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Chapter 8
The Fundamental Group
and Covering Spaces
In the first seven chapters we have dealt with point-set topology. This chapter provides an introduction to algebraic topology. Algebraic topology may be regarded as
the study of topological spaces and continuous functions by means of algebraic
objects such as groups and homomorphisms. Typically, we start with a topological
space (X, ) and associate with it, in some specific manner, a group GX (see Appendix I). This is done in such a way that if (X, ) and (Y, ) are homeomorphic, then
GX and GY are isomorphic. Also, a continuous function f : X → Y “induces” a homomorphism f * : GX → GY, and, if f is a homeomorphism, then f* is an isomorphism.
Then the “structure” of GX provides information about the “structure” of (X, ).
One of the purposes of doing this is to find a topological property that distinguishes
between two nonhomeomorphic spaces. The group that we study is called the fundamental group.
8.1 Homotopy of Paths
In Section 3.2 we defined paths and the path product of two paths. This section studies an equivalence relation defined on a certain collection of paths in a topological
space. Review Theorems 2.15 and 2.27, because these results about continuous functions will be used repeatedly in this chapter and the next. The closed unit interval
[0, 1] will be denoted by I.
Definition. Let (X, ) and (Y, ) be topological spaces and let f, g: X → Y be continuous functions. Then f is homotopic to g, denoted by f g, if there is a continuous function H: X I → Y such that H(x, 0) f(x) and H(x, 1) g(x) for all
x X. The function H is called a homotopy between f and g . 265
266 Chapter 8
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The Fundamental Group and Covering Spaces
In the preceding definition, we are assuming that I has the subspace topology determined by the usual topology on . Thus X I has the product topology determined by and .
We think of a homotopy as a continuous one-parameter family of continuous
functions from X into Y . If t denotes the parameter, then the homotopy represents a
continuous “deformation” of the function f to the function g as t goes from 0 to 1.
The question of whether f is homotopic to g is a question of whether there is a continuous extension of a given function. We think of f as being a function from X {0}
into Y and g as being a function from X {1} into Y, so we have a continuous function from X {0, 1} into Y and we want to extend it to a continuous function from
X I into Y.
EXAMPLE 1. Let X and Y be the subspaces of defined by
X {(x, y) : x2 y2 1} and Y {(x, y) : (x 1)2 y2 1
or (x 1)2 y2 1}. Define f, g: X → Y by f(x, y) (x 1, y) and g(x, y) (x 1, y). (See Figure 8.1).
g
X
Y
f
Figure 8.1
Then f is not homotopic to g .
EXAMPLE 2. Let X be the subspace of defined by X {(x, y) :
x2 y2 1} and let Y X X. Define f, g: X → Y by f(x, y) ((1, 0), (x, y))
and g (x, y) ((0,1), (x, y)). Then the function H: X I → Y defined by
H((x, y), t) ((1 t2, t), (x, y)) is a homotopy between f and g so f g .
Notice that X is a circle and Y is a torus, f “wraps” the circle X around
{(1, 0)} X, and g “wraps” the circle X around {(0, 1)} X (see Figure 8.2). Let A
denote the arc from ((1, 0), (1, 0)) to ((0, 1), (1, 0)) (see Figure 8.2). Then H maps
X I onto A X.
8.1 Homotopy of Paths
267
A
(0, 1)
g
((0, 1), (1, 0))
(1, 0)
(( 3/2, 1/2), (1, 0))
H |(X 3 {1/2})
f
((1, 0), (1, 0))
Figure 8.2
THEOREM 8.1. Let (X, ) and (Y, ) be topological spaces. Then is an
equivalence relation on C(X, Y) (the collection of continuous functions that map X
into Y ).
Proof. Let f C(X, Y). Define H: X I → Y by H(x, t) f(x) for each (x, t) X I. Then H is continuous and H(x, 0) H(x, 1) f(x), so f f .
Suppose f, g C(X, Y) and f g . Then there is a continuous function
H: X I → Y such that H(x, 0) f(x) and H(x, 1) g(x) for all x X. Define
K: X I → Y by K(x, t) H(x, 1 t). Then K is continuous, and K(x, 0) H(x, 1) g(x) and K(x, 1) H(x, 0) f(x) for all x X. Therefore g f .
Suppose f, g, h C(X, Y), f g, and g h. Then there are continuous functions F, G: X I → Y such that F(x, 0) f(x), F(x, 1) g(x), G(x, 0) g(x), and
G(x, 1) h(x) for all x X. Define H: X I → Y by
H(x, t) {
F(x, 2t),
G(x, 2t 1),
for all x X and 0 t 12
for all x X and 12 t 1.
Then H is continuous, and H(x, 0) F(x, 0) f(x) and H(x, 1) G(x, 1) h(x)
for all x X. Therefore f h. If and are paths in a topological space, we define a relation between them
that is stronger than homotopy.
