods U of x and V of y such that
U ∩ V = ∅ and U, V ⊆ E. Since y is a limit point of A0 , there exists some z ∈ V ∩ A
not equal to y. Then z 6= x since U and V are disjoint, which proves that x ∈ A0 .
Therefore A0 is closed in X.
Theorem 29. [Problem 2-9] Let X be a discrete space, Y be a space with the trivial
topology, and Z be any topological space. Any maps f : X → Z and g : Z → Y
are continuous, and if Z is Hausdorff, then the only continuous maps h : Y → Z are
constant 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 topological
spaces, 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 for
the topology of Y .
Proof. Every set in f (B) is open since f is an open map. Let U be an open set in
Y and let y ∈ U . Since f is surjective we have y = f (x) for some x, so f −1 (U ) is
a neighborhood of x. There exists some B ∈ B with x ∈ B and B ⊆ f −1 (U ), so
f (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 of
X whose union equals X. Let T be the collection of all unions of finite intersections of
elements of B.
(1) T is a topology. (It is called the topology generated by B, and B is called a
subbasis 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 open
sets. If
[
[
(Uα,1 ∩ · · · ∩ Uα,mα ) and
(Uβ,1 ∩ · · · ∩ Uβ,nβ )
α∈A
β∈B
are elements of T , then their intersection is
[
Uα,1 ∩ · · · ∩ Uα,mα ∩ Uβ,1 ∩ · · · ∩ Uβ,nβ ,
(α,β)∈A×B
which is also in T . This proves that T is a topology. To prove (2), we only need to
show that for every topology T 0 containing B, we have T ⊆ T 0 . But this is clear from
the fact that T 0 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 order
topology.
(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)), which
shows that (a, b) is open and [a, b] is closed. For (2), let a and b be distinct points in
X. If there exists some c with a < c < b then we have the neighborhoods L(c) of a and
R(c) of b with L(c) ∩ R(c) = ∅. Otherwise, the neighborhoods L(b) of a and R(a) of b
satisfy L(b) ∩ R(a) = ∅. This shows that X is Hausdorff. For (3), if E is a closed set
containing (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 is
a 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 takes
convergent 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 neighborhood
basis 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 show
that the sequence {pn } converges to p. Let U be a neighborhood of p. There exists a
BN ∈ 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
that pn 6= p for all n, for otherwise p ∈ A and we are
of p then there exists an integer N such that pn ∈ U
pN ∈ U ∩ A and pN 6= p by our previous assumption,
point of A.
with pn → p. We may assume
done. If U is a neighborhood
for all n ≥ N . In particular,
which proves that p is a limit
Theorem 34. [Problem 2-16] If X is a second countable topological space, then every
collection 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 disjoint
nonempty open subsets. For each α, choose a point xα ∈ Uα ; there exists an element
Bα ∈ B with xα ∈ Bα and Bα ⊆ Uα . Since Bα 6= Bβ whenever α 6= β, we have a
bijection 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+ .
Theorem 36. Let X and Y be topological spaces with the property that every point in
X has a neighborhood homeomorphic to an open set in Y . Then X is first countable if
Y 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 a
neighborhood of x then f (U ∩ N ) is a neighborhood of f (x), so there exists a B ∈ B
with B ⊆ f (U ∩ N ). Then f −1 (B) ⊆ U ∩ N , which shows that {f −1 (B) : B ∈ B} is a
countable neighborhood basis of x.
Corollary 37. [Problem 2-21] All locally Euclidean spaces and metric spaces are first
countable.
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Example 38. [Problem 2-22] Let X = R2 as a set, but with the topology determined
by 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 a
subset of B. For if U = {(1, y) : a < y < b} ∈ B contains (1, 0) then there are no values
of a, b for which U is a subset of B. The map R2 → X is not continuous either because
no 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 Vp
is 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−1
p (B) ⊆ Uα ∩ Np . Since ϕp |Bp,α : Bp,α → B is a homeomorphism, Bp,α is a coordinate
ball. By our construction, the set B = {Bp,α : (p, α) ∈ M × A and p ∈ Uα } is a basis
for M .
Theorem 40. [Problem 2-24] Suppose X is locally Euclidean of dimension n, and
f : X → Y is a surjective local homeomorphism. Then Y is also locally Euclidean of
dimension n.
Proof. Let y ∈ Y so that y = f (x) for some x ∈ X. There exists a neighborhood N of x
and homeomorphism ϕ : N → U where U is open in Rn , and there exists a neighborhood
N 0 of x such that f (N 0 ) is open and f |N 0 : N 0 → f (N 0 ) is a homeomorphism. Then
f (N ∩ N 0 ) is a neighborhood of y and
ϕ|N ∩N 0 ◦ (f |N ∩N 0 )−1 : f (N ∩ N 0 ) → ϕ(N ∩ N 0 )
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 Int M is an n-manifold and ∂M is an (n − 1)-manifold (without boundary).
Proof. If p ∈ Int M then for some homeomorphism ϕ : N → U where p ∈ N and U is
an open subset of Hn we have ϕ(p) ∈ Int Hn . There exists an open ball B ⊆ Int Hn ∩ U
around ϕ(p), so ϕ|ϕ−1 (B) is a homeomorphism from the neighborhood ϕ−1 (B) of p to
the open subset B of Rn . This shows that Int M is an n-manifold. Similarly, if p ∈ ∂M
then 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 discards
the 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 neighborhood
of 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 have
Uα = A ∩ Vα for set Vα open in X. Then
[
[
[
Uα =
A ∩ Vα = A ∩
Vα
α∈A
α∈A
α∈A
S
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 subset. Then the subspace topology on A is the same as the metric topology obtained by
restricting the metric of M to points in A.
(X)
Proof. Denote the open ball in a metric space X of radius r around a point x by Br (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
(X)
(X)
(A)
some open ball Br (p) contained in V , and by definition A ∩ Br (p) = Brn (p) is an
o
(A)
open ball around p contained in A. This shows that the set of all open balls Br (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 subsets
of 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). For
part (2), we have B = A ∩ U for some U open in X, so B is open in X since A is open
in 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 .
S
Proof. It is clear that B = X1 ×· · ·×Xn . Let U = U1 ×· · ·×Un and U 0 = U10 ×· · ·×Un0
be elements of B. Since
U ∩ U 0 = (U1 ∩ U10 ) × · · · × (Un ∩ Un0 )
and each Ui ∩ Ui0 is open in Xi , we have U ∩ U 0 ∈ B. This proves that B is a basis. Theorem 46. [Exercise 3.26] The product topology on Rn = R × · · · × R is the same
as 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 product
topology on Rn . Note first that every open ball is open in the product topology. If
U ⊆ Rn is open in the product topology and p = (p1 , . . . , pn ) ∈ U , then there exist sets
U1 , . . . , Un open in R such that p ∈ U1 × · · · × Un and U1 × · · · × Un ⊆ U . For each
i, choose an open ball (p − ri , p + ri ) contained in Ui . Let r = min {r1 , . . . , rn }; then
Br (p) is an open ball around p that is contained in U . This proves that the set of all
open 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 every
Ui is open in Xi .
(2) The product topology is “associative” in the sense that the three product topologies 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
by
fi (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 subspace
topology 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 Theorem
3.27. To show that πi is an open map, let U1 × · · · × Un be an element of the standard
basis 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 fi−1 (U1 × · · · × Un ) is open
whenever each Ui is open in Xi . If xj ∈ Uj for all j 6= i then fi−1 (U1 × · · · × Un ) = Ui ,
which is open. Otherwise, fi−1 (U1 × · · · × Un ) = ∅, which is also open. For (4), it
is clear that every set in B is open. Let U be an open set in X1 × · · · × Xn and let
x = (x1 , . . . , xn ) ∈ U . There exists some U1 × · · · × Un ⊆ U such that x ∈ U1 × · · · × Un
and 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 , which
proves 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 Vi
open 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 in
the subspace topology and let x ∈ U . We have U = (A1 × · · · × An ) ∩ V for some V
open in X1 × · · · × Xn . There exists some V1 × · · · × Vn ⊆ V such that x ∈ V1 × · · · × Vn
and 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 subspace
topology.
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 such
that U1 ∩ U2 = ∅. Then the sets X1 × · · · × Xi−1 × Uj × Xi+1 × · · · × Xn for j = 1, 2 are
disjoint 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 each
i.
(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 πi−1 (Ai ) =
X1 × · · · × Ai × · · · × Xn is closed in X1 × · · · × Xn and
n
\
A1 × · · · × An =
πi−1 (Ai )
i=1
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 U
be a neighborhood of x, and let U1 × · · · × Un ⊆ U be a neighborhood of x with each
Ui open in Xi . Since Ui is a neighborhood of xi ∈ Ai , it contains a point x0i 6= xi in
Ai . Thus x0 = (x01 , . . . , x0n ) 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 nonempty
topological spaces {Xα }α∈A . Declare a subset of the disjoint union X = α∈A Xα 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 \ E
is 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 is
nonempty 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 are
neighborhoods of x1 and x2 respectively with Xα ∩ Xβ = ∅. Therefore X is Hausdorff.
Next, we prove that S
X is second countable. For each Xα , let Bα be a countable base
for 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 easy
to see that X is locally Euclidean of dimension n since each Xα is locally Euclidean of
dimension n. Therefore X is an n-manifold.
Theorem 50. [Exercise 3.46] Let (X, T ) be a topological space, Y be any set, and
π : 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 collection
of sets in TY then
!
[
[
Uα =
π −1 (Uα ) ∈ T
π −1
α∈A
−1
since each π (Uα ) is in T , so
S
α∈A
α∈A
Uα ∈ TY . Similarly, if U1 , . . . , Un ∈ TY then
π −1 (U1 ∩ · · · ∩ Un ) = π −1 (U1 ) ∩ · · · ∩ π −1 (Un ) ∈ T
since 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 quotient
map if and only if it takes saturated open sets to open sets, or saturated closed sets to
closed sets.
Proof. Suppose that π takes saturated open sets to open sets. If U is open in Y then
π −1 (U ) is open in X since π is continuous. Also, if U ⊆ Y and π −1 (U ) is open in X
then U = π(π −1 (U )) is open since π −1 (U ) is saturated. Conversely, if π is a quotient
map and U = π −1 (V ) is a saturated open set then π(U ) = π(π −1 (V )) = V , which is
open. Finally, π takes saturated open sets to open sets if and only if π takes saturated
closed sets to closed sets, since π −1 (U ) = X \ π −1 (Y \ U ) for any U open in Y .
Theorem 52. [Exercise 3.63]
(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
→ Yα }α∈A is an indexed family of quotient maps, then the map q :
` {qα : Xα`
α Xα →
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. The
`
case for the restriction of q to a closed set is similar. For (5), let U ⊆ α Yα . We have
a
U ⊆
Yα ⇔ U ∩ Yα ⊆ Yα for every α ∈ A
open
α
open
17
⇔ qα−1 (U ) ⊆ Xα for every α ∈ A
open
−1
⇔ q (U ) ∩ Xα ⊆ Xα for every α ∈ A
open
a
⇔ q −1 (U ) ⊆
Xα .
open
α
Theorem 53. [Exercise 3.85] Any subgroup of a topological group is a topological group
with the subspace topology. Any finite product of topological groups is a topological group
with 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, and
let 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 is
also a topological group with the subspace topology. Now let G1 , . . . , Gn be topological
groups with operations µi : Gi × Gi → Gi and ιi : Gi → Gi for i = 1, . . . , n. We want
to show that the maps
µ : (G1 × · · · × Gn ) × (G1 × · · · × Gn ) → G1 × · · · × Gn
((g1 , . . . , gn ), (g10 , . . . , gn0 )) 7→ (g1 g10 , . . . , gn gn0 )
and
ι : G1 × · · · × Gn → G1 × · · · × Gn
(g1 , . . . , gn ) 7→ (g1−1 , . . . , gn−1 )
are continuous. But
ϕ : (G1 × · · · × Gn ) × (G1 × · · · × Gn ) → (G1 × G1 ) × · · · × (Gn × Gn )
((g1 , . . . , gn ), (g10 , . . . , gn0 )) 7→ ((g1 , g10 ), . . . , (gn , gn0 ))
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 sets
A0 = {0}, Ai = [1/(i + 1), 1/i] for i = 1, 2, . . . , show that the gluing lemma (Lemma
3.23) 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 all
i ≥ 1 and x ∈ Ai . By the gluing lemma, there is a unique continuous map f : X → R
such that f (0) = 1 and f (x) = 0 for all x ∈ (0, 1]. However, this map is clearly not
continuous at 0.
18
Theorem 55. [Problem 3-4] Any closed ball in Rn is an n-dimensional manifold with
boundary.
Proof. Let Bn = B1 (0) be the closed unit ball in Rn , let N = (0, . . . , 0, 1) be the “north
pole”, and let T = {0}n−1 × [0, 1] be the vertical line connecting 0 and N . Define
σ : Bn \ T → Hn given by
x1
xn−1
−1
σ(x1 , . . . , xn ) =
,...,
, kxk − 1 ,
kxk − xn
kxk − xn
with its inverse given by
1
(2u1 , . . . , 2un−1 , ke
uk2 − 1)
σ −1 (u1 , . . . , un ) =
2
(un + 1)(ke
uk + 1)
where u
e = (u1 , . . . , un−1 ). Since both maps are continuous, this proves that Bn \ T is
homeomorphic to Hn . Furthermore, we can repeat the same argument by taking the
vertical line T = {0}n−1 × [−1, 0] instead, giving us Euclidean neighborhoods of every
point in Bn except for 0. But the identity map on the open unit ball Bn is a suitable
coordinate 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 product
of 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 ⊆ U
such that x ∈ U1 × · · · × Un and each Ui is open in Xi . For every i, fi (Ui ) is open since
fi 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 discrete
space, 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 is
the subset 4 = {(x, x) : x ∈ X} ⊆ X × X. Then X is Hausdorff if and only if 4 is
closed in X × X.
Proof. Suppose X is Hausdorff and let (x1 , x2 ) ∈ (X × X) \ 4. Since x1 6= x2 , there
exist 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 open
and 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 discrete
topology.
(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 =
S
α∈A (aα , bα ), which is open in X. Therefore
[
U ×V =
{u} × V
u∈U
is open in X. Also, every element {(c, y) : a < y < b} in the basis of X is clearly open
in M . Part (1) then follows from Lemma 22. From Theorem 47 we know that M
is locally Euclidean (of dimension 2) and Hausdorff. To show that M is not second
countable, suppose B is a countable basis for M . Consider the collection C of sets of
the 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 the
uncountability of R.
Theorem 59. [Problem 3-10]
`
(1) (Characteristic Property of Disjoint Union Topologies) Let X = α∈A Xα be
a disjoint union space. For any topological space B, a map f : X → B is
continuous if and only if each fα = f ◦ iα is continuous, where iα : Xα → X is
the canonical injection:
X
f
iα
Xα
fα
B
(2) (Uniqueness of the Disjoint Union Topology) Let`{Xα }α∈A be a collection of
topological spaces. The disjoint union topology on α∈A Xα is the unique topology that satisfies the characteristic property.