Definition. Let and be paths in a topological space (X, ). Then is path
homotopic to , denoted by p , if and have the same initial point x0
and the same terminal point x1 and there is a continuous function H: I I → X
such that H(x, 0) (x) and H(x, 1) (x) for all x X, and H(0, t) x0
and H(1, t) x1 for all t I. The function H is called a path homotopy between
f and g . 268 Chapter 8
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The Fundamental Group and Covering Spaces
The question of the existence of a path homotopy between two paths is again a
question of obtaining a continuous extension of a given function. This time we have
a continuous function on (I {0, 1}) ({0, 1} I) and we want to extend it to a
continuous function on I I (see Figure 8.3).
X
I3I
x0
x1
Figure 8.3
EXAMPLE 3. Let X {(x, y) 2: 1 x2 y2 4} (see Figure 8.4), let
: I → X be the path that maps I in a “linear” fashion onto the arc A from (1, 0)
to (1, 0), and let : I → X be the path that maps I in a “linear” fashion onto the arc
B from (1, 0) to (1, 0). Then is homotopic to (by a homotopy that transforms
into in the manner illustrated in Figure 8.5), but is not path homotopic to because we cannot “get from” to and keep the “end points” fixed without “crossing the hole” in the space.
If (X, ) is a topological space and x0 X, let (X, x0) {: I → X: is continuous and (0) (1) x0}. Observe that if , (X, x0), then the product
(defined in Section 3.2) * (X, x0). It is, however, easy to see that * is not
associative on (X, x0). Each member of (X, x0) is called a loop in X at x0.
The proof of the following theorem is similar to the proof of Theorem 8.1 and
hence it is left as Exercise 1.
A
(21, 0) (1, 0)
B
Figure 8.4
8.1 Homotopy of Paths
(I)
(I)
269
(I)
(I)
(I)
(I)
(I)
(I)
(I)
(I)
(I)
(I)
Figure 8.5
THEOREM 8.2. Let (X, ) be a topological space, and let x0 X. Then p is an
equivalence relation on (X, x0). Let (X, ) be a topological space and let x0 X. If (X, x0), we let []
denote the path-homotopy equivalence class that contains . Then we let 1(X, x0)
denote the set of all path-homotopy equivalence classes on (X, x0), and we define
an operation on 1(X, x0) by [] [] [ * ]. The following theorem tells us
that is well-defined.
THEOREM 8.3. Let (X, ) be a topological space and let x0 X. Furthermore,
let 1, 2, 1, 2 (X, x0) and suppose 1 p 2 and 1 p 2. Then 1 * 1 p
2 * 2.
Proof. Since 1 p 2 and 1 p 2, there exist continuous functions F, G: I I → X such that F(x, 0) 1(x), F(x, 1) 2(x), G(x, 0) 1(x), and G(x, 1) 2(x) for all x I, and F(0, t) F(1, t) G(0, t) G(1, t) x0 for all t I.
Define H: I I → X by
H(x, t) {
F(2x, t),
if 0 x 12 and 0 t 1
G(2x 1, t), if 12 x 1 and 0 t 1.
Then H is continuous,
H(x, 0) F(2x, 0),
if 0 x 12
1(2x),
if 0 x 12
(1 * 1)(x)
G(2x 1, 0) if 12 x 1
1(2x 1), if 12 x 1
{
ˇ
{
270 Chapter 8
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The Fundamental Group and Covering Spaces
and
H(x, 1) F(2x, 1),
if 0 x 12
2(2x),
0 x 12
(2 * 2)(x)
G(2x 1, 1) if 12 x 1
2(2x 1), 12 x 1
{
{
ˇ
for all x X, and H(0, t) F(0, t) x0 and H(1, t) G(1, t) x0 for all t I.
Therefore 1 * 1 p 2 * 2. The following three theorems show that if (X, ) is a topological space and
x0 X, then (1(X, x0), ) is a group.
THEOREM 8.4. Let (X, ) be a topological space, let x0 X, and let [], [],
[] 1(X, x0). Then ([] []) [] [] ([] []) .
Proof. We show that ( * ) * p * ( * ). First we observe that for each
x I,
( * )(2x),
[( * ) * ](x) (2x 1),
{
(4x),
if 0 x 12
(4x 1),
if 12 x 1
(2x 1),
{
if 0 x 14
if 14 x 12
if 12 x 1
and
(2x),
[ * ( * )](x) (4x 2),
(4x 3),
{
if 0 x 12
if 12 x 34
if 34 x 1.
Next we draw a picture to illustrate how we arrive at the definition of the desired
path homotopy H.
( 12 , 1)
(0, 1)
( 34 , 1)
(1, 1)
L
t
(0, 0)
( 14 , 0)
( 12 , 0)
(1, 0)
8.1 Homotopy of Paths
271
If t I, then the intersection of the horizontal line L and the line segment joining
(14, 0) and (12, 1) has coordinates ((1 t)4, t), and the intersection of L and the line
segment joining (12, 0) and (34, 1) has coordinates ((2 t)4, t). Therefore for each
t I, we want H to be defined in terms of if x (1 t)4, in terms of if
(1t)/4 x (2 t)4, and in terms of if x (2 t)4. Also for each t I, we
want H(0, t) H((1 t)4, t) H((2 t)4, t) H(1, t) x0. Using elementary
analytic geometry, we arrive at the definition of H. Define H: I I → X by
{
( )
4x
,
1t
H(x, t) (4x 1 t),
(
1t
4
1t
2t
if 0 t 1 and
x
4
4
2t
if 0 t 1 and
x 1.
4
if 0 t 1 and 0 x )
4x 2 t
,
2t
Thus H is continuous, and it is a routine check to show that H(x, 0) [( * ) * ](x) and H(x, 1) [ * ( * )](x) for all x I, and H(0, t) H(1, t)
x0 for all t I. We have proved that is associative on 1(X, x0). Now we prove that
(1(X, x0), ) has an identity element.
THEOREM 8.5. Let (X, ) be a topological space, let x0 X, and let e: I → X be
the path defined by e(x) x0 for each x I. Then [] [e] [e] [] [] for each
[] 1(X, x0).
Proof. We prove that if [] 1(X, x0) then * e p and e * p .
Let [] 1(X, x0). We draw a picture to illustrate how we arrive at the definition of the path homotopy H to show that * e p .
(0, 1)
(1, 1)
L
t
(0, 0)
e
( 12 , 0)
(1, 0)
272 Chapter 8
The Fundamental Group and Covering Spaces
■
For each t I, the intersection of the line L with the line segment joining (12, 0) and
(1, 1) has coordinates ((1 t)2, t). Therefore for each t I, we want the homotopy
defined in terms of if x (1 t)2. Using analytic geometry, we arrive at the
formula
H(x, t) {
( )
2x
,
1t
x0,
if 0 x if
1t
2
1t
x 1.
2
It is a routine check to see that H(x, 0) ( * e)(x) and H(x, 1) (x) for each
x I and H(0, t) H(1, t) x0 for each t I.
The proof that e * p is left as Exercise 2. We have proved that [e] is the identity element of 1(X, x0). We complete the
proof that 1(X, x0), ) is a group by showing that each member of 1(X, x0) has an
inverse in 1(X, x0).
THEOREM 8.6. Let (X, ) be a topological space, let x0 X, and let [] 1(X, x0). Then there exists [] 1(X, x0) such that [] [] [] [] [e].
Proof. We prove that there is an (X, x0) such that * p e and * p e.
Define : I → X by (x) (1 x) for each x X. We draw a picture to
illustrate how we arrive at the definition of the path homotopy H to show that
* p e.
(0, 1)
(1, 1)
S1
L
S2
t
(0, 0)
( 12 , 0)
(1, 0)
For each t I, we map the part of L that lies between S1 and S2 into the point
f(1 t), we map the segment [(0, t), ((1 t)2, t)] in the same manner that maps
the interval [0, 1 t], and we map the segment [((1 t)2, t), (1, t)] in the same
8.1 Homotopy of Paths
273
manner that maps the interval [t, 1]. Observing that 2((1 t)2) 1 t and
2((1 t)2) 1 t, we arrive at the formula
{
(2x),
H(x, t) (1 t),
(2x 1),
1t
2
1t
1t
x
if
2
2
1t
x 1.
if
2
if 0 x It is easy to see that H(x, 0) ( * )(x) and H(x, 1) e(x) for all x I and
H(0, t) H(1, t) x0 for each t I.
The proof that * p e is left as Exercise 3. We have proved that (1(X, x0), ) is a group.
Definition. Let (X, ) be a topological space and let x0 X. Then (1(X, x0), ) is
called the fundamental group of X at x0. The fundamental group was introduced by the French mathematician Henri
Poincaré (1854–1912) around 1900. During the years 1895–1901, he published a
series of papers in which he introduced algebraic topology. It is not true that algebraic topology developed as an outgrowth of general topology. In fact, Poincaré’s
work on algebraic topology preceded Hausdorff’s definition of a topological space.
Neither was Poincaré’s work influenced by Cantor’s theory of sets.
Poincaré was surpassed only by Leonard Euler (1707–1783) as the most prolific
writer of mathematics. Poincaré published 30 books and over 500 papers on mathematics, and he is regarded as the founder of combinatorial topology.
Throughout the remainder of this text, we denote (1(X, x0), ) by 1(X, x0).
The natural question to ask is whether the fundamental group of a topological space
depends upon the base point x0. This and other questions will be answered in the
remaining sections of this chapter.
We conclude this section with the definition of a contractible space and two
theorems whose proofs are left as exercises.
Definition. A topological space (X, ) is contractible to a point x0 X if there is a
continuous function H: X I → X such that H(x, 0) x and H(x, 1) x0 for each
x X and H(x0, t) x0 for each t I. Then (X, ) is contractible if there exists
x0 X such that (X, ) is contractible to x0. THEOREM 8.7. Let (X, ) be a topological space and let (Y, ) be a contractible
space. If f, g: X → Y are continuous functions, then f is homotopic to g . THEOREM 8.8. Let (X, ) be a contractible space and let (Y, ) be a pathwise
connected space. If f, g: X → Y are continuous functions, then f is homotopic to g. 274 Chapter 8
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The Fundamental Group and Covering Spaces
EXERCISES 8.1
1.
Let (X, ) be a topological space and let x0 X. Prove that p is an equivalence relation on (X, x0).
2.
Let (X, ) be a topological space, let x0 X, and define e: I → X by
e(x) x0 for each x X. Show that if (X, x0), then e * p .