Proof. We have
f is continuous ⇔ f −1 (U ) ⊆ X for all U ⊆ B
open
⇔f
−1
open
(U ) ∩ Xα ⊆ Xα for all α ∈ A and U ⊆ B
open
open
20
⇔ fα−1 (U ) ⊆ Xα for all α ∈ A and U ⊆ B
open
open
⇔ fα is continuous for all α ∈ A,
`
which proves (1). For (2), suppose that Xg = α∈A Xα has some other topology that
satisfies the characteristic property, and write Xd for the usual disjoint union space on
`
α∈A Xα . By applying the characteristic property to the identity map, we see that
every 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 is
a 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 · · · q Xn ) × 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 in
Xi × 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 in
Y . But V × W is also open in (X1 q · · · q Xn ) × Y , which shows that U is open in
(X1 q· · ·qXn )×Y . Conversely, let U be open in (X1 q· · ·qXn )×Y and let (x, y) ∈ U .
We have (x, y) ∈ V × W ⊆ U for some V open in X1 q · · · q Xn and some W open in
Y . Since (V × W ) ∩ (Xi × Y ) = (V ∩ Xi ) × W is open in Xi × Y for every i, V × W ⊆ U
is 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 and
open maps. See Theorem 47.
Example 62. [Problem 3-11] Proposition 3.62(d) showed that the restriction of a quotient 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 an
open 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 a
quotient 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, let
Vk = {[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−1
k (p1 , . . . , pbk , . . . , pn+1 )
where pk is omitted. This map is well-defined since [p1 , . . . , pn+1 ] = [q1 , . . . , qn+1 ] implies
that pi = λqi for every i. Furthermore, π is a homeomorphism. By choosing at least
two 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 complex
subspaces of Cn+1 , called n-dimensional complex projective space. Topologize
CPn as the quotient (Cn+1 \{0})/C∗ , where C∗ is the group of nonzero complex numbers
acting 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 . Define
an equivalence relation on X by declaring (x, 0) ∼ (x, 1) if x 6= 0. Then the quotient
space X/ ∼ is locally Euclidean and second countable, but not Hausdorff. (This space
is 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 neighborhood N of p homeomorphic to some open interval in R. If p = [(0, 0)] then let
N = {[(x, 0)] : −1 < x < 1} and define ϕ : N → (−1, 1) by [(x, 0)] 7→ x. It is easy to
check that this map is a well-defined homeomorphism. So X/ ∼ is locally Euclidean and
therefore second countable by Proposition 3.56. But X/ ∼ is not Hausdorff, since any
two 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 implies
that 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 G
is a topological group if and only if the map θ : G × G → G given by (x, y) 7→ xy −1 is
continuous.
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 and
defines a homeomorphism of G/Γ with itself.
(2) A topological space X is said to be homogeneous if for any x, y ∈ X, there is
a 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/Γ
satisfies (π ◦ Lg )(g 0 ) = gg 0 Γ and is constant on the fibers of π, there exists a unique
eg : G/Γ → G/Γ satisfying L
eg ◦ π = π ◦ Lg . Furthermore, L
eg−1 is a
continuous map L
eg since
continuous inverse of L
eg−1 ◦ L
eg ◦ π = L
eg−1 ◦ π ◦ Lg = π ◦ Lg−1 ◦ Lg = π
L
eg ◦ L
eg−1 ◦ π = π. This shows that L
eg is a homeomorphism of G/Γ with
and similarly L
eg0 g−1 is a homeomorphism that
itself. For (2), if gΓ and g 0 Γ are cosets in G/Γ then L
0
takes gΓ to g Γ.
Theorem 69. [Problem 3-22] Let G be a topological group acting continuously on a
topological 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, that
π −1 (π(U )) = {g · x : g ∈ G, x ∈ U } is open. Since the group action α : G × X → X is
continuous, 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 in
X/G, and apply Theorem 57.
Theorem 70. [Problem 3-23] If Γ is a normal subgroup of the topological group G then
the 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, π
is also a quotient map in the topology sense. Let µ : G × G → G and ι : G → G be the
product and inverse maps respectively. By Theorem 56, π × π : G × G → G/Γ × G/Γ is
a continuous, surjective and open map, and therefore a quotient map. Since Γ is normal
in G, π ◦ µ is constant on the fibers of π × π and there exists a unique continuous map
µ
e : G/Γ × G/Γ → G/Γ satisfying µ
e ◦ (π × π) = π ◦ µ. Similarly, there exists a unique
23
continuous map e
ι : G/Γ → G/Γ satisfying e
ι ◦ π = π ◦ ι. The group axioms are then
easily checked for µ
e and e
ι.
Chapter 4. Connectedness and Compactness
Theorem 71. In a topological space X, the path connectivity relation ∼p is an equivalence relation.
Proof. For any point p ∈ X, the path f : [0, 1] → X with f (t) = p is a path from p to
itself. 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] → X
is a path from q to r, then by the gluing lemma the map h : [0, 1] → X given by
(
f (2t)
if 0 ≤ t ≤ 21 ,
h(t) =
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 component
is 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 Proposition 4.21. For (2), suppose B and C are path components both containing points of
A. 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 countable collection of nonempty T
closed subsets of X that are nested, which means that
Fn ⊇ Fn+1 for each n. Then n Fn is nonempty.
T
S
Proof. Suppose that n Fn is empty. Then X = n X \ Fn , so there is a finite subcover
S
{X \ Fn1 , . . . , X \ Fnk } where we take n1 < · · · < nk . But ki=1 X \ 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 subsequence
that converges to a point x, and it is well-known that a Cauchy sequence converges if
it has a convergent subsequence. Therefore xn → x.
24
Theorem 76. [Exercise 4.67] Any finite product of locally compact spaces is locally
compact.
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 x
and V of y. Then U × V is a precompact neighborhood of (x, y), since U × V = U × V
is compact.
Theorem 77. [Exercise 4.73] Suppose A is an open cover of X such that each element
of A intersects only finitely many others. Then A is locally finite. This need not be
true when the elements of A are not open.
Proof. Let x ∈ X and choose some A ∈ A such that x ∈ A. There are finitely
many elements of A that intersect A, so A is the required neighborhood of x. For a
counterexample when the elements of A are not required to be open, take X = R and
A = {{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. Then
A, B are compact, so by Lemma 4.34 there exist disjoint open subsets U, V ⊆ X such
that 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 such
that A ⊆ U and B ⊆ V . So U ∩ E and V ∩ E are disjoint open subsets of E such that
A ⊆ U ∩ E and B ⊆ V ∩ E.
Theorem 79. [Exercise 4.87] Every compact manifold with boundary is homeomorphic
to a subset of some Euclidean space.
Proof. If M is a compact manifold with boundary then the double D(M ) of M is a compact manifold (without boundary), and is homeomorphic to a subset of some Euclidean
space. The restriction of this homeomorphism to M is the desired homeomorphism. Theorem 80. [Problem 4-1]
(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 a
separation 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-manifold
and an n-manifold for any n > 1.
Proof. Let M be a nonempty topological space that is both a 1-manifold and an nmanifold for some n > 1. Choose some p ∈ M and let ϕ1 : U1 → V1 and ϕ2 : U2 → V2
be homeomorphisms where U1 and U2 are neighborhoods of p, V1 is open in R, and
V2 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−1
2 )(B) \ {ϕ1 (p)}, but W2 is
disconnected 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 coordinate
charts (U, ϕ) and (V, ψ) such that U, V are neighborhoods of p, ϕ(U ) is open in Int H1 ,
ψ(V ) is open in H1 , ϕ(p) > 0 and ψ(p) = 0. Let W = U ∩ V ; then ϕ(W ) is homeomorphic to ψ(W ). But this is impossible, for ϕ(W ) \ {ϕ(p)} is disconnected while
ψ(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) : z 2 = x2 + y 2
Proof. Let M be the union of the x-axis and the y-axis, and suppose that M is a manifold. By Theorem 20, M cannot be a 0-manifold. Now suppose M is a 1-manifold
and 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 removing (0, 0) from B produces four connected components (the left, top, right and bottom
parts of the cross shape). Therefore M must be a n-manifold for n > 1. But this
implies that the positive x-axis is both a 1-manifold and an n-manifold, contradicting
Theorem 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 the
same property.
Proof. The result for the first two properties follow from Theorem 30. Suppose that X
is locally compact and let B be a basis of precompact open sets in X. Let y ∈ Y so
that y = f (x) for some x ∈ X. Let U be a neighborhood of x such that f (U ) is open
and 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 Y
is locally compact.
Theorem 85. [Problem 4-9] Any n-manifold is a disjoint union of countably many
connected n-manifolds.
Proof. Any n-manifold M is the disjoint union of its connected components, which are
open by Proposition 4.25 and therefore are n-manifolds by Proposition 2.53. Furthermore, 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 shows
that X is connected. Suppose that X is path-connected and let γ : [0, 1] → X be
path connecting the points (0, 0) and (2/π, 1). Choose a δ > 0 such that kγ(t)k < 1/2
whenever 0 ≤ t < δ. This is impossible, for (0, 1) is a limit point of B. Also, X is
not locally connected since any neighborhood of (0, 0) is disconnected. The two path
components 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 hU i generated by U is open and closed
in G.
(3) For any connected subset U ⊆ G containing 1, hU i 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
S (1), let H be an
open subgroup of G. Then every coset gH is open in G, and G\H = g∈G\H gH is open.
For (2), let x ∈ hU i.SSince µ(x, ·) is a homeomorphism, µ(x, U ) ⊆ hU i is a neighborhood
of x. Then hU i = x∈hU i µ(x, U ) is open. For (3), let x ∈ hU i. Then µ(x, U ) ⊆ hU i
is a connected set containing x, and µ(µ(x,S
U ), U ) ⊆ hU i is a connected set containing
both x and 1. By Proposition 4.9, hU i = x∈hU i µ(µ(x, U ), U ) is connected. For (4),
if U is a neighborhood of 1 then hU i is both open and closed in G by (2). Therefore
hU i = G.
Theorem 88. Every σ-compact space is Lindelöf.
27
Proof. Let X be a σ-compact space and let A be a countable collection of compact
subsets that cover X. Let U be an open cover of X. Each A ∈ A is covered by finitely
many sets from U, so there is a countable subcover.
Theorem 89. [Problem 4-16] A locally Euclidean Hausdorff space is a topological manifold if and only if it is σ-compact.
Proof. Let M be a locally Euclidean Hausdorff space. If M is a topological manifold
then Proposition 4.60 shows that
S there is a countable collection B of regular coordinate
balls that cover M , i.e. M = B∈B B. Conversely, suppose that M is σ-compact. Then
M is Lindelöf by Theorem 88, so the argument of Proposition 4.60 shows that M is
second countable.
Theorem 90. [Problem 4-17] Suppose M is a manifold of dimension n ≥ 1, and
B ⊆ 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 0 of B and a homeomorphism ϕ : B 0 → Br0 (0) that takes
B to Br (0) for some r0 > r > 0. In particular, B 0 \ B is homeomorphic to Br0 (0) \ Br (0).
For any x ∈ ∂B there is a homeomorphism ψ : U → V where U is a neighborhood of
ϕ(x) in Br0 (0) \ Br (0), V is open in Hn , and ψ(ϕ(x)) = 0. Then ψ ◦ ϕ is the required
coordinate map from a neighborhood of x ∈ ∂B in B 0 \ 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 Mi0 = Mi \ Bi . Choose a homeomorphism f : ∂M20 →
∂M10 (such a homeomorphism exists by Problem 4-17). Let M1 #M2 (called a connected
sum of M1 and M2 ) be the adjunction space M10 ∪f M20 .
(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 then
M10 and M20 are still connected. Let e : M10 → M10 ∪f M20 and f : M20 → M10 ∪f M20 be
the canonical embeddings. By Theorem 3.79 we have e(M10 ) ∩ f (M20 ) 6= ∅, so M1 #M2
is connected. If M1 and M2 are compact then M10 and M20 are closed and therefore
compact. This implies that M1 #M2 = e(M10 ) ∪ f (M20 ) 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 that
Ui ≈ 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 let
Ci ⊇ Bi be the larger coordinate balls around pi . Let ji : Mi \ Bi → M1 #M2 be the
injections. 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 if
and only if n ∈ A implies −n ∈ A. Then Z with this topology is second countable and
limit point compact but not compact.
Proof. Let Bi = {−i, i}. Then B = {B0 , B1 , . . . } is a countable basis for Z. Now let U
be a subset of Z with at least two elements, and choose some nonzero n ∈ U . Then −n
is a limit point of U , since any neighborhood of −n must contain n. In particular, Z
is limit point compact. However, the open cover B of Z has no finite subcover, so Z is
not compact.
Theorem 94. [Problem 4-21] Let V be a finite-dimensional real vector space. A norm
on 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 the
same 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 two
norms on V , and write Vi for V equipped with the metric induced by |·|i . Consider the
unit sphere S1 = {v ∈ V : |v|1 = 1}, which is compact. The image of S1 under |·|2 is
therefore 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 open
ball 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
|v − x|2 = |v − x|1 |v − x|
< r
|x|2
1 2
v−x
|v − x|
≤r
1 2
since (v − x)/ |v − x|1 ∈ S1 . This shows that v ∈ B2 . A similar argument shows that
every open ball in V1 around a point x ∈ V contains an open ball in V2 around x. By
Theorem 23, V1 and V2 have the same topology.
Theorem 95. [Problem 4-23] Let X be a locally compact Hausdorff space. The onepoint 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)
(2)
(3)
(4)
(5)
T is a topology.
X ∗ is a compact Hausdorff space.
X is dense in X ∗ if and only if X ∗ is noncompact.
X is open and and has the subspace topology.
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 ∗ \SUα is a compact subset of X (and therefore ∞ ∈ Uα ). If A2 is empty then
S
α∈A1 Uα is open in X since each Uα is open in X. If A2 is not empty then
α∈A Uα =
!
!
[
\
\
X∗ \
Uα =
X ∗ \ Uα ∩
X ∗ \ Uα
α∈A
α∈A1
α∈A2
!
\
=
X \ Uα
!
∩
α∈A1
\
X ∗ \ Uα
α∈A2
is a compact subset of X since S
each X \ Uα is closed for α ∈ A1 and each X ∗ \ Uα is
compact 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∗ \
Uα =
X ∗ \ Uα
α∈A
is compact since A is finite, so
\
α∈A
Uα =
α∈A
T
Uα ∈ T . If A1 is not empty then
!
!
\
\
Uα ∩
X ∗ \ (X ∗ \ Uα )
α∈A
α∈A1
α∈A2
30
!
=
\
α∈A1
Uα
!
∩
\
X \ (X ∗ \ Uα )
α∈A2
is open since for each α ∈ A2 the set X ∗ \ Uα is compact and therefore closed since X
is Hausdorff. This proves that T is a topology.