3.
Let (X, ) be a topological space, let x0 X, and define e: I → X by
e(x) x0 for each x X. Let (X, x0), and define : I → X by (x) (1 x) for each x X. Prove that * p e.
4.
Let (X, ), (Y, ) and (Z, ) be a topological spaces, let f1, f2: X → Y be
continuous functions such that f1 f2, and let g1, g2: Y → Z be continuous
functions such that g1 g2. Prove that g1 f1 g2 f2.
5.
Let n . subset of n is convex if for each x, y and t I,
(1 t)x ty X. Let X be a convex subset of n and let and be paths
in X such that (0) (0) and (1) (1). Prove that is path homotopic to .
6.
Let (X, ) be a topological space and let f, g: X → I be continuous functions.
Prove that f is homotopic to g .
7.
Let (X, ) be a pathwise connected space and let , : I → X be continuous
functions. Prove that is homotopic to .
8.
(a) Prove that (with the usual topology) is contractible.
ˇ
(b) Prove that every contractible space is pathwise connected.
9.
Prove Theorem 8.7.
10.
Prove Theorem 8.8.
11. Let X be a convex subset of n and let x0 X. Prove that X is contractible
to x0.
12.
Let (G, , ) be a topological group (see Appendix I), and let e denote the
identity of (G, ). For , (G, e), define by ( )(x) (x) (x).
(a) Prove that is an operation on (G, e) and that ((G, e), ) is a
group.
(b) Prove that the operation on (G, e) induces an operation on
1(G, e) and that (1(G, e), ) is a group.
(c) Prove that the two operations and on 1(G, e) are the same.
(d) Prove that 1(G, e) is abelian.
8.2 The Fundamental Group
275
8.2 The Fundamental Group
In this section we establish some properties of the fundamental group. The first result
is that if (X, ) is a pathwise connected space and x0, x1 X, then 1(X, x0) is isomorphic to 1(X, x1). Before we begin the proof, we draw a picture (Figure 8.6) to
illustrate the idea. It is useful to look at this picture as you read the proof.
X
x1
x0
Figure 8.6
THEOREM 8.9. Let (X, ) be a pathwise connected space, and let x0, x1 X.
Then 1(X, x0) is isomorphic to 1(X, x1).
Proof. Let : I → X be a continuous function such that (0) x0 and (1) x1,
and define : I → X by (x) (1 x) for each x X. Then define
: 1 (X, x0) → 1 (X, x1) by ([]) [( * ) * ] for each [] 1 (X, x0). We
want to prove that is an isomorphism. First observe that if
(X, x0), then ( * ) * (X, x1). In order to show that is welldefined, it is sufficient to show that if and are members of (X, x0) and P ,
then ( * ) * P ( * ) * . The proof of this fact is left as Exercise 1(a).
Now let [], [] 1(X, x0). Then ([] []) ([ * ]) [( *
( * )) * ], and ([]) ([]) [( * ) * ] [( * ) * ] [( * ) *
) * (( * ) * )]. Thus in order to show that is a homomorphism, it is sufficient
to show that ( * ( * )) * p (( * ) * ) * (( * ) * ). The proof of this
fact is left as Exercise 1(b).
Now we show that is one-to-one. Suppose [] , [] 1(X, x0) and
([]) ([]). Then [( * ) * ] [( * ) * ], and so ( * ) * p
( * ) * . In order to show that [] [], it is sufficient to show that p , and
we leave the proof of this fact as Exercise 1(c).
Now we show that maps 1(X, x0) onto 1(X, x1). Let [] 1(X, x1). Then
(X, x1) and so * ( * ) (X, x0). Then ([ * ( * ]) [( * ( *
( * )) * ]. Thus it is sufficient to show that ( * ( * ( * ))) * p , and
the proof of this fact is left as Exercise 1(d). 276 Chapter 8
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The Fundamental Group and Covering Spaces
EXAMPLE 4. Let X be a convex subset (see Exercise 5 of Section 8.1) of n. Then
X is pathwise connected because if x, y , the function : I → X defined by
(t) (1 t)x ty is a path from x to y. Thus up to isomorphism the fundamental
group of X is independent of the base point. Let x0 X and let (X, x0). Define
H: I I → X by H(x, t) tx0 (1 t)(x). Then H(x, 0) (x) and H(x, 1) x0 for each x X. Also H(0, t) tx0 (1 t)(0) x0 and H(1, t) tx0 (1 t)(1) x0. Therefore is path homotopic to the path e: I → X defined by
e(x) x0 for each x X. Hence the fundamental group of X at x0 is the trivial group
(that is, the group whose only member is the identity element), and, by Theorem 8.9,
if x is any member of X, then 1(X, x) is the trivial group.
Let (X, ) be a topological space, let x0 X, and let C be the path component
of X that contains x0. Since the continuous image of a pathwise connected space is
pathwise connected (see Exercise 7 of Section 3.2), we have the following results. If
(X, x0), then is also a member of (C, x0). Also if H: I I → X is a continuous function such that H(x, t) x0 for some (x, t) I I, then H(I I) C.