We now prove that X ∗ is compact Hausdorff. Since X is already Hausdorff, it suffices
to check that ∞ and any x ∈ X can be separated by neighborhoods. Let E be a
precompact 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 of
X ∗ . Choose some U ∈ U containing ∞ so that X ∗ \ U is a compact subset of X. Then
there exists some finite subcover {U1 , . . . , Uk } of X ∗ \ U , and {U1 , . . . , Uk , U } is a finite
subcover of X ∗ .
Suppose that X is noncompact and let U be a nonempty open set in X ∗ . If U contains
no points from X then U = {∞}. Then X ∗ \ U = X is compact, which contradicts our
assumption that X is noncompact. This proves that X is dense in X ∗ . Conversely, if
X 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 let
U be a neighborhood of ∞. Then X ∗ \ U is compact and contains finitely many values
of qn , so U contains infinitely many values of qn . Therefore qn → ∞. Conversely, if
qn → ∞ and K ⊆ X is compact then X ∗ \K is a neighborhood of ∞, so X ∗ \K contains
all 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, as
defined in Example 3.6. Then σ extends to a homeomorphism of Sn with the one-point
compactification 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 a
continuous map taking ∞ to ∞. Therefore σ ∗ is a homeomorphism. From Theorem 97
we 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, and
let p ∈ M . Then M is homeomorphic to the one-point compactification of M \ {p}.
c = (M \ {p})∗ and define ϕ : M → M
c by
Proof. Let M
(
x if x =
6 p,
ϕ(x) =
∞ if x = p.
31
c. If ∞ ∈
c \ U is compact,
Let U be open in M
/ U then ϕ−1 (U ) is open. If ∞ ∈ U then M
so
c \ U ) = M \ (M
c \ U)
ϕ−1 (U ) = M \ ϕ−1 (M
c \ U is closed. Therefore ϕ is continuous. By Lemma 4.50, ϕ is a
is open since M
homeomorphism.
Theorem 98. [Problem 4-27] If X and Y are noncompact, locally compact Hausdorff
spaces, 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 ∗ (∞) = ∞. Let
U be open in Y ∗ . If ∞ ∈
/ U then (f ∗ )−1 (U ) is open since f is continuous. If ∞ ∈ U
then 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 continuous
map 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 locally
finite closed cover of X. If for each α ∈ A we are given a continuous map fα : Xα → Y
such that fα |Xα ∩Xβ = fβ |Xα ∩Xβ for all α and β, then there exists a unique continuous
map 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. Let
x ∈ X and choose a neighborhood U of x that intersects with finitely many elements
Aα1 , . . . , Aαk ∈ {Aα }. Let E be a closed subset of f (U ) so that E = F ∩ f (U ) for some
F closed in Y . Then
(f |U )−1 (E) = f −1 (E) ∩ U
= f −1 (F ) ∩ U
=
k
[
fα−1
(F ) ∩ U
i
i=1
fα−1
(F )
i
which is closed in U since each
is closed. This shows that every point of X has a
neighborhood 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 that
for every open cover of X, there exists a partition of unity subordinate to it. Then X
is paracompact.
Proof. Let U = {Uα }α∈A be an open cover X. By our hypothesis, there exists a partition
of unity {ψα }α∈A subordinate to U. For each α, let Vα = ψα−1 ((0, 1]). We want to show
that {Vα } is a locally finite open refinement of U. Each Vα is open in X since (0, 1] is
always open in ψα (X). Furthermore, Vα ⊆ supp ψα ⊆ 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 of
sets in {supp ψα }. In particular, N intersects with a finite number of sets in {Vα } since
Vα ⊆ supp ψα for every α.
Chapter 5. Cell Complexes
Theorem 101. [Exericse 5.3] Suppose X is a topological space whose topology is coherent with a family B of subspaces.
(1) If Y is another topological space, then a map f : X → Y is continuous if and
only 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 quotient
map.
Proof. If f is continuous then it is clear that every f |B is continuous. Suppose that
every 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 an
adjunction 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 and
form the adjunction space (X \ e0 ) ∪φ D0 . Define ψ : (X \ e0 ) q D0 → X as being equal
to inclusion on X \ e0 and to φ on D0 . Then ψ makes the same identifications as the
quotient map defining the adjunction space, so it remains to show that ψ is a quotient
map. The argument is identical to that of Proposition 5.18.
Theorem 103. [Exercise 5.34] If K is a Euclidean simplicial complex, then the collecb consisting of the interiors of the simplices of K is a regular CW decomposition
tion K
of |K|.
33
b are disjoint, for the intersection of two simplices is
Proof. First note that the sets in K
either empty or a face of each, and if a point is in the interior of a simplex σ, it cannot
b is a partition of |K|. Proposition 5.32 shows
be in a face of σ. This shows that K
that for each σ ∈ K we have a homeomorphism φ : 4k → σ where 4k is the standard
k-simplex. Since φ restricts to a homeomorphism from Int 4k to the interior of σ and
φ(∂4k ) maps into the boundary of σ, we can take φ as a characteristic map for the
b must also be locally finite. By
interior of σ. Furthermore, since K is locally finite, K
b
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 of
a simplex of K, {f0 (v0 ), . . . , f0 (vk )} are the vertices of a simplex of L (possibly with
repetitions). Then there is a unique simplicial map f : |K| → |L| whose vertex map is
f0 . It is a simplicial isomorphism if and only if f0 is a bijection satisfying the following
additional 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 subsets
K0 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 there
exists a unique map fα : σ → Rm that is the restriction of an affine map, takes vi to
f0 (vi ) for each i, and takes σ onto a simplex of L. Since the maps fα agree on the
intersection of their domains and K is locally finite, Theorem 99 shows that there is
a unique (simplicial) map f : |K| → |L|. If in addition f0 is a bijection satisfying the
given condition then a similar process gives an inverse to f .
Theorem 105. [Problem 5-1] Suppose D and D0 are closed cells (not necessarily of the
same dimension).
(1) Every continuous map f : ∂D → ∂D0 extends to a continuous map f : D → D0 ,
with F (Int D) ⊆ Int D0 .
(2) Given points p ∈ Int D and p0 ∈ Int D0 , F can be chosen to take p to p0 .
(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 → D0 such that ϕ(Sn−1 ) =
∂D and ψ(Sm−1 ) = ∂D0 . Define fe : Sn−1 → Sm−1 by fe = ψ −1 ◦ f ◦ ϕ. Define
Fe : Bn → Bm by Fe(0) = 0 and
x
e
e
(*)
F (x) = kxk f
kxk
for x 6= 0. Then Fe is a continuous extension of fe, and F = ψ ◦ Fe ◦ ϕ−1 is a continuous
extension of f . If f is a homeomorphism then Fe is easily seen to be a bijection, and by
34
Lemma 4.25, Fe is a homeomorphism. For part (2), Proposition 5.1 shows that there
exist homeomorphisms G : Bn → Bn and G0 : Bm → Bm such that G(0) = ϕ−1 (p) and
G0 (0) = ψ −1 (p0 ). Modify the previous definitions by setting fe = (G0 )−1 ◦ ψ −1 ◦ f ◦ ϕ ◦ G
and F = ψ ◦ G0 ◦ Fe ◦ G−1 ◦ ϕ−1 . Then
F (p) = (ψ ◦ G0 ◦ Fe ◦ G−1 )(ϕ−1 (p))
= (ψ ◦ G0 ◦ Fe)(0)
= (ψ ◦ G0 )(0)
= p0
as desired.
Theorem 106. [Problem 5-2] Suppose D is a closed n-cell, n ≥ 1.
(1) Given any point p ∈ Int D, there is a continuous function F : D → [0, 1] such
that 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 Int D.
Proof. For (1), apply Theorem 105 to the map f : ∂D → [−1, 1] satisfying f (x) = 1 for
all x ∈ D, obtaining a map F : D → [−1, 1] with F (p) = 0. From (*) it is easy to see
that F −1 ({1}) = ∂D and F −1 ({0}) = {p}. For (2), repeat the proof of Theorem 105
but set Fe(0) = 1 and
x
e
e
F (x) = 1 − kxk f
kxk
for x 6= 0.
Theorem 107. Let X be a connected topological space and let ∼ be an equivalence
relation 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 neighborhood
U of q such that q1 ∼ q2 for every q1 , q2 ∈ U . In particular, for every r ∈ U we have
p ∼ q and q ∼ r which implies that p ∼ r, and U ⊆ S. This shows that S is open. If
q ∈ X \ S then there is a neighborhood U of q such that q1 ∼ q2 for every q1 , q2 ∈ U . If
p ∼ 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. Theorem 108. [Problem 5-3]
(1) Given any two points p, q ∈ Bn , there is a homeomorphism ϕ : Bn → Bn such
that ϕ(p) = q and ϕ|∂Bn = Id∂Bn .
35
(2) For any topological manifold X, every point of X has a neighborhood U with
the property that for any p, q ∈ U , there is a homeomorphism from X to itself
taking 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 and
choose a regular coordinate ball B around x. If p, q ∈ B then it follows from part (1)
that there
existsa 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) follows
by applying Theorem 107 to the equivalence relation defined by p ∼ q if and only if
there 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, then
there 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 that
the 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. For
each 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 dim M > 1
and (p1 , . . . , pk ) and (q1 , . . . , qk ) are two ordered k-tuples of distinct points in M , then
there 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 regular
coordinate balls E1 , . . . , Ek around p1 , . . . , pk and regular coordinate
balls F1 , . . . , Fk
S
around q1 , . . . , qk . For each 1 ≤ j ≤ k the manifold Mj = M \ i6=j Ei ∪ Fi is connected,
so Theorem 108 shows that there exists a homeomorphism ϕj : Mj → Mj that takes
pj to qj . By the gluing lemma, we can extend ϕj to a homeomorphism ψj : M → M
such that ψj |Ei ∪Fi is the identity for every i 6= j. Then ψ = ψ1 ◦ · · · ◦ ψk is the desired
homeomorphism taking pj to qj for j = 1, . . . , k.
Theorem 111. [Problem 5-5] Suppose X is a topological space and {Xα } is a family
of 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 ,→ X
are continuous.
Proof. Suppose that the topology T of X is coherent with {Xα } and let T 0 be some
topology for which the inclusion maps iα : Xα ,→ X are continuous. If U ∈ T 0 then
36
every i−1
α (U ) = U ∩ Xα is open, so U is open in T since T is coherent with {Xα }. This
shows that T 0 ⊆ T . Conversely, suppose that T is the finest topology for which the
inclusion maps iα are continuous. Let U be a subset of X such that U ∩ Xα ∈ T for all
α. If U is not open in T then by defining a new topology T 0 ⊇ T such that U ∈ T 0 ,
we obtain a finer topology such that every iα is continuous. This is a contradiction, so
U must be open in T .
Theorem 112. [Problem 5-6] Suppose X is a topological space. The topology of X is
coherent 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 ∩ Uα
is open for every α ∈ A. Then
[
[
U =U ∩X =U ∩
Uα =
U ∩ Uα
α∈A
α∈A
is open, which proves (1). Now let {Eα }α∈A be a locally finite closed cover of X and
let E be a subset of X such that E ∩ Eα is closed for every α ∈ A. Let x be a limit
point of E; we want to show that x ∈ E. Since {Eα } is locally finite, there exists a
neighborhood 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 some
y ∈ U ∩ V not equal to x suchSthat y ∈ E. But then y ∈ Eαi for some i, which shows
that x is a limit point of E ∩ ni=1 Eαi . Since
n
n
[
[
E ∩ Eαi
E∩
Eαi =
i=1
is closed, x ∈ E ∩
Sn
i=1
i=1
Eαi ⊆ E. This proves (2).
Theorem 113. [Problem 5-7] Suppose X is a topological space whose topology is coherent with a collection {Xα }α∈A of subspaces of X, and for each α ∈ A we are given
a continuous map fα : Xα → Y such that fα |Xα ∩Xβ = fβ |Xα ∩Xβ for all α and β. Then
there 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). This
makes f a well-defined map since fα |Xα ∩Xβ = fβ |Xα ∩Xβ for all α and β. Let U be open
in Y . Then for every α ∈ A we have that fα−1 (U ) = f −1 (U ) ∩ Xα is open in Xα , so
f −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 coherent
with 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 closed
for 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 is
locally 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 that
intersects with finitely many elements e1 , . . . , en . Then U is a precompact neighborhood
of x, since U is a closed subset of the compact set e1 ∪ · · · ∪ en . Conversely, suppose
that 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 many
cells of X.
Theorem 117. [Problem 5-12] Let Pn be n-dimensional projective space. The usual
inclusion Rk+1 ⊆ Rn+1 for k < n allows us to consider Pk as a subspace of Pn . Then
Pn has a CW decomposition with one cell in each dimension 0, . . . , n such that the
k-skeleton is Pk for 0 < k < n.
Proof. We use induction on n. The result is clearly true for n = 0, so assume that the
result is true for Pn−1 and consider Pn . Define the map
F : Bn → Rn+1 \ {0}
q
2
2
(x1 , . . . , xn ) 7→ x1 , . . . , xn , 1 − |x1 | − · · · − |xn |
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−1
and 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 upper
hemisphere 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 , and
that 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, . . . , 2n
such that the 2k-skeleton is CPk for 0 < k < n.
38
Proof. Proceed as in Theorem 117, replacing Bn with the complex unit ball
q
2
2
n
Bn = (z1 , . . . , zn ) ∈ C : |z1 | + · · · + |zn | = 1
which has dimension 2n.
Theorem 119. [Problem 5-14] Every nonempty compact convex subset D ⊆ Rn is a
closed cell of some dimension.
Proof. Let x + S be an affine subspace of minimal dimension k containing D and let
A = [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 otherwise
D would be contained in the affine subspace spanned by A, which has dimension less
than 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 contained
in D. Therefore ` = k, and considering D as a subset of x + S, the interior of D is
nonempty. 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 presentation produces a topologically equivalent presentation.
Proof. We only prove the result for subdividing and reflecting. Let P = hS | W1 , . . . , Wk i
be a polygonal presentation and let P 0 = hS, e | W10 , . . . , Wk0 i be the presentation formed
by replacing every occurrence of a by ae and every occurrence of a−1 by e−1 a−1 ,
taking each word Wi to Wi0 . First assume P has words of length 3 or more. Let
P1 , . . . , Pk be polygonal regions for P, let P10 , . . . , Pk0 be the polygonal regions for
`
`
P, and let π : ki=1 Pi → M and π 0 : ki=1 Pi0 → M 0 be the quotient maps. Let
`
`
f : ki=1 Pi → ki=1 Pi0 be a map that takes edges labeled with a to the two corresponding edges labeled a and e (such that the preimage of a has length 1/2), and similarly
for a−1 . This map can be chosen to be a homeomorphism by Theorem 105. Since
π 0 ◦ f makes the same identifications as π, M and M 0 are homeomorphic. If P has
a single word of length 2, it is easy to check that the individual cases are homeomorphic to the sphere or projective plane (as is noted in Example 6.9). For reflection,
`
`
define f : ki=1 Pi → ki=1 Pi0 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 projective
planes contains a subspace that is homeomorphic to the Möbius band.