Thus 1(X, x0) 1(C, x0). Therefore 1(X, x0) depends only on the path component C of X that contains x0, and it does not give us any information about X C.
Let (X, ) be a pathwise connected space, and let x0, x1 X. While Theorem
8.9 guarantees that 1(X, x0) is isomorphic to 1(X, x1), different paths from x0 to x1
may give rise to different isomorphisms.
If (X, ) is a pathwise connected space, it is common practice to speak of the
fundamental group of X without reference to the basepoint.
We show that the fundamental group is a topological property by introducing a
homomorphism induced by a continuous function.
Definition. A topological pair (X, A) is an ordered pair whose first term is a topological space (X, ) and whose second term is a subspace A of X. We do not distinguish between the topological pair (X, ) and the topological
space (X, ). If x0 X, we let (X, x0) denote the topological pair (X, {x0}).
Definition. If (X, A) and (Y, B) are topological pairs, a map f : (X, A) → (Y, B) is a
continuous function f : X → Y such that f(A) B.
THEOREM 8.10. Let (X, ) and (Y, ) be topological spaces, let x0 X and
y0 Y, and let h: (X, x0) → (Y, y0) be a map. Then h induces a homomorphism
h * : 1(X, x0) → 1(Y, y0).
Proof. Let [] 1(X, x0). Then : I → X is a continuous function such that
(0) (1) x0, so h : I → Y is a continuous function such that
(h )(0) (h )(1) y0. Therefore [h ] 1(Y, y0). Define h * : 1(X, x0)
→ 1(Y, y0) by h * ([]) [h ]. In order to show that h * is well-defined, it is sufficient to show that if , (X, x0) and p , then h p h . Suppose
8.2 The Fundamental Group
277
, (X, x0) and p . Then there is a continuous function F: I I → X such
that F(x, 0) (x) and F(x, 1) (x) for all x I and F(0, t) F(1, t) x0 for
all t I. Define G: I I → Y by G(x, t) (h F)(x, t). Then it is easy to see that
G(x, 0) (h )(x) and G(x, 1) (h )(x) for all x I and G(0, t) G(1, t) y0. Therefore h p h , and so h * is well defined.
Now we show that h * is a homomorphism. Let [], [] 1(X, x0). Then
h * ([] []) h * ([ * ]) [h ( * )] and h * ([]) h * ([]) [h ] [h ]
[(h ) * (h )]. Since
(h ( * ))(x) {
(h )(2x),
(h )(2x 1),
if 0 x 12
((h ) * (h ))(x),
if 12 x 1
h * is a homomorphism. In the remainder of this text, whenever we speak of the homomorphism
induced by a continuous function, we mean the homomorphism defined in the
proof of Theorem 8.10.
Definition. If (X, A) and (Y, B) are topological pairs and f, g: (X, A) → (Y, B) are
maps such that f A g A, then f is homotopic to g relative to A, denoted by
f g rel A, if there is a continuous function H: X I → Y such that H(x, 0) f(x) and H(x, 1) g(x) for all x X and H(a, t) f (a) for all a A and t I. If x0 X, then we write f g rel x0 rather than f g rel {x0}.
THEOREM 8.11. Let (X, ) and (Y, ) be topological spaces, let x0 X and
y0 Y, and let h, k: (X, x0) → (Y, y0) be maps such that h k rel x0.Then h * k * .
Proof. Since h k rel x0, there is a continuous function H: X I → Y such that
H(x, 0) h(x) and H(x, 1) k(x) for all x X and H(x0, t) y0 for all t I. Let
[] 1(X, x0). Then h * ([]) [h ] and k * ([]) [k ]. Define G: I I → Y
by G(x, t) H((x), t) for all (x, t) I I. By Theorem 2.27, G is continuous. Note
that G(x, 0) H( (x), 0) (h )(x) and G(x, 1) ((x), 1) (k )(x) for
each x I and G(0, t) H((0), t) H(x0, t) y0 and G(1, t) H((1), t) H(x0, t) y0 for each t I. Therefore h p k , and hence [h ] [k ].
Thus h * k * . THEOREM 8.12. Let (X, ), (Y, ), and (Z, ) be topological spaces, let
x0 X, y0 Y, and z0 Z, and let h: (X, x0) → (Y, y0) and k: (Y, y0) → (Z, z0) be
maps. Then (k h) * k * h * .
Proof. Let [] 1(X, x0). Then (k h) * ([]) [(k h) ] and (k * h * )[]) k * ([h ]) [k (h )]. Since the composition of functions is associative (see Exercise 21 in Appendix C), (k h) k (h ). Therefore (k h) * k * h * . 278 Chapter 8
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The Fundamental Group and Covering Spaces
Definition. Let (X, ) and (Y, ) be topological spaces and let x0 X and y0 Y .
Then (X, x0) and (Y, y0) are of the same homotopy type if there exist maps
f : (X, x0) → (Y, y0) and g: (Y, y0) → (X, x0) such that f g iY rel y0 and g f iX
rel x0. THEOREM 8.13. Let (X, ) and (Y, ) be topological spaces and let x0 X and
y0 Y . If (X, x0) and (Y, y0) are of the same homotopy type, then 1(X, x0) is isomorphic to 1(Y, y0).