39
Since P2 #P2 is homeomorphic to the Klein bottle and the Klein bottle contains a copy
of the Möbius band.
Example 122. [Problem 6-2] Note that both a disk and a Möbius band are manifolds
with boundary, and both boundaries are homeomorphic to S1 . Show that it is possible
to obtain a space homeomorphic to a projective plane by attaching a disk to a Möbius
band along their boundaries.
We have the presentation ha, b, c, d, e | abcdeci for a Möbius band and the presentation
ha, b, c, d | abe−1 d−1 i for a (closed) disk with its edges identified with the boundary of
the Möbius band. Attaching the disk to the Möbius band gives the presentation
a, b, c, d, e | abe−1 d−1 , abcdec ≈ a, b, c, d, e | abe−1 d−1 , decabc
≈ ha, b, c | abcabci
≈ ha | aai ,
which is the projective plane.
Example 123. [Problem 6-3] Show that the Klein bottle is homeomorphic to a quotient
obtained by attaching two Möbius bands together along their boundaries.
This corresponds to the presentation
a, b, c, d | abcb, a−1 dc−1 d ≈ a, b, c, d | bcba, a−1 dc−1 d
≈ b, c, d | bcbdc−1 d
≈ b, c, d, e | bce, e−1 bdc−1 d
≈ b, c, d, e | ebc, c−1 de−1 bd
≈ b, d, e | ebde−1 bd
≈ e, f | ef e−1 f
≈ e, f | f ef e−1 ,
which is the Klein bottle.
Theorem 124. [Problem 6-5] Every compact 2-manifold with boundary is homeomorphic 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 a
`
homeomorphism ϕ : ∂M → ki=1 S1 . Let M 0 be the compact 2-manifold formed by
attaching half of a sphere to each ϕ−1 (S1 ); then M is homeomorphic to M 0 with the
interiors of the half-spheres removed.
Example 125. [Problem 6-6] For each of the following surface presentations, compute
the Euler characteristic and determine which of our standard surfaces it represents.
40
(1) ha, b, c | abacb−1 c−1 i
(2) ha, b, c | abca−1 b−1 c−1 i
We run the classification algorithm on each of the presentations:
a, b, c | abacb−1 c−1 ≈ a, b, c | cb−1 c−1 aba
≈ a, b, c, d | cb−1 c−1 ad, d−1 ba
≈ a, b, c, d | dcb−1 c−1 a, a−1 b−1 d
≈ b, c, d | ddcb−1 c−1 b−1
≈ b, c, d, e | ddcb−1 e, e−1 c−1 b−1
≈ b, c, d, e | eddcb−1 , bce
≈ hc, d, e | ddcceei ,
so (1) is homeomorphic to P2 #P2 #P2 with Euler characteristic −1;
a, b, c | abca−1 b−1 c−1 ≈ a, b, c | c−1 abca−1 b−1
≈ a, b, c, d | c−1 ad, d−1 bca−1 b−1
≈ a, b, c, d | dc−1 a, a−1 b−1 d−1 bc
≈ b, c, d | dc−1 b−1 d−1 bc
≈ b, c, d, e | dc−1 b−1 e, e−1 d−1 bc
≈ b, c, d, e | edc−1 b−1 , bce−1 d−1
≈ d, e | ede−1 d−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 topological
space, and A be any subset of X. Then any two continuous maps f, g : X → B that
agree on A are homotopic relative to A.
Proof. The straight-line homotopy between f and g is in fact a homotopy relative to
A.
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 stationary
on {0, 1}.
41
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 connected
if and only if every element of π1 (X) is the identity, i.e. f · g ∼ cp for all paths f, g
from p to q.
Theorem 129. [Exercise 7.15] Every convex subset of Rn is simply connected, and Rn
itself is simply connected.
Proof. This follows from Theorem 126.
Theorem 130. [Exercise 7.23] The path homotopy relation is preserved by composition
with continuous maps. That is, if f0 , f1 : I → X are path-homotopic and ϕ : X → Y is
continuous, 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 seen
to be a path homotopy from ϕ ◦ f0 to ϕ ◦ f1 .
Theorem 131. [Exercise 7.27]
(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 is
continuous. Let r1 : B → A and r2 : X → B be retractions. Then r1 ◦ r2 : X → A is
also 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 therefore
simply 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 the
class of all topological spaces.
42
Proof. It is clear that homotopy equivalence is reflexive and symmetric. Let ϕ : X → Y
and ψ : Y → Z be homotopy equivalences with homotopy inverses ϕ : Y → X and
ψ : 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-point
space {p} ⊆ X then X is simply connected, so there is a path γ : I → X from p to any
point q ∈ X. Let H : X × I → X be a deformation retraction to {p}; then
(
H(x, 2t) if 0 ≤ t ≤ 1/2,
H 0 (x, t) =
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 to
an isomorphism that respects the injections.
Proof. Let (S, (iα )) and (S 0 , (i0α )) be two coproducts for a family of objects (Xα ). By
the universal property, there exist unique morphisms f : S → S 0 and f 0 : S 0 → S such
that i0α = f ◦ iα and iα = f 0 ◦ i0α for every α. Then iα = f 0 ◦ f ◦ iα , so by uniqueness we
have f 0 ◦ f = IdS . Similarly, i0α = f ◦ f 0 ◦ i0α implies that f ◦ f 0 = IdS 0 . Therefore f is
an 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 continuous
maps. 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 · · · q Xn → Y be continuous
maps. 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. Now
suppose that Hj : Y × I → Xj are homotopies from fj to gj in (1). Then the map
H : 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 from
fj to gj in (2). There exists a unique continuous map H : (X1 × I) q · · · q (Xn × I) → Y
such that H|Xj ×I = Hj for every j. Let
ι : (X1 q · · · q Xn ) × I → (X1 × I) q · · · q (Xn × I)
be the identity map, which is continuous by Theorem 60. Then the map H ◦ ι is a
homotopy from f to g.
Theorem 137. [Problem 7-1] Suppose f, g : X → Sn are continuous maps such that
f (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)
;
k(1 − t)f (x) + tg(x)k
we must check that the denominator is never zero. If (1 − t)f (x) + tg(x) = 0 and
0 < t < 1 then
(1 − t)f (x)
−tg(x)
f (x) =
=
= −g(x)
k(1 − t)f (x)k
k−tg(x)k
since kf (x)k = kg(x)k = 1. We are given that this cannot happen, so H is a homotopy
from f to g.
Theorem 138. [Problem 7-2] Suppose X is a topological space, and g is any path in
X from p to q. Let φg : π1 (X, p) → π1 (X, q) denote the group isomorphism defined in
Theorem 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)
ψ∗
π1 (Y, ψ(p))
φg
π1 (X, q)
φψ◦g
ψ∗
π1 (Y, ψ(q)).
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 let
p, q ∈ X. Then π1 (X, p) is abelian if and only if all paths from p to q give the same
isomorphism 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 are
trivial. If g1 and g2 are paths from p to q then by 138,
φg1 = φg2 ⇔ φ−1
g2 ◦ 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 and
only if every inner automorphism of π1 (X, p) is trivial. One direction is immediate, for
φg1 ·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
(
2s(1 − t, t)
if s ∈ [0, 1/2],
F (s, t) =
(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→ g2 g1−1 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 assume
that g = 1 for part (2). Let f and g be loops based at 1 ∈ G and define F : I × I → G
by (s, t) 7→ f (s)g(t). Then f · g ∼ g · f by Theorem 140, so [f ] · [g] = [g] · [f ]. This
shows 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 0 = fn ·f1 ·· · ··fn−1 .
Let fe and fe0 be the circle representatives of f and f 0 respectively. Then fe is (freely)
homotopic to fe0 .
Proof. Let µk : I → S1 be given by s 7→ exp((k + s)2πi/n). By reparametrizing fe
and fe0 , we may assume that fk = fe ◦ µk−1 = fe0 ◦ µk for each k = 1, . . . , n. Define
H : S1 × I → X by H(z, t) = fe(e−2πit/n z) where z is taken to be a complex number.
Then H is a homotopy from fe to fe0 .
Theorem 143. [Problem 7-6] For any path-connected space X and any base point
p ∈ X, the map sending a loop to its circle representative induces a bijection between
the set of conjugacy classes of elements of π1 (X, p) and [S1 , X] (the set of free homotopy
classes of continuous maps from S1 to X).
Proof. Let C be the set of conjugacy classes of elements of π1 (X, p) and denote the
conjugacy class of an element [f ] ∈ π1 (X, p) by [[f ]]. Let ϕ : C → [S1 , X] be the map
that sends an element [[f ]] ∈ C to the free homotopy class of the circle representative
fe of f . We first check that ϕ is well-defined. Let [g] ∈ [[f ]] so that [g] = [h] · [f ] · [h] for
some [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 ge ' fe. Let α : S1 → X
be a continuous map, let ω : I → X be given by ω(s) = α(e2πis ), and let γ be a path
from p to ω(0). We have ω ' γ · γ · ω and by Lemma 142 the circle representative
of 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 fe is
homotopic to ge. Then [f ] = [g], and in particular we have [[f ]] = [[g]]. This shows that
ϕ is bijective.
Theorem 144. [Problem 7-7] Suppose (M1 , d1 ) and (M2 , d2 ) are metric spaces. A map
f : M1 → M2 is said to be uniformly continuous if for every ε > 0 there exists a
δ > 0 such that for all x, y ∈ M1 , d1 (x, y) < δ implies d2 (f (x), f (y)) < ε. If M1 is
compact, 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 cover
U = 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 that
U ∩ V = ∅. Now U ∩ r−1 (V ∩ A) is a neighborhood of x contained in X \ A, since
r(A ∩ U ∩ r−1 (V ∩ A)) ⊆ A ∩ U ∩ V ∩ A = ∅.
Theorem 146. [Problem 7-9] Suppose X and Y are connected topological spaces, and
the fundamental group of Y is abelian. If F, G : X → Y are homotopic maps such that
F (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 a
loop based at x. Consider the map H ◦ (α × IdI ) : I × I → Y . By Lemma 7.17 we
have 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-homotopic
then 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 → X
be a loop based at x. Consider the map H ◦ (α × IdI ) : I × I → Y . By Lemma 7.17 we
have f · g ∼ h · k where f = F ◦ α, g(s) = h(s) = H(x, s), and k is a constant path. We
have 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 is
contractible, 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 c
then f = f ◦ IdX ' f ◦ c, and f ◦ c is a constant map. Similarly, if IdY is homotopic to
a constant map c then f = IdY ◦f ' c ◦ f , and c ◦ f is a constant map.
Theorem 149. [Problem 7-11] The Möbius band is homotopy equivalent to S1 .
47
Proof. We define the Möbius band B to be the geometric realization of the presentation
ha, b, c | abcbi. In other words, it is the quotient space formed from I × I by identifying
the edge {0} × I with {1} × I. Let q : I × I → B be the associated quotient map and
define H : (I × I) × I → B by
1
1
H((x, y), t) = q x, + t y −
.
2
2
e : B × I → B and in fact H
e is a (strong)
Then H descends to a continuous map H
1
deformation retraction from B to q(I × {1/2}) ≈ S .
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
(
x/(1 − t) if 0 ≤ t < 1,
H(x, t) =
(0, 0)
if t = 1
is a strong deformation retraction from X to {(0, 0)}. Furthermore,
(
H(x, 2t)
if 0 ≤ t ≤ 1/2,
H 0 (x, t) =
(2t − 1, 0) if 1/2 < t ≤ 1.
is a deformation retraction from X to {(1, 0)}. However, {(1, 0)} cannot be a strong
deformation retract of X since it is a limit point of X but no retraction is possible
directly towards (1, 0).
Theorem 151. [Problem 7-14] Let M be a compact connected surface that is not homeomorphic to S2 . Then there is a point p ∈ M such that M \ {p} is homotopy equivalent
to 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, the
`
disjoint union space α Xα is their coproduct in the category Top.
Proof. This follows from Theorem 59.
Theorem 153. [Problem 7-17] The wedge sum is the coproduct in the category Top∗ .
W
Proof. Let ((Xα , pα ))α∈A be a family of pointed
spaces,
let
X
=
and define
`
` α∈A Xα W
jα : Xα → X by jα = q ◦ iα where iα : Xα → α∈A Xα and q : α∈A Xα → α∈A Xα is
the quotient map. Note that jα (pα ) = jβ (pβ ) for all α, β; denote this common value by
p. 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∈A Xα → W such that fα W
= f ◦ iα for all
α. Since f (pα ) = r for all α, there exists a unique continuous map g : α∈A Xα → 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
L of abelian groups. The direct
sum, together with the obvious injections iα : Gα ,→
α Gα , is the coproduct of the
Gα ’s in the category Ab.
Proof.
L Let X be an abelian group and
P let fα : Gα → X be homomorphisms. Let
f : α Gα → X be given by (gα ) 7→ α∈A fα (gα ), which is well-defined since gα = 0
for all but finitely many α. It is clear that fα = f ◦ iα for all α. SupposeL
that f 0 is
another homomorphism such that fα = f 0 ◦ iα for all α. Then for all (gα ) ∈ α Gα we
have
!
X
X
X
0
0
f ((gα )) = f
iα (gα ) =
(f 0 ◦ iα )(gα ) =
fα (gα ),
α∈A
α∈A
α∈A
0
which shows that f = f .
Remark 155. [Problem 7-19] The direct sum does not yield the coproduct in the category
Grp. Take G1 = G2 = Z, H = GL(2, R) and the maps fk : Gk → H defined by
1 n
1 0
f1 (n) =
and f2 (n) =
.
0 1
n 1
There is no homomorphism f : Z ⊕ Z → H such that fk = f ◦ ik for k = 1, 2 since
1 1 1 0
2 1
f ((1, 1)) = f ((1, 0) + (0, 1)) =
=
0 1 1 1
1 1
but
1 0
f ((1, 1)) = f ((0, 1) + (1, 0)) =
1 1
1 1
1 1
=
.
0 1
1 2
Chapter 8. The Circle
Theorem 156. [Exercise 8.7] A rotation of S1 is a map ρ : S1 → S1 of the form
ρ(z) = eiθ z for some fixed eiθ ∈ S1 . If ρ is a rotation, then N (ρ ◦ f ) = N (f ) for every
loop f in S1 .
Proof. Define ge : S1 → R by z 7→ θ/2π + fe(z). Then ge is a lift of ρ ◦ f , so
N (ρ ◦ f ) = ge(1) − ge(0)
= θ/2π + fe(1) − θ/2π − fe(0)
= fe(1) − fe(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 traverses
the circle ∂B counterclockwise once has a winding number of 1. If γ is null-homotopic
in 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 from
Corollary 7.38.
Theorem 158. [Problem 8-2] A nonempty topological space cannot be both a 2-manifold
and 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 nmanifold for some n > 2. Choose some p ∈ M and let ϕ1 : U1 → V1 and ϕ2 : U2 → V2
be homeomorphisms where U1 and U2 are neighborhoods of p, V1 is open in R2 , and
V2 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−1
2 )(B)\{ϕ1 (p)}, but W2 is not simply
connected 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 coordinate
charts (U, ϕ) and (V, ψ) such that U, V are neighborhoods of p, ϕ(U ) is open in Int H2 ,
ψ(V ) is open in H2 , and ψ(p) ∈ ∂H2 . Let W = U ∩ V ; then ϕ(W ) is homeomorphic to
ψ(W ). But this is impossible, for ϕ(W ) \ {ϕ(p)} is not simply connected by Theorem
157 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 ◦ ω
has a winding number of zero if and only if ϕ extends to a continuous map from B2 to
S1 .
Theorem 161. [Problem 8-5] Every nonconstant polynomial in one complex variable
has a zero.
50
Proof. Suppose that p(z) = z n + an−1 z n−1 + · · · + a1 z + 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) = z n + an−1 εz n−1 + · · · + a1 εn−1 z + a0 εn
so that pε (z) = εn p(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→ z n to pε |S1 . Therefore pε |S1 has a
winding number of n, and the map φ : S1 → C \ {0} given by φ(z) = p(z/ε) = ε−n pε (z)
also has a winding number of n. But φ is homotopic to the constant loop cp(0) = ca0 by
the 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)
;
kx − f (x)k
by Theorem 160, ϕ|S1 has degree zero. Define H : S1 × I → S1 by
H(x, t) =
x − (1 − t)f (x)
.
kx − (1 − t)f (x)k
If t = 0 then the denominator is never zero. Otherwise,
kx − (1 − t)f (x)k ≥ kkxk − (1 − t) kf (x)kk ≥ t > 0
for t ∈ (0, 1] since kxk = 1 and kf (x)k ≤ 1. This shows that H is a well-defined
homotopy 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
fe : I → R of f such that fe(0) = 0 and fe(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
e : I → R be the lift of α given by s 7→ ns. By Corollary 8.5, there exists
e
a lift f of f such that fe(0) = 0 = α
e(0), and since α ∼ f , Corollary 8.6 shows that
e
f (1) = α
e(1) = n.
Theorem 164. [Problem 8-7] If ϕ : S1 → S1 is continuous and deg ϕ 6= ±1, then ϕ is
not injective.
Proof. The result is clear when deg ϕ = 0, so assume otherwise; since deg ϕ 6= ±1, we
have |deg ϕ| > 1. Let ε : I → S1 be the map s 7→ e2πis . By Lemma 163, there exists
a lift ϕ
e : I → R of ϕ ◦ ε such that ϕ(0)
e
= 0 and |ϕ(1)|
e
= |deg ϕ| > 1. Choose a point
0 < s < 1 such that ϕ(s)
e = ±1. Noting that ε ◦ ϕ
e = 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 different
degrees. Then there is a point z ∈ S1 where ϕ(z) = −ψ(z).
Proof. This follows from Theorem 137.
Theorem 166. [Problem 8-9] Suppose f : I → C \ {0} is a continuously differentiable
loop. Then its winding number is given by
ˆ 1 0
1
f (s)
ds.
N (f ) =
2πi 0 f (s)
Proof. Let fe : I → R be a lift of f / |f |; then exp(2πife(s)) = f (s)/r(s) for all s ∈ I,
where r = |f |. Now
ˆ 1 0
ˆ 1
1
f (s)
1
2πir(s)fe0 (s) exp(2πife(s)) + r0 (s) exp(2πife(s))
ds =
ds
2πi 0 f (s)
2πi 0
r(s) exp(2πife(s))
ˆ 1
0
1
r
(s)
0
=
2πife (s) +
ds
2πi 0
r(s)
i1
1 h
2πife(s) + log r(s)
=
2πi
0
1
=
[2πife(1) − 2πife(0) + log r(1) − log r(0)]
2πi
= fe(1) − fe(0)
= N (f ).
n
n
Theorem 167. [Problem 8-10] A vector field on R is a continuous map V : R →
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 neighborhood
containing 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 winding
number of V around f , denoted by N (V, f ), to be the winding number of the loop
V ◦ 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 standard
counterclockwise loop around the unit circle. This integer is called the index of
V at p, and is denoted by Ind(V, p).
(3) Now assume V has finitely many singular points in the open unit disk. Then
the index of V around the loop ω is equal to the sum of the indices of V at the
interior singular points.
52
γ
C
ω
pn
β
pn
α
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 by
H(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. The
result is clear when n = 1. Assuming the result for n − 1, suppose we have n singular
points p1 , . . . , pn . Let γ be a keyhole path in B2 \{p1 , . . . , pn } that traverses the unit disk
counterclockwise and traverses a small circle C around pn clockwise, such that γ encloses
p1 , . . . , pn−1 (see Figure 0.1). Then N (V,
P γ) encloses p1 , . . . , pn−1 , and ignoring pn , γ is
homotopic to ω. Therefore N (V, γ) = n−1
i=1 Ind(V, pi ) by the induction hypothesis. Let
γ
b be the path γ with the gap closed, so that γ
b is homotopic to both γ and ω·α·β·α where
α is a path from 1 ∈ S1 to a point x on C and β is a clockwise loop around C based at
x. By applying Theorem 166 we have N (V, γ
b) = 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
b) − N (V, β)
= N (V, γ) − N (V, β)
n
X
=
Ind(V, pi ).
i=1
Theorem 168. [Problem 8-11] For each continuous map ϕ : T2 → T2 there is a 2 × 2
integer 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 such
that 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 over
the integers.
Proof. We can consider π1 (T2 , (1, 1)) as a free Z-module with basis {[ω1 ], [ω2 ]} where
ωj is the standard loop in the jth copy of S1 . For a continuous map ϕ : T2 → T2 such
that ϕ(1, 1) = (1, 1), define D(ϕ) to be the unique 2 × 2 integer matrix that represents
the 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 projections
then
ν11 (1) ν12 (1)
(*)
D(ϕ) =
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 degrees
of the component maps ϕjk . Part (1) follows immediately since (ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗ . If
ϕ 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 unique
continuous maps ϕk : S1 → T2 such that ϕjk = pj ◦ϕk for each j. Define ϕ : T2 → T2 by
ϕ(s1 , s2 ) = ϕ1 (s1 )ϕ2 (s2 ) where the multiplication is the component-wise multiplication
of complex numbers; then D(ϕ) = E. More explicitly, we have
E11 E12 s1
ϕ11 (s1 )ϕ12 (s2 )
s s
ϕ
=
= 1E21 2E22 .
s2
ϕ21 (s1 )ϕ22 (s2 )
s1 s2
If ϕE : T2 → T2 denotes the above map for the matrix E then ϕEF = ϕE ◦ ϕF by direct
computation. 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 , so
D(ψ −1 ◦ ϕ) = D(ψ −1 )D(ϕ) = I and D(ϕ) is invertible. Conversely, if D(ϕ) is invertible
then 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)
φ
i
S
ϕ
H
Proof. For each σ ∈ S there is a unique homomorphism ϕσ : F (σ) → H such that
ϕσ (σ) = ϕ(σ). If iσ : F (σ) → F (S) is the canonical injection, by Theorem 9.5 there is
a unique homomorphism φ : F (S) → H such that ϕσ = φ ◦ iσ for every σ ∈ S. Since
φ(σ) = (φ ◦ iσ )(σ) = ϕ(σ), the map φ extends ϕ. Furthermore, if φ0 : F (S) → H is
another homomorphism that extends ϕ then
ϕσ (σ n ) = ϕ(σ)n = [(φ0 ◦ iσ )(σ)]n = (φ0 ◦ iσ )(σ n )
for every σ ∈ S, so φ0 = φ.
Theorem 170. [Exercise 9.11] The free group on S is the unique group (up to isomorphism) satisfying the characteristic property of Theorem 169.
Proof. Let G and G0 be two groups satisfying the characteristic property, and let i :
S → G and i0 : S → G0 be the inclusion maps. There exist unique homomorphisms
ϕ : G0 → G and ϕ0 : G → G0 such that i = ϕ◦i0 and i0 = ϕ0 ◦i. Then ϕ◦ϕ0 ◦i = ϕ◦i0 = i,
so ϕ ◦ ϕ0 = IdG since both ϕ ◦ ϕ0 and IdG satisfy the diagram with G in place of F (S)
and H. Similarly, ϕ0 ◦ ϕ = IdG0 . This shows that G ∼
= G0 .
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 unique
homomorphism φ : ZS → H extending ϕ.
(2) The free abelian group Z {σ1 , . . . , σn } on a finite set is isomorphic to Zn via the
map (k1 , . . . , kn ) 7→ k1 σ1 + · · · + kn σn .
Proof. If φ : ZS → H is a homomorphism extending ϕ then
φ(k1 σ1 + · · · + kn σn ) = k1 φ(σ1 ) + · · · + kn φ(σn )
= k1 ϕ(σ1 ) + · · · + kn ϕ(σn )
for all k1 , . . . , kn ∈ Z and σ1 , . . . , σn ∈ S, which shows that φ is unique. But the above
equation 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 isomorphism between the free abelian group on S and the direct sum of infinite cyclic groups
generated by elements of S:
M
ZS ∼
Z {σ} .
=
σ∈S
Proof. This is clear from the definition of the direct sum.
Theorem 173. [Exercise 9.17]
(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 isomorphic.
Proof. If G is a free abelian group then there is a subset S ⊆ G such that the inclusion
S ,→ G induces an isomorphism ϕ : ZS → G. Then {ϕ(σ) : σ ∈ S} is clearly a
basis for G. Conversely, if an abelian group G has a basis S then the induced map
ϕ : ZS → G has trivial kernel and is surjective. For (2), let G and G0 be two free abelian
groups with bases B and B 0 respectively; assume that there is a bijection ϕ : B → B 0 .
Since G ∼
= ZB 0 , it suffices to show that ZB ∼
= ZB 0 . If i : B → ZB
= ZB and G0 ∼
0
0
0
and i : B → ZB are the canonical injections, there exist unique homomorphisms
φ : ZB 0 → ZB and φ0 : ZB → ZB 0 such that i0 ◦ ϕ = φ0 ◦ i and i ◦ ϕ−1 = φ ◦ i0 . Then
φ ◦ φ0 ◦ i = φ ◦ i0 ◦ ϕ = i ◦ ϕ−1 ◦ ϕ = i, so φ ◦ φ0 = IdZB by uniqueness in Theorem 171.
Similarly, φ0 ◦ φ = IdZB 0 . This shows that ZB ∼
= ZB 0 .
Theorem 174. [Problem 9-1] The free product of two or more nontrivial groups is
infinite and nonabelian.
Proof. Let G1 and G2 Q
be nontrivial groups. Choose non-identity elements g1 ∈ G1 and
n
g2 ∈ G2 ; then wn =
i=1 g1 g2 produces an infinite sequence of distinct elements in
Q
n
0
G1 ∗ G2 . If wn = g1 i=1 g2−1 g1−1 then wn0 wn = g1 but wn wn0 6= g1 .
Theorem 175. [Problem 9-2] A free group on two or more generators has center consisting of the identity alone.
Proof. We can assume that there are exactly two generators x and y. Let xm wy n ∈
F (x, y) where m 6= 0 or n 6= 0 and w is some word. Suppose m 6= 0. Then
(xm wy n )(y −n w−1 x−m y) = y
but
(y −n w−1 x−m y)(xm wy n ) 6= y.
Similarly, if n 6= 0 then xm wy n and xy −n w−1 x−m do not commute.
56
Theorem 176. [Problem 9-3] A group G is free if and only if it has a generating set
S ⊆ G such that every element g ∈ G other than the identity has a unique expression
as a product of the form
g = σ1n1 · · · σknk
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, since
any expression of the indicated form is a reduced word. Conversely, let S ⊆ G. If S
generates G then the induced homomorphism φ : F (S) → G is surjective; if S also
satisfies 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 be
group homomorphisms for j = 1, 2.
(1) There exists a unique homomorphism f1 ∗ f2 : G1 ∗ G2 → H1 ∗ H2 such that the
following diagram commutes for j = 1, 2:
G1 ∗ G2
f1 ∗ f2
H1 ∗ H2
i0j
ij
Gj
fj
Hj
where ij : Gj → G1 ∗ G2 and i0j : 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 hS1 ∪ S2 | R1 ∪ R2 i is a presentation of the free product group
hS1 | R1 i ∗ hS2 | R2 i.
Proof. Consider the homomorphisms i0j ◦ fj : Gj → H1 ∗ H2 for j = 1, 2. By the
characteristic property there exists a unique homomorphism f1 ∗ f2 : G1 ∗ G2 → H1 ∗ H2
such that i0j ◦ 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 ),
`i : 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 ◦ `i . Since S1 and S2 are disjoint, there
57
is a unique map k : S1 ∪ S2 → F (S1 ) ∗ F (S2 ) satisfying k ◦ ji = `i ◦ ki . This induces a
unique homomorphism ψ : F (S1 ∪ S2 ) → F (S1 ) ∗ F (S2 ) satisfying k = ψ ◦ j. Now
ϕ ◦ ψ ◦ j ◦ ji = ϕ ◦ k ◦ ji = ϕ ◦ `i ◦ ki = mi ◦ ki = j ◦ ji ,
so ϕ ◦ ψ ◦ j = j and ϕ ◦ ψ = IdF (S1 ∪S2 ) by uniqueness. Similarly,
ψ ◦ ϕ ◦ `i ◦ ki = ψ ◦ mi ◦ ki = ψ ◦ j ◦ ji = k ◦ ji = `i ◦ ki ,
so ψ ◦ ϕ ◦ `i = `i and ψ ◦ ϕ = IdF (S1 )∗F (S2 ) by uniqueness. This proves that ϕ is an
isomorphism. 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 homomorphisms
fi : F (Si )/Ri → F (S1 ∪ S2 )/R1 ∪ R2 such that fi ◦ πi = π ◦ mi . Then there is a unique
homomorphism 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 ◦ `i . Since R1 ∪ R2 ⊆ ker(p ◦ ψ), there is a unique homomorphism
g : F (S1 ∪ S2 )/R1 ∪ R2 → (F (S1 )/R1 ) ∗ (F (S2 )/R2 ) satisfying g ◦ π = p ◦ ψ. Now
g ◦ f ◦ n i ◦ πi = g ◦ f i ◦ πi = g ◦ π ◦ m i
= p ◦ ψ ◦ mi = p ◦ ψ ◦ ϕ ◦ `i = p ◦ `i = 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 ◦ `i
= f ◦ p ◦ `i = 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
hS1 ∪ S2 | R1 ∪ R2 i ∼
= hS1 | R1 i ∗ hS2 | R2 i .
Theorem 178. [Problem 9-5] Let S be a set, let R and R0 be subsets of the free group
F (S), and let π : F (S) → hS | Ri be the projection onto the quotient group. Then
hS | R ∪ R0 i is a presentation of the quotient group hS | Ri /π(R0 ).
58
Proof. We want to show that
F (S)/R ∪ R0 ∼
= (F (S)/R)/π(R0 ).