Proof. Suppose (X, x0) and (Y, y0) are of the same homotopy type. Then there exist
maps f : (X, x0) → (Y, y0) and g: (Y, y0) → (X, x0) such that f g iY rel y0 and
g f iX rel x0. By Theorem 8.11, (f g) * : 1(Y, y0) → 1(Y, y0) and (g f) * :
1(X, x0) → 1(X, x0) are the identity homomorphisms. By Theorem 8.12, (f g) * f * g * and (g f) * g * f * . By Theorem C.2, f * : 1(X, x0) → 1(Y, y0) is an
isomorphism. THEOREM 8.14. Let (X, ) and (Y, ) be topological spaces, let h: X → Y be a
homeomorphism, let x0 X, and let y0 h(x0). Then 1(X, x0) is isomorphic to
1(Y, y0).
Proof. Since h is a homeomorphism, h1: Y → X is a homeomorphism. Also note
that x0 h1(y0). Since h1 h iX and h h1 iY, (X, x0) and (Y, y0) are of the
same homotopy type. Therefore by Theorem 8.13, 1(X, x0) is isomorphic to
1(Y, y0). EXERCISES 8.2
1.
Let (X, ) be a topological space, let x0, x1 X, let : I → X be a continuous
function such that (0) x0 and (1) x1, and let , (X, x0).
(a) Prove that if , then ( * ) * p ( * ) * .
(b) Prove that ( * ( * )) * p (( * ) * ) * (( * ) * ).
(c) Prove that if ( * ) * p ( * ) * , then p .
(d) Prove that ( * ( * ( * ))) * p .
2.
A pathwise connected space (X, ) is simply connected provided that for
each x0 X, 1(X, x0) is the trivial group. Let (X, ) be a simply connected
space and let , : I → X be paths such that (0) (0) and (1) (1).
Prove that p .
3.
Prove that every contractible space is simply connected.
8.3 The Fundamental Group of the Circle
4.
279
A topological space (X, ) is weakly contractible if there exists x0 X and a
continuous function H: X I → X such that H(x, 0) x and H(x, 1) x0
for each x X.
(a) Give an example of a weakly contractible space that is not contractible.
(b) Prove that every weakly contractible space is simply connected.
5.
A subspace A of a topological space (X, ) is a retract of X if there exists a
continuous function r: X → A such that r(a) a for each a A. The function r is called a retraction of X onto A. Let A be a subspace of a topological
space (X, ), let r: X → A be a retraction of X onto A, and let a0 A. Prove
that r * : 1(X, a0) → 1(A, a0) is a surjection. Hint: Consider the map
j: (A, a0) → (X, a0), defined by j(a) a for each a A.
6.
Let (X, ) be a pathwise connected space, let x0, x1 X, let (Y, ) be a
topological space, let h: X → Y be a continuous function, let y0 h(x0) and
y1 h(x1), and for each i 0, 1, let hi denote the map from the pair (X, xi)
into the pair (Y, yi) defined by hi(x) h(x) for each x X. Prove that there
are isomorphisms : 1(X, x0) → 1(X, x1) and : 1(Y, y0) → 1(Y, y1)
such that (h1) * (h0) * .
8.3 The Fundamental Group of the Circle
Our goal in this section is to show that the fundamental group of the circle is isomorphic to the group of integers. In order to do this we need several preliminary results.
We begin with the definition of the function p that maps onto S1. Define p: → S1
by p(x) ( cos 2x, sin 2x). One can think of p as a function that wraps around
S1. In particular, note that for each integer n, p is a one-to-one map of [n, n 1) onto
S1. Furthermore, if U is the open subset of S1 indicated in Figure 8.7, then p1(U) is
the union of a pairwise disjoint collection of open intervals.
(
U
(
Figure 8.7
280 Chapter 8
■
The Fundamental Group and Covering Spaces
To be specific, let U {(x, y) S1: x 0 and y 0}. Then if x p1(U),
1
cos 2x and sin 2x are both positive, so p1(U) n(n, n 4 ). Moreover, for
1
each n , p[n, n 4] is a one-to-one function from the closed interval [n, n 14] onto
U. Thus, since [n, n 14] is compact, p[n, n 14] is a homeomorphism of [n, n 14] onto
U. Therefore p(n, n 14) is a homeomorphism of (n, n 14) onto U. Throughout the
remainder of this section, p denotes the function defined above.
The following result is known as the Covering Path Property. You may wish to
consider Example 5 before reading the proof.
THEOREM 8.15. Let : I → S1 be a path and let x0 such that p(x0) (0).
Then there is a unique path : I → such that (0) x0 and p .
Proof. Since is continuous, for each t I there is a connected neighborhood Ut of t
such that (Ut) is a proper subset of S1. Since I is compact, the open cover {Ut: t I}
of I has a finite subcover {U1, U2, ..., Um}. If Ui I for some i, then (I) is a connected
proper subset of S1. If Ui I for any i, choose Ui so that 0 Ui. There exists t I
such that t 0, [0, t] Ui, and t Ui Uj for some j i. If Ui Uj I,
([0, t] ) and ([t, 1]) are connected proper subsets of S1. If Ui Uj I, choose
t I such that t t, [t, t] Uj and t Uj U for some k (i k j).