Let
π 0 : F (S) → F (S)/R ∪ R0 ,
π 00 : F (S)/R → (F (S)/R)/π(R0 )
be the quotient maps. Since R ⊆ ker(π 0 ), there is a unique homomorphism f1 :
F (S)/R → F (S)/R ∪ R0 such that π 0 = f1 ◦ π. But π(R0 ) ⊆ ker(f1 ) since R0 ⊆ ker(π 0 ),
so there is a unique homomorphism f : (F (S)/R)/π(R0 ) → F (S)/R ∪ R0 such that
f1 = f ◦ π 00 . Since R ∪ R0 ⊆ ker(π 00 ◦ π), there exists a unique homomorphism
g : F (S)/R ∪ R0 → (F (S)/R)/π(R0 ) such that π 00 ◦ π = g ◦ π 0 . Now
f ◦ g ◦ π 0 = f ◦ π 00 ◦ π = f1 ◦ π = π 0
and
g ◦ f ◦ π 00 ◦ π = g ◦ f1 ◦ π = g ◦ π 0 = π 00 ◦ π,
so f ◦ g = IdF (S)/R∪R0 and g ◦ f = Id(F (S)/R)/π(R0 ) by uniqueness.
Theorem 179. [Problem 9-6]
(1) The free group on generators α1 , . . . , αn has the presentation
F (α1 , . . . , αn ) ∼
= hα1 , . . . , αn | ∅i .
In particular, Z has the presentation hα | ∅i.
(2) The group Z × Z has the presentation hβ, γ | βγ = γβi.
(3) The cyclic group Zn has the presentation
Zn ∼
= hα | αn = 1i .
(4) The group Zm × Zn has the presentation
Zm × Zn ∼
= hβ, γ | β m = 1, γ n = 1, βγ = γβi .
Proof. Part (3) follows from Theorem 178. Let ϕ : Z × Z → hβ, γ | βγ = γβi be the
isomorphism in (2). We have
Zm × Zn ∼
= (Z × Z)/(mZ × nZ)
∼
= hβ, γ | βγ = γβi /ϕ(mZ × nZ).
∼
= hβ, γ | βγ = γβi /{β m , γ n }
∼
= hβ, γ | β m = 1, γ n = 1, βγ = γβi
by Theorem 178.
59
Theorem 180. [Problem 9-7] The free abelian group on a set S is uniquely determined
up 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. Then
every basis of G is finite.
Proof. Let n be the rank of G. If G has an infinite basis then it has a linearly independent set S with n + 1 elements, and ZS is a free abelian group of rank n + 1 contained
in G. This contradicts Proposition 9.19.
Theorem 182. [Problem 9-9] Suppose C is a concrete category with faithful functor
F : C → Set. If S is a set, a free object on S in C is an object F ∈ Ob(C) together
with 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 diagram
commutes:
F(F )
F(φ)
i
S
ϕ
F(Y )
(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 ) and
i2 : S → F(F2 ). By definition, there exist unique morphisms φ1 : F2 → F1 and
φ2 : 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 implies
that φ1 ◦ φ2 = IdF1 . Similarly, φ2 ◦ φ1 = IdF2 . This proves (1). Part (2) follows from
Theorem 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 two
vertices v, v 0 ∈ Γ, there is an edge path from v to v 0 . In a connected graph, any two
vertices can be connected by a simple edge path.
Proof. One direction is evident. Suppose that Γ is connected. Apply Theorem 107 with
the equivalence relation on Γ defined by x ∼ y if and only if there is an edge path
containing x and y; this proves the converse.
In a connected graph, we can connect any two vertices by an edge path. If the edge
path 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 edge
path between v0 and vk . By repeating the process we eventually create a simple edge
path between v0 and vk . Note that if v0 = vk then we can let the edge path be the
trivial 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−1 b−1 g −1 = (gag −1 )(gbg −1 )(gag −1 )−1 (gbg −1 )−1 .
−1
If x ∈ [G, G] then x = x1 · · · xn for elements xi = ai bi a−1
i bi , 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] is
clearly trivial since aba−1 b−1 = e for all a, b ∈ G. If [G, G] is trivial and a, b ∈ G then
aba−1 b−1 ∈ [G, G], so aba−1 b−1 = e and therefore ab = ba. For (3), let a[G, G] and
b[G, G] be elements of G/[G, G]. We have
ab[G, G] = baa−1 b−1 ab[G, G] = ba[G, G]
since a−1 b−1 ab ∈ [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 any
homomorphism ϕ : G → H, there exists a unique homomorphism ϕ
e : Ab(G) → H such
that the following diagram commutes:
61
G
ϕ
H
ϕ
e
Ab(G)
Proof. We have
ϕ(aba−1 b−1 ) = ϕ(a)ϕ(b)ϕ(a)−1 ϕ(b)−1 = e
for all a, b ∈ G since H is abelian. Therefore [G, G] ⊆ ker(ϕ) and there exists a unique
homomorphism ϕ
e : Ab(G) → H such that ϕ = ϕ
e ◦ π where π : G → Ab(G) is the
quotient 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 } and
V = Sn \ {S}. Then U ∪ V = Sn , U and V are open in Sn , and the sets U , V and U ∩ V
are 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 both
trivial. 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 the
line segment {0} × {0} × [−1, 1]. Compute π1 (X, N ), where N = (0, 0, 1) is the north
pole, 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 loop
i ◦ f is a loop in S2 \ P , which is simply connected. Therefore i∗ π1 (U ∩ B) is trivial and
π1 (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 pole
down 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 simple
edge path.
Proof. Suppose that v and w are two vertices in a tree that are joined by two different
simple edge paths P and Q. Let Q be the edge path from w to v obtained by reversing
Q, 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 · Q
contains a cycle.
Theorem 189. [Problem 10-4] Every vertex u in a finite tree T is a strong deformation
retract of the tree.
Proof. If the tree has no edges, there is nothing to prove. Suppose that u is incident
with exactly one edge. The proof follows as in Theorem 10.10, but we use the fact that
every tree with at least two vertices contains two vertices each incident with exactly one
edge. By always choosing v 6= u in the proof, we obtain a strong deformation retraction
onto u. Now suppose that u is incident with two edges e, e0 which have endpoints v, v 0
(not equal to u). We have T \ (e ∪ u ∪ e0 ) = U ∪ U 0 where U and U 0 are disjoint trees
containing v and v 0 respectively, and where v and v 0 are both incident with at most one
edge. Choose strong deformation retractions from U to v and U 0 to v 0 . By combining
these with the obvious strong deformation retraction from e ∪ u ∪ e0 , we have a strong
deformation 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 fact
that S2 \ {pn+1 } is homeomorphic to R2 .
Example 191. [Problem 10-5] Compute the fundamental group of the complement of
the three coordinate axes in R3 , giving explicit generator(s).
Using the map x 7→ x/ kxk 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 these
loops generate π1 (X).
Theorem 192. [Problem 10-6] Suppose M is a connected manifold of dimension at
least 3, and p ∈ M . Then the inclusion M \ {p} ,→ M induces an isomorphism
π1 (M \ {p}) ∼
= π1 (M ).
63
Proof. Let B be a coordinate ball around p and let U = B and V = M \{p} in Corollary
10.5. Choose some base point q in B \ {p}. Then the inclusion M \ {p} ,→ M induces
an 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 Corollary
7.38, so π1 (M, q) ∼
= π1 (M \ {p} , q).
Theorem 193. [Problem 10-7] Suppose M1 and M2 are connected n-manifolds with
n ≥ 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 base
point q in U1 ∩ U2 . Since Rn \ {0} is simply connected for n ≥ 3, by Corollary 10.4 and
Theorem 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, connected
2-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 2manifold, it follows from Theorem 10.22 that it suffices to prove that any two connected
sums 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 that
are centered at the points {0, 2, 4, . . . , 2n − 2} in C, which are pairwise tangent to each
other along the x-axis. Prove that π1 (Xn , 1) is a free group on n generators, and describe
explicit 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 n
loops representing the generators of π1 (Xn , 1).
Theorem 196. [Problem 10-10] For any finitely presented group G, there is a finite
CW complex whose fundamental group is isomorphic to G.
Proof. Let hα1 , . . . , αn | r1 , . . . , rm i be a presentation of G. Let Γ be a graph with a
single vertex v and n loops at v; denote these loops by α
e1 , . . . , α
en . By Theorem 10.12,
∼
π1 (Γ, v) = hα1 , . . . , αn | ∅i. Let Γ0 = Γ. For each i = 1, . . . , m, write ri = αkp11 · · · αkp`` ,
64
αkp`` ] ∈ π1 (Γ, v) and let βe : S1 → Γ be the
let β : I → Γ be a representative of [e
αkp11 ] · · · [e
circle representative of β. Attach a 2-cell Di = B2 to Γi−1 along the attaching map βe
to obtain a new CW complex Γi ; then
π1 (Γi , v) ∼
= hα1 , . . . , αn | r1 , . . . , ri i
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 presentation
of the fundamental group together with a specific loop representing each generator.
(1)
(2)
(3)
(4)
A closed disk with two interior points removed.
The projective plane with two points removed.
A connected sum of n tori with one point removed.
A connected sum of n tori with two points removed.
Denote these spaces by X1 , . . . , X4 .
(1) π1 (X1 ) ∼
= hα, β | ∅i where α is a loop around the first interior point and β is a
loop around the second interior point.
Example 198. [Problem 10-12] Give a purely algebraic proof that the groups hα, β | αβαβ −1 i
and hρ, γ | ρ2 γ 2 i 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 , where
πG : G → Ab(G) and πH : H → Ab(H) are the quotient maps. If G = H and f = IdG
then Ab(f ) ◦ π = π, so Ab(f ) = IdAb(G) by uniqueness. If K is a group and g : H → K
is 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 a
functor.
Theorem 200. [Problem 10-17] Ab(G) is the unique group (up to isomorphism) that
satisfies the characteristic property expressed in Theorem 10.19, for any group G.
65
Proof. Suppose Ab(G) and Ab(G)0 are two groups that satisfy the characteristic property. Let π : G → Ab(G) and π 0 : G → Ab(G)0 be the canonical maps. There exist
unique homomorphisms ϕ : Ab(G)0 → Ab(G) and ϕ0 : Ab(G) → Ab(G)0 satisfying
ϕ◦π 0 = π and ϕ0 ◦π = π 0 . Since ϕ◦ϕ0 ◦π = ϕ◦π 0 = π and ϕ0 ◦ϕ◦π 0 = ϕ0 ◦π = π 0 , we have
ϕ ◦ ϕ0 = IdAb(G) and ϕ0 ◦ ϕ = IdAb(G)0 by uniqueness. Therefore Ab(G) ∼
= Ab(G)0 . 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 homomorphisms
mi : Ab(Gi ) → Ab(G1 ∗ G2 ) satisfying mi ◦ αi = α ◦ ki , so there exists a unique
homomorphism 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 rank
n, 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 group
F (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)) and
ψ : Ab(F (S)) → ZS such that ϕ ◦ j = ` ◦ k and ψ ◦ ` ◦ k = j. Then
ϕ ◦ ψ ◦ ` ◦ k = ϕ ◦ j = ` ◦ k,
66
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 characteristic of Γ is χ(Γ) = V − E, where V is the number of vertices and E is the number of
edges. The fundamental group of Γ is a free group on 1 − χ(Γ) generators, and therefore
χ(Γ) is a homotopy invariant.
Proof. Let T be a spanning tree in Γ. Since the number of vertices in a tree is always
one more than the number of edges, we have χ(T ) = 1. By Theorem 10.12, there is one
generator of π1 (Γ) for each edge of Γ not in T . Therefore χ(Γ) = 1 − n where n is the
number of generators of π1 (Γ), and the result follows.
Theorem 205. [Problem 10-21]
(1) If a pushout of a pair of morphisms exists, it is unique up to isomorphism in
the category C.
(2) The amalgamated free product is the pushout of two group homomorphisms with
the same domain.
(3) Let S1 and S2 be sets with nonempty intersection. In the category of sets, the
pushout of the inclusions S1 ∩ S2 ,→ S1 and S1 ∩ S2 ,→ S2 is S1 ∪ S2 together
with 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, the
pushout always exists.
Proof. Let fi : A0 → Ai for i = 1, 2 be a pair of morphisms. Suppose there are two
pushouts, P and P 0 , with morphisms gi : Ai → P and gi0 : Ai → P 0 . Then there exist
unique morphisms h : P 0 → P and h0 : P → P 0 such that h ◦ gi0 = gi and h0 ◦ gi = gi0 .
Since
h ◦ h0 ◦ gi = h ◦ gi0 = gi and h0 ◦ h ◦ gi0 = h0 ◦ gi = gi0 ,
we have h ◦ h0 = IdP and h0 ◦ h = IdP 0 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 usual
inclusions, and choose gi = π ◦ ji . Let
C = (j1 ◦ f1 )(a)(j2 ◦ f2 )(a)−1 : a ∈ A0
67
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 pair
of homomorphisms such that h1 ◦ f1 = h2 ◦ f2 . There exists a unique homomorphism
b
h : A1 ∗ A2 → B such that b
h ◦ ji = hi . For a ∈ A0 we have
b
h((j1 ◦ f1 )(a)(j2 ◦ f2 )(a)−1 ) = (b
h ◦ j1 ◦ f1 )(a)(b
h ◦ j2 ◦ f2 )(a)−1
= (h1 ◦ f1 )(a)(h2 ◦ f2 )(a)−1
= e,
so C ⊆ ker(b
h) and there exists a unique homomorphism h : A1 ∗A0 A2 → B such that
h◦π =b
h. Since
h ◦ gi = h ◦ π ◦ ji = b
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 . Define
g : S1 ∪ S2 → T by letting g|Si = fi ; this is well-defined due to the conditions on f1 and
f2 . For (4), let g1 : X → Z and g2 : Y → Z be continuous maps such that g1 ◦ f = g2 |A .
Let j1 : X → X qY and j2 : Y → X qY be the canonical injections. Let gb : X qY → Z
be the unique continuous map such that gb ◦ j1 = g1 and gb ◦ j2 = g2 ; then gb descends to a
unique continuous map g : X ∪f Y → Z satisfying g ◦ q = gb where q : X q Y → X ∪f Y
is the quotient map. Since g ◦ (q ◦ ji ) = gb ◦ 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 as
the smallest equivalence relation such that f1 (x) ∼ f2 (x) for every x ∈ A0 . Take the
pushout P to be the quotient space (A1 q A2 )/ ∼; the proof is then similar to (4). Chapter 11. Covering Maps
Theorem 206. [Exercise 11.2]
(1) Every covering map is a local homeomorphism, an open map, and a quotient
map.
(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.
68
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, open
subset 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 in
Example 195. Define a map q : X3 → X2 by letting A, B, and C denote the unit circles
centered at 0, 2, and 4, respectively, and defining


z ∈ A;
z,
2
q(z) = 2 − (z − 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 \ B ε (1) are evenly covered
open 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 is
not 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 components 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 a
G-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 be
an evenly covered neighborhood of x and let σ : U → E be a local section of q so that
q ◦ σ = IdU . Then f |U = (f |U ◦ q) ◦ σ which is continuous since f ◦ q is continuous. By
Proposition 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.
69
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 neighe be the sheet of q −1 (U ) containing x so that q| e : U
e →U
borhood of q(x) and let U
U
e then we are done since U
e is Hausdorff, and if y ∈
e
is a homeomorphism. If y ∈ U
/ U
e is a neighborhood of x and E \ U
e is a neighborhood of y. Part (2) follows from
then U
Proposition 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 Example
11.6 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+ = πi−1 ((0, ∞)) and let
Sni− = πi−1 ((−∞, 0)). Let Pni = Pn \ q(πi−1 ({0})); then q −1 (Pni ) is the disjoint union of
Sni+ and Sni− while q|Sni+ and q|Sni− are homeomorphisms. Since the sets Pn1 , . . . , Pnn cover
Pn , 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 Chapter
1, with the origin removed.) Show that the projection π1 : C2 → C onto the first
coordinate 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 of
two sheets. The sets C \ R cover C \ {0} as R varies, which shows that q is a covering
map.
Example 214. [Problem 11-4] Show that there is a two-sheeted covering of the Klein
bottle by the torus.
See Figure 0.2.
70
Theorem 215. [Problem 11-5] Let M and N be connected manifolds of dimension
f → M is a k-sheeted covering map. Then there is a connected
n ≥ 2, and suppose q : M
f#N · · · #N
sum M #N that admits a k-sheeted covering by a manifold of the form M
f with k disjoint copies of N ).
(connected sum of M
Proof. Let U be some evenly covered neighborhood in M and let B be a regular coordinate ball that lies inside U . Then q −1 (B) is the disjoint union of k coordinate
e1 , . . . , B
ek in M
f. Choose a regular coordinate ball C in N . Form the connected
balls B
f#N · · · #N by
sum M #N by identifying ∂B with ∂C, and form the connected sum M
ei with Ci for i = 1, . . . , k, where Ci ⊆ Ni is C in the ith copy Ni of N .
identifying B
f#N · · · #N → M #N by setting q 0 | f e = q| f e and
Define a covering map q 0 : M
M \Bi
M \Bi
q 0 |Ni \Ci = IdN \C for each i.
Theorem 216. [Problem 11-6] Every nonorientable compact surface of genus n ≥ 1
has 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 = 2
follows from Example 214. Assume that the result holds for n = 2, . . . , k, and consider
a nonorientable compact surface (P2 )#k+1 . By the induction hypothesis, there is a twosheeted covering map q : (T2 )#k−2 → (P2 )#k−1 . By Theorem 215, there exists a twosheeted covering map q 0 : (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 00 : (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 is
proper, q −1 ({y}) is also compact. This implies that q −1 ({y}) is a finite discrete set since
q is a local homeomorphism. Write q −1 ({y}) = {p1 , . . . , pn }; by Lemma 109, we can
choose pairwise disjoint open sets U1 , . . . , Un with pi ∈ Ui . By shrinking each set,Swe can
assume that q|Ui : Ui → q(Ui ) is a homeomorphism forTi = 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) ∈ V
then q(x) ∈
/ q(X \ U ), so x ∈ U . Therefore q −1 (V ) is the disjoint union of the open sets
−1
q (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 finitesheeted.
Proof. Every covering map is a local homeomorphism, so one direction follows from the
argument 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}) = {e
x1 , . . . , x
ek } and let W be
−1
f
f
fi
an evenly covered neighborhood of x so that q (W ) = W1 ∪ · · · ∪ Wk where each W
f
fi . For each i
is connected and open, q|W
ei ∈ W
fi : Wi → W is a homeomorphism, and x
Tk
fi ); then Vx is a neighborhood
choose a set Ui ∈ U containing x
ei . Let Vx = i=1 q(Ui ∩ W
−1
of x and q (Vx ) is covered by U1 , . . . , Un . Since B is compact, there are finitely many
such 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) is
compact.
Theorem 219. [Problem 11-11] Let q : E → X be a covering map. Then E is compact
if 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 and
therefore compact. Since q is a local homeomorphism, q −1 ({x}) must be a finite discrete
set. The converse follows from Theorem 218.
Theorem 220. [Problem 11-12] A continuous map f : S1 → S1 is said to be odd if
f (−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 the
following diagram commutes:
ϕ
G
H
ϕ
e
Ab(G)
(2) If deg f is also even, then g lifts to a map ge : S1 → S1 such that p2 ◦ ge = g.
Furthermore, ge ◦ p2 and f are both lifts of g ◦ p2 that agree at either 1 or −1, so
they 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(z 2 ) = 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 ge 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 ◦ ge ◦ p2 = g ◦ p2 = p2 ◦ f ,
so ge ◦ p2 and f are both lifts of g ◦ p2 . Also, since ge(1)2 = g(1) = f (1)2 we have
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(e
g ◦ p2 )(1) = (e
g ◦ p2 )(−1) = ±f (1), so ge ◦ p2 and f agree at either 1 or −1. By Theorem
11.12, ge ◦ p2 = f . This proves (2). But f is odd while ge ◦ p2 is even, which is a
contradiction. 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(z 2 ) = 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 point
x ∈ 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) =
.
kF (x) − F (−x)k
Let g : S1 × I → S2 be given by
√
√
((x, y), t) 7→
1 − t2 x, 1 − t2 y, t ;
then f ◦ g is a homotopy from f |S1 to a constant map. But f |S1 is odd and has odd
degree by Theorem 220, so f |S1 cannot be null-homotopic.
Theorem 223. [Problem 11-15] Given three disjoint, bounded, connected open subsets
U1 , U2 , U3 ⊆ R3 , there exists a plane that simultaneously bisects all three, in the sense
that 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 real
number λ such that the plane through λx and orthogonal to x bisects U . Fix some i and
define V : R → [0, ∞) by λ 7→ Vol(U ∩ Hλ+ ) where Hλ+ = {v ∈ R3 : hx, vi > hx, λxi}.
Since V is monotonic and continuous and 0, Vol(U ) ∈ V (R), there exists a unique
λ ∈ R such that V (λ) = Vol(U )/2. For x ∈ S2 and an open set U ⊆ R3 , denote
this value of λ by ΛU (x). Note that ΛU (−x) = −ΛU (x). Define F : S2 → R2 by
x 7→ (ΛU1 (x) − ΛU2 (x), ΛU2 (x) − ΛU3 (x)); by Theorem 222 there exists some x ∈ S2 such
that 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 is
simply connected. Therefore Y is simply connected. Let A be the semicircular arc, let
γ : I → A be a path that traces out A from (0, 1) to (2/π, 1) and let ω : [0, 1] → S1 be
the 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
fe : Y → R is a lift of f then fe((0, 1)) = fe((2/π, 1)), but this is impossible for fe|A ◦ γ
would be a lift of a loop with winding number 1 while (fe|A ◦ γ)(0) = (fe|A ◦ γ)(1).
Theorem 225. [Problem 11-18] If X is a topological space that has a universal covering
space then X is semilocally simply connected.
Proof. Let q : E → X be a covering map with E simply connected. Let x ∈ X, let U
e be the
be an evenly covered neighborhood of x, choose some x
e ∈ q −1 ({x}) and let U
−1
component of q (U ) containing x
e. If γ : I → U is a loop based at x then by Corollary
11.14 there is a path γ
e : I → E such that γ = q ◦ γ
e and γ
e(0) = x
e. But γ
e(I) ⊆ q −1 (U )
e ; the fact that γ
is connected, so γ
e(I) ⊆ U
e(1) ∈ q −1 ({x}) implies that γ
e(1) = x
e. Since
E is simply connected, γ
e is null-homotopic in E and therefore γ is null-homotopic in
X. This shows that U is relatively simply connected.
Example 226. [Problem 11-19] For each n ∈ N, let Cn denote the circle inSR2 with
center (1/n, 0) and radius 1/n. The Hawaiian earring is the space H = n∈N Cn ,
with the subspace topology.
(1) H is not semilocally simply connected, and therefore has no universal covering
space.
(2) The cone on H is simply connected and semilocally simply connected, but not
locally simply connected.
The point (0, 0) ∈ H does not have a relatively simply connected neighborhood, for any
neighborhood of (0, 0) must contain some circle Cn for n sufficiently large. The cone
CH = (H × I)/(H × {0}) on H is contractible and is therefore simply connected and
semilocally simply connected. Consider the open set U = H × (1/2, 1] as a subset of
CH; the point (0, 0, 1) has no simply connected neighborhood in U . Therefore CH is
not locally simply connected.
Theorem 227. [Problem 11-20] Suppose X is a connected space that has a contractible
universal covering space. For any connected and locally path-connected space Y , a
continuous map f : Y → X is null-homotopic if and only if for each y ∈ Y , the induced
homomorphism f∗ : π1 (Y, y) → π1 (X, f (y)) is the trivial map. This result need not hold
if the universal covering space is not contractible.
Proof. We can assume that Y is nonempty. Let q : E → X be a covering map with E
contractible. 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 fe : Y → E such that f = q ◦ fe and fe(y) = e. Let G : E × I → E be a
deformation retraction of E to a point; then q ◦ G ◦ (fe × IdI ) is a homotopy from f to
a constant map. The converse follows from Theorem 147.
Example 228. [Problem 11-21] For which compact, connected surfaces M do there
exist continuous maps f : M → S1 that are not null-homotopic? Prove your answer
correct.
If M = P2 # · · · #P2 = (P2 )#k (with k copies of the projective plane) then we have an
induced homomorphism f∗ : π1 ((P2 )#k ) → Z. If k = 1 then the only such homomorphism 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. If
M = (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 of
Autq (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 and
q|U is a homeomorphism. If ϕ ∈ Autq (E) and x ∈ U ∩ ϕ(U ) then x = ϕ(y) for some
y ∈ U . By Proposition 12.1 we have that y ∈ q −1 ({q(x)}), so y = x since q|U is a
homeomorphism. Applying Proposition 12.1 again shows that ϕ must be the identity
map.
Theorem 230. [Exercise 12.13] Given a covering space action of a group Γ on a
topological space E, the restriction of the action to any subgroup of Γ is a covering
space action.
Proof. Obvious.
Theorem 231. [Problem 12-1] Suppose q1 : E → X1 and q2 : E → X2 are normal
coverings. There exists a covering X1 → X2 making the obvious diagram commute if
and only if Autq1 (E) ⊆ Autq2 (E).
Proof. Choose any e ∈ E, let x1 = q1 (e) and let x2 = q2 (e). If there is a covering
q : 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 by
Theorem 12.14 we have a chain of covering maps
π
qe2
1
E −→
E/ Autq1 (E) −
→ X2
such that q2 = qe2 ◦ π1 . Since there is a homeomorphism ψ : X1 → E/ Autq1 (E) such
that ψ ◦ q1 = π1 by Theorem 236, the map qe2 ◦ 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) = z 2 .
(c) f (z) = 2 − z.
(d) f (z) = 2 − z 2 .
It is easy to check manually that q is a normal covering. Example 11.17 shows that
if we write π1 (X3 , 1) = hω1 , ω2 , ω3 | ∅i where ωi goes counterclockwise around the ith
circle and similarly π1 (X2 , 1) = ha, b | ∅i then q∗ π1 (X3 , 1) = ha, b2 , bab−1 i. Therefore
Autq (X3 ) ∼
= ha, b | a, b2 , bab−1 i. For (3), we have the following images of π1 (S1 , 1) under
f∗ : hai, ha2 i, hbi, hb2 i. 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 109, 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

z,
z ∈ A,



2 − (2 − z)2 , z ∈ B,
q(z) =

(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 described 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, starting
that 1, 1, 3 and 5. Let c1 be the lower half of B and let c2 be the lower half of
C. Then π1 (X4 , 1) is the free group on {[ω1 ], [ω2 ], [c1 · ω3 · c1 ], [c1 · c2 · ω4 · c2 · c1 ]}, and
q∗ π1 (X4 , 1) = G = ha, b2 , ba2 b−1 , bab−1 a−1 b−1 i. But G is not normal in π1 (X2 , 1) since
b ∈ a−1 b−1 Gba 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 let
X 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 by
translating 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) = ha, b | ∅i, where a represents a loop that traverses the top circle counterclockwise and b traverses the bottom circle counterclockwise. Since X is homotopic
to 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 centered at 2πk +i counterclockwise,
returns to (0, 0). It is clear that [q ◦ωk ] = bk ab−k ,
k and
−k
so q∗ π1 (X, 0) is generated by b ab : 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 topology
is 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
e be the component of q −1 (U ) containing ϕ(x). Then F −1 (U
e ) = {ϕ} is open,
and let U
which shows that Autq (E) is discrete.
Theorem 236. [Problem 12-7] Suppose q : E → X is a covering map (not necessarily
normal). Let E 0 = E/ Autq (E) be the orbit space, and let π : E → E 0 be the quotient
map. Then there is a covering map q 0 : E 0 → X such that q 0 ◦ π = q.
Proof. Since q is constant on the fibers of π, it descends to a continuous map q 0 : E 0 → X
such that q 0 ◦ π = q, and it remains to show that q 0 is a covering map. Let x ∈ X and
let U be an evenly covered neighborhood of x. We have (q 0 )−1 (U ) = π(q −1 (U )), so
it remains to show that π(q −1 (U )) is the disjoint union of open sets that are mapped
homeomorphically onto U by q 0 . But Proposition 12.1 shows that each element of
Autq (E) permutes the components of q −1 (U ), so it is clear that π(q −1 (U )) is the disjoint
e is component of π −1 (U 0 ) then π| e
union of open sets. If U 0 is one of these open sets and U
U
and q|Ue are homeomorphisms, so (q 0 )|U 0 = q|Ue ◦ (π|Ue )−1 is also a homeomorphism. Example 237. [Problem 12-8] Consider the action of Z on Rm \ {0} defined by n · x =
2n x.
(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), define
f : Rm \ {0} → Sm−1 × S1 by
x
x 7→
, exp (2πi log2 kxk) .
kxk
77
Since f is constant on each orbit, it descends to a continuous map fe : (Rm \ {0})/Z →
Sm−1 × S1 such that f = fe ◦ q. Let θ : S1 → [0, 1] be an inverse to s 7→ exp(2πis) such
that 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 ≥ 2
then Rm+1 \ {0} is simply connected, so the universal covering space of Sm−1 × S1 is
homeomorphic 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.
f and a two-sheeted covering map q : M
f→
(2) Prove that there exists a manifold M
M.
We have π1 (M ) ∼
= Z2 ∗Z2 which has the presentation hS | Ri, where S = {α, β, γ, δ} and
−1 −1
R = {αβα β , γδγ −1 δ −1 }. But Z ∼
= hS | R ∪ {β, γ, δ}i, so Z2 ∼
= hS | R ∪ {α2 , β, γ, δ}i.