Since {U1, U2, ..., Um} is a finite cover of I, after a finite number of steps, we will obtain
t0, t1, t2, ..., tn such that 0 t0, tn 1, ti1 ti and ([ti1, ti]) is a connected proper
subset of S1 for each i 1, 2, ..., n.
For each i 0, 1, 2, ..., n, let Pi be the statement: There is a unique continuous
function i: [0, ti] → such that i(0) x0 and p i [0, ti]. In order to prove the
theorem, it is sufficient to show that Pi is true for each i. It is clear that P0 is true. Suppose 0 i n and Pi1 is true. Let Vi1 denote the component of p1([ti1, ti]))
that contains i1(ti1), and let pi1 pVi 1. Then pi1: Vi1 → ([ti1, ti]) is a
homeomorphism. Define i: [0, ti] → by
i(t) {
i1(t),
(pi1)1((t)),
if 0 t ti1
if ti1 t ti .
Then i is continuous, i(0) x0, and p i [0, ti]. We must show that i is
unique. Suppose i: [0, ti ] → is a continuous function such that i (0) x0 and
p i [0, ti ]. If 0 t ti1, then i(t) i(t) since Pi1 is true. Suppose ti1 t ti. Since (p i) (t) (t) ([ti1, ti]), i(t) p1(([ti1, ti])). Therefore
i(t) Vi1 since i is continuous and Vi1 is a component of p1(([ti1, ti])).
Therefore i(t), i(t) Vi1 and (pi1 i)(t) (pi1 i)(t). Hence i(t) i(t)
since pi1 is a homeomorphism. EXAMPLE 5. Let : I → S1 be a homeomorphism that maps I onto
{(x, y) S1: x 0 and y 0} and has the property that (0) (1, 0) and
(1) (0, 1) and let x0 2. Then the function : I → given by Theorem 8.15 is
8.3 The Fundamental Group of the Circle
281
a homeomorphism of I onto [2, 2.25], which has the property that (0) 2 and
(1) 2.25.
The following result is known as the Covering Homotopy Property.
THEOREM 8.16. Let (X, ) be a topological space, let f : X → be a continuous
function, and let H: X I → S1 be a continuous function such that H(x, 0) (p f)(x) for each x X. Then there is a continuous function F: X I → such
that F(x, 0) f(x) and (p F)(x, t) H(x, t) for each (x, t) X I.
Proof. For each x X, define x: I → S1 by x(t) H(x, t). Now (p f)(x) x(0),
and hence, by Theorem 8.15, there is a unique path x: I → such that
x(0) f(x) and p x x. Define F: X I → R by F(x, t) x(t). Then
(p F)(x, t) (p x)(t) H(x, t) and F(x, 0) x(0) f(x).
We must show that F is continuous. Let x0 X. For each t I, there is a neighborhood Mt of x0 and a connected neighborhood Nt of t such that H(Mt Nt) is
contained in a proper connected subset of S1. Since I is compact, the open cover
n M . Then M is a
{Nt: t I} of I has a finite subcover {Nt1, Nt2..., Ntn}. Let M i1 ti
neighborhood of x0, and, for each i 1, 2, ....., n, H(M Nti) is contained in a
proper connected subset of S1. Thus there exist t0, t1, ..., tm such that t0 0, tm 1,
and, for each j 1, 2, ..., m, tj1 tj and H(M [tj1, tj]) is contained in a proper
connected subset of S1.
For each j 0, 1, ..., m, let Pj be the statement: There is a unique continuous function Gj : M [0, tj] → such that Gj(x, 0) f(x) and (p Gj)(x, t) H(x, t). We
want to show that Pj is true for each j. It is clear that P0 is true. Suppose 0 j m and
Pj1 is true. Let Uj1 be a connected proper subset of S1 that contains H(M Ij1),
let Vj1 denote the component of p1(Uj1) that contains Gj1(M {tj1}), and
let pj1 pVj 1. Then pj1: Vj1 → Uj1 is a homeomorphism. Define Gj: M [0, tj] → by
Gj(x, t) {
Gj1(x, t),
(pj1)1 (H(x, t)),
if 0 t tj1
if tj1 t tj.
Then Gj is continuous, Gj(x, 0) f(x), and (p Gj)(x, t) H(x, t). We must show
that Gj is unique. Suppose Gj: M [0, tj] → is a continuous function such that
Gj(x, 0) f(x) and (p Gj)(x, t) H(x, t). If 0 t tj1, then Gj(x, t) Gj(x, t)
because Pj1 is true. Suppose tj1 t tj. Since (p Gj)(x, t) H(x, t) Uj1, Gj(x, t) p1(Uj1). Therefore, since Gj is continuous and Vj1 is a component of p1(Uj1), Gj(x, t) Vj1. Hence Gj(x, t) and Gj(x, t) are members of Vj1
and (pj1 Gj)(x, t) (pj1 Gj)(x, t). Thus, since pj1 is a homeomorphism,
Gj(x, t) Gj(x, t). We have proved that there is a unique continuous function
G: M I → such that G(x, 0) f(x) and (p G)(x, t) H(x, t). Therefore
G F(M I). Since M is a neighborhood of x0 in X, F is continuous at (x0, t) for each
282 Chapter 8
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The Fundamental Group and Covering Spaces
t I. Since x0 is an arbitrary member of X, F is continuous. This proof also shows
that F is unique. If is a loop in S1 at (1, 0), then p(0) (0), and hence, by Theorem 8.15,
there is a unique path : I → 1 such that (0) 0 and p . Since
(p )(1) (1) (1, 0), (1) p1(1, 0), and hence (1) is an integer. The
integer (1) is called the degree of the loop , and we write deg() (1).