It follows from Theorem 178 that the index of the normal closure G of {α2 , β, γ, δ} in
f → M such that
π1 (M ) is 2. By Theorem 12.18, there is a covering map q : M
f) ∼
f) has order 2. This implies that q is
q ∗ π1 ( M
= G, and by Corollary 12.8, Autq (M
two-sheeted.
Example 239. [Problem 12-14] Give an example to show that a subgroup of a finitely
generated nonabelian group need not be finitely generated.
See Example 234.
Theorem 240. [Problem 12-15] Suppose X is a topological space that has a universal
covering space. Let x ∈ X, and write G = π1 (X, x). Let CovX denote the category
whose 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 whose
morphisms are G-equivariant maps. Define a functor F : CovX → SetG as follows: for
any covering q : E → X, F(q) is the set q −1 ({x}) with its monodromy action; and for
any 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 such
that the conjugacy class of q∗ π1 (E) is the same as the isotropy type of S, and by
Theorem 11.29, the isotropy type of F(q) is equal to the isotropy type of S. Then
F(q) is isomorphic to S by Proposition 11.26. Furthermore, Proposition 12.1 shows
that F : HomCovX (q1 , q2 ) → HomSetG (F(q1 ), F(q2 )) is injective. Suppose qi : Ei → X
are covering maps for i = 1, 2 and f : F(q1 ) → F(q2 ) is G-equivariant. Choose
78
e1 ∈ F(q1 ) and let e2 = f (e1 ) ∈ F(q2 ); then Ge1 ⊆ Ge2 by Proposition 11.24, and
q∗ π1 (E1 , e1 ) ⊆ q∗ π1 (E2 , e2 ) by Theorem 11.29. Applying Theorem 11.37 shows that
there is a covering homomorphism q : E1 → E2 from q1 to q2 taking e1 to e2 . Since
F(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 continuously
on a Hausdorff space E. If the map α : G × E → E defining the action is a proper
map, 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 a
proper map. Let π : E × E → E be the projection onto the first coordinate. Suppose
L ⊆ E × E is compact; then π(L) is also compact, and α−1 (π(L)) is compact since
α 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 ) be
the homomorphism defined by setting f# σ = f ◦ σ for each singular p-simplex σ. If
c ∈ 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 that
f# Bp (X) ⊆ Bp (Y ). Let πX : Zp (X) → Hp (X) and πY : Zp (Y ) → Hp (Y ) be the
quotient maps. Since πY ◦ f# : Zp (X) → Hp (Y ) satisfies Bp (X) ⊆ ker(πY ◦ f# ), there
is a unique homomorphism f∗ : Hp (X) → Hp (Y ) such that the following diagram
commutes:
Zp (X)
f#
πX
Hp (X)
Zp (Y )
πY
f∗
Hp (Y )
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 of
X 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).
79
Thus the pth singular homology group defines a covariant functor from the category of
topological spaces to the category of abelian groups.
Proof. Part (1) follows from the fact that (IdX )# = IdCp (X) and that IdHp (X) satisfies
the 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. Then
for 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 . By
Theorem 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, then
F∗ = 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 let
πC , 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
F
G
0 → C∗ −
→ D∗ −
→ E∗ → 0
be a short exact sequence of chain maps. Then for each p there is a connecting homomorphism ∂∗ : Hp (E∗ ) → Hp−1 (C∗ ) such that the following sequence is exact:
(*)
∂
F
G
∂
F
∗
∗
∗
∗
∗
· · · −→
Hp (C∗ ) −→
Hp (D∗ ) −→
Hp (E∗ ) −→
Hp−1 (C∗ ) −→
··· .
Proof. Consider the diagram
80
0
Cp+1
F
∂
0
Cp
Cp−1
F
Cp−2
Dp
Dp−1
G
Dp−2
0
Ep
0
∂
G
∂
F
Ep+1
∂
∂
F
∂
0
G
∂
∂
0
Dp+1
Ep−1
0
∂
G
Ep−2
0.
Every square in this diagram commutes, and the horizontal rows are exact. Suppose
e ∈ Ep with ∂p e = 0. Since G is surjective, there is some d ∈ Dp such that e = Gd.
Then G∂p d = ∂p Gd = ∂p e = 0, so ∂p d ∈ ker G and ∂p d = F c for some c ∈ Cp−1 by
exactness at Dp−1 . We have F ∂p−1 c = ∂p−1 F c = ∂p−1 ∂p d = 0, so ∂p−1 c = 0 since F is
injective. Therefore we can define a map ∂# : ker ∂p → Hp−1 (C∗ ) by setting ∂# e = πC c
where πC : ker ∂p → Hp−1 (C∗ ) is the quotient map. We first ensure that ∂# is welldefined. Suppose we have d0 ∈ Dp and c0 ∈ Cp−1 with e = Gd0 and ∂p d0 = F c0 . Then
d − d0 ∈ ker G, so d − d0 = F a for some a ∈ Cp . Now F (c − c0 ) = ∂p (d − d0 ) = ∂p F a =
F ∂p a, so c − c0 = ∂p a since F is injective. We have πC c = πC c0 , which shows that ∂# is
well-defined. Next, we show that ∂# is a homomorphism. If e0 ∈ Ep then e0 = Gd0 and
∂p d0 = F c0 for some d0 ∈ Dp and c0 ∈ Cp−1 . So e+e0 = G(d+d0 ) and ∂p (d+d0 ) = F (c+c0 ),
which implies that ∂# (e + e0 ) = ∂# e + ∂# e0 . Finally, we show that im ∂p+1 ⊆ ker ∂# .
Suppose that e = ∂p+1 e0 for some e0 ∈ Ep+1 ; then e0 = Gd0 for some d0 ∈ Dp+1 . We
have G∂p+1 d0 = ∂p+1 Gd0 = ∂p+1 e0 = e = Gd, so d − ∂p+1 d0 ∈ ker G and d − ∂p+1 d0 = F a
for some a ∈ Cp . Now F c = ∂p d = ∂p (d − ∂p+1 d0 + ∂p+1 d0 ) = ∂p F a = F ∂p a, so c = ∂p a
since F is injective. Then πC c = 0, which shows that im ∂p+1 ⊆ ker ∂# . Therefore ∂#
descends to a homomorphism ∂∗ : Hp (E∗ ) → Hp−1 (C∗ ) such that ∂∗ πE = ∂# , where
πE : ker ∂p → Hp (E∗ ) is the quotient map.
Now we prove exactness of the sequence (*). Given a cycle c, we write [c] for the
homology class of c. Suppose [c] ∈ Hp (C∗ ) with [c] = ∂∗ [e] for some [e] ∈ Hp (E∗ ). From
the definition of ∂∗ there is some d ∈ Dp+1 such that ∂p+1 d = F c, so F∗ [c] = [F c] = 0.
Therefore im ∂∗ ⊆ ker F∗ . Conversely, if F∗ [c] = [F c] = 0 then F c = ∂p+1 d for some
d ∈ Dp+1 . We have ∂p+1 Gd = G∂p+1 d = GF c = 0, and ∂∗ [Gd] = [c] from the definition
of ∂∗ . This shows that ker F∗ ⊆ im ∂∗ , and proves exactness at Hp (C∗ ). Next, we prove
exactness at Hp (D∗ ). Since GF = 0 and G∗ F∗ = 0, it is immediate that im F∗ ⊆ ker G∗ .
Suppose G∗ [d] = 0, i.e. Gd = ∂p+1 e for some e ∈ Ep+1 . Since G is surjective, there
is some d0 ∈ Dp+1 such that e = Gd0 . Then G∂p+1 d0 = ∂p+1 Gd0 = ∂p+1 e = Gd, so
d − ∂p+1 d0 ∈ ker G and d − ∂p+1 d0 = F c for some c ∈ Cp . Since F∗ [c] = [F c] = [d], this
81
shows that ker G∗ ⊆ im F∗ and proves exactness at Hp (D∗ ). Finally, we prove exactness
at Hp (E∗ ). Suppose [e] ∈ Hp (E∗ ) such that [e] = G∗ [d] = [Gd] for some d ∈ Dp with
∂p d = 0. Let [c] = ∂∗ [e] = ∂∗ [Gd]. From the definition of ∂∗ we have F c = ∂p d = 0,
so c = 0 since F is injective. This shows that im G∗ ⊆ ker ∂∗ . Conversely, suppose
[e] ∈ Hp (E∗ ) with ∂∗ [e] = 0. This means that e = Gd and ∂p d = F c for some
d ∈ Dp and some boundary c ∈ Cp−1 . Choose c0 ∈ Cp such that c = ∂p c0 . Then
∂p d = F ∂p c0 = ∂p F c0 , so ∂p (d − F c0 ) = 0 and G(d − F c0 ) = Gd = e. Therefore
G∗ [d − F c0 ] = [Gd] = [e], so ker ∂∗ ⊆ im G∗ . This proves exactness at Hp (E∗ ).
Theorem 247. [Exercise 13.39] The induced cohomology homomorphism satisfies the
following 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→ H p (X; G), f 7→ f ∗ define a contravariant functor
from 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 integer
p ≥ 0 the map f ∗ : H p (Y ; G) → H p (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 )# = IdC p (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 , p2
be nondegenerate base points. For i = 1, 2, let Wi be a neighborhood of pi that admits
a strong deformation retraction onto {pi }. Let q : X1 q X2 → X1 ∨ X2 be the quotient
map, 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 q X2 , the sets U and V are open in X1 ∨ X2 . Applying
Theorem 13.16 and noting that U ∩ V is contractible, we have an exact sequence
k −l
∗
∗
0 → Hp (U ) ⊕ Hp (V ) −−
−→
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 deformation
retract 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)
82
(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. Let
Br (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 sequence
reduces 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 Theorem
80 and Theorem 157. Suppose U ⊆ Rn is open with n > 2 and ϕ : U → Rm is a
homeomorphism. Choose any x ∈ U ; then U \ {x} ≈ Rm \ {ϕ(x)}, so Hn−1 (U \ {x}) ∼
=
m
Hn−1 (R \ {ϕ(x)}) = 0. This contradicts part (1).
Theorem 250. [Problem 13-3] A nonempty topological space cannot be both an mmanifold 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 can
assume that n > 2. Let M be a nonempty topological space that is both a m-manifold
and an n-manifold for m > n > 2. Choose some p ∈ M and let ϕ1 : U1 → V1 and
ϕ2 : U2 → V2 be homeomorphisms where U1 and U2 are neighborhoods of p, V1 is
open in Rm , and V2 is open in Rn . Let B be an open ball around ϕ1 (p) contained in
ϕ1 (U1 ∩ U2 ). Then B ≈ (ϕ2 ◦ ϕ−1
1 )(B), but this contradicts Theorem 249.
Theorem 251. [Problem 13-4] Suppose M is an n-dimensional manifold with boundary. Then the interior and boundary of M are disjoint (cf. Theorem 159).
Proof. Suppose p ∈ M is both an interior and boundary point. Choose coordinate
charts (U, ϕ) and (V, ψ) such that U, V are neighborhoods of p, ϕ(U ) is open in Int Hn ,
ψ(V ) is open in Hn , and ψ(p) ∈ ∂Hn . Let W = U ∩ V ; then ϕ(W ) is homeomorphic
to ψ(W ). But this is impossible, for Hn−1 (ϕ(W ) \ {ϕ(p)}) 6= 0 by Theorem 249 while
Hn−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. Theorem
160).
Proof. Define a homotopy H : Sn ×I → Sn from a constant map to f by H(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.
83
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 Corollary
13.4.
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 a
continuous map ϕ : Bn → Sn−1 by
x − f (x)
ϕ(x) =
.
kx − f (x)k
This contradicts Theorem 252, since we can define a homotopy H : Sn−1 × I → Sn−1
from ϕ|Sn−1 to IdSn−1 by
x − (1 − t)f (x)
H(x, t) =
kx − (1 − t)f (x)k
using an argument identical to Theorem 162.
Theorem 255. [Problem 13-8] If n is even then Z2 is the only nontrivial group that
can act freely on Sn by homeomorphisms.
Proof. Suppose G acts freely on Sn by homeomorphisms. For any g ∈ G, write ϕg for
the homeomorphism x 7→ g · x. Then deg ϕg = ±1 by Proposition 13.25, so deg defines
a homomorphism from G to {±1}. If deg ϕg = 1 and ϕg 6= IdSn then ϕg has no fixed
point since G acts freely. But deg ϕ = −1 by Theorem 13.29, which is a contradiction.
This shows that deg is an injective homomorphism, so G is either the trivial group or
is isomorphic to Z2 .
Example 256. [Problem 13-9] Use the CW decomposition of Theorem 117 and the
results of this chapter to compute the singular homology groups of the 3-dimensional
real 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, so
H2 (P2 ) = 0.
Let ϕ = (q◦F )|∂B3 as defined in Theorem 117, and consider P2 as a subspace of P3 . Since
H2 (P2 ) = 0, the kernel K of ϕ∗ : H2 (∂B3 ) → H2 (P2 ) is H2 (∂B3 ) = H2 (S2 ) = Z and the
image L is 0. It is clear that H0 (P3 ) = Z. By Proposition 13.33, if p = 1 or p > 3 then
the inclusion P2 ,→ P3 induces an isomorphism. Therefore H1 (P3 ) = H1 (P2 ) = Z2 and
Hp (P3 ) = 0 for p > 3. If p = 2 then we have a short exact sequence
0 → L ,→ H2 (P2 ) → H2 (P3 ) → 0.
84
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

Z,





Z2 ,
3
Hp (P ) = 0,



Z,



0,
p = 0,
p = 1,
p = 2,
p = 3,
p > 3.
Theorem 257. [Problem 13-11] For any field F of characteristic zero, the functor
G 7→ Hom(G, F), f 7→ f ∗ is exact.
Proof. Suppose that the sequence
f
g
··· → A →
− B→
− C → ···
is exact at B; we want to show that
f∗
g∗
· · · ← Hom(A, F) ←− Hom(B, F) ←
− 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 βf a = 0 for all a ∈ A. We define a map
γ ∈ Hom(im g, F) by setting γ(c) = βb for any c = gb ∈ im g. To show that γ is welldefined, suppose b, b0 ∈ B with gb = gb0 . Then b − b0 ∈ ker g, so b − b0 = f a for some
a ∈ A, and βb − βb0 = β(b − b0 ) = βf a = 0. It is clear that γ is a homomorphism since
g and β are homomorphisms. By Lemma 13.42 there is an extension γ 0 ∈ Hom(C, F)
of γ, so β = g ∗ γ 0 and ker f ∗ ⊆ im g ∗ as desired.
Theorem 258. [Problem 13-12] Let X be a topological space and let U, V ⊆ X be
open subsets whose union is X. Then there is an exact Mayer-Vietoris sequence for
cohomology with coefficients in a field F of characteristic zero:
· · · → H p−1 (U ∩ V ; F) → H p (X; F) → H p (U ; F) ⊕ H p (V ; F) → H p (U ∩ V ; F) → · · · .
Proof. This follows from Theorem 257.
85
References
[1] John M. Lee. Introduction to Smooth Manifolds. Springer, 2nd edition, 2013.