EXAMPLE 6. Define : I → S1 by (x) ( cos 4x, sin 4x). Then “wraps” I
around S1 twice in a clockwise direction. The unique path : I → such that
(0) 0 and p given by Theorem 8.15 is defined by (x) 2x for each
x I. Therefore deg() (1) 2.
ˇ
THEOREM 8.17. Let 1 and 2 be loops in S1 at (1, 0) such that 1 p 2. Then
deg(1) deg(2).
Proof. Since 1 p 2, there is a continuous function H: I I → S1 such that
H(x, 0) 1(x) and H(x, 1) 2(x) for each x I and H(0, t) H(1, t) (1, 0)
for each t I. By Theorem 8.15, there is a unique path 1: I → such that 1(0) 0
and (p 1) 1. Thus, by Theorem 8.16, there is a unique continuous funtion
F: → such that (p F)(x, t) H(x, t) for each (x, t) I I and
F(x, 0) 1(x) for each x I. Define : I → by (t) F(0, t) for each t I.
Then is continuous, and (p )(t) (p F)(0, t) H(0, t) (1, 0) for each
t I. Therefore (I) p1(1, 0), and, since (I) is connected and 1(1, 0) is a discrete subspace of , is a constant function. Thus F(0, t) (t) (0) F(0, 0) 1(0) 0 for each t I. Define 2: I → by 2(x) F(x, 1) for each x I. Then
2(0) F(0, 1) 0 and (p 2)(x) (p F)(x, 1) H(x, 1) 2(x) for each
x I. By definition, deg(1) 1(1) and deg(2) 2(1). Now define a path
: I → by (t) F(1, t) for each t I. Again (p )(t) (p F)(1, t) H(1, t) (1, 0) for each t I, and hence (I) p1(1, 0). Therefore is a constant
function, and hence F(1, t) (t) (0) F(1, 0) 1(0) 0 for each t I.
Therefore 2(1) F(1, 1) F(1, 0) 1(1), and hence deg(1) deg(2). We are ready to prove that the fundamental group of the circle is isomorphic to
the group of integers. Since S1 is pathwise connected, the fundamental group of S1 is
independent of the base point.
THEOREM 8.18. 1(S1, (1, 0)) is isomorphic to the group of integers.
Proof. Define : 1(S1, (1, 0)) → by ([]) deg(). By Theorem 8.17, is
well-defined. Let [1], [2] 1(S1, (1, 0)). By Theorem 8.15, there are unique paths
1, 2: I → such that 1(0) 2(0) 0, p 1 1, and p 2 2. By definition, deg(1) 1(1) and deg(2) 2(1). Define : I → by
(x) {
1(2x),
1(1) 2(2x1),
if 0 x 12
if 12 x 1.
8.4 Covering Spaces 283
Since 2(0) 0, is continuous. Now (0) 1(0) 0, and, since
p(1(1) 2(2x1)) p(2(2x1)),
(p 1)(2x),
if 0 x 12
(p )(x) (p 2)(2x 1), if 12 x 1
(2x),
if 0 x 12
1
(1 * 2)(x).
2(2x 1), if 12 x 1
{
{
Therefore ([1] [2]) ([1 * 2]) deg(1 * 2) (1) 1(1) 2(1)
deg(1) deg(2) ([1]) ([2]), and so is a homomorphism.
Now we show that maps 1(S1, (1, 0)) onto . Let z , and define a path
1: I → by (t) zt for each t I. Then (0) 0 and (1) z, so p : I → S1
is a loop in S1 at (1, 0). Therefore [p ] 1(S1, (1, 0)), and, by definition,
deg(p ) (1) z. Therefore ([p ]) deg(p ) z.
Finally, we show that is one-to-one. Let [1], [2] 1(S1, (1, 0)) such that
([1]) ([2]). Then deg(1) deg(2). By Theorem 8.15, there are unique
paths 1, 2: I → such that 1(0) 2(0) 0, p 1 1, and p 2 2. By
definition, deg(1) 1(1) and deg(2) 2(1). So 1(1) 2(1). Define F: I I → by F(x, t) (1 t)1(x) t2(x) for each (x, t) I I. Then F(x, 0) 1(x) and F(x, 1) 2(x) for each x I and F(0, t) 0 and F(1, t) 1(1) 2(1) for each t I. Therefore p F: I I → S1 is a continuous function such that
(p F)(x, 0) (p 1)(x) 1(x) and (p F)(x, 1) (p 2)(x) 2(x) for each
x I and (p F)(0, t) p(0) (1, 0) and (p F)(1, t) (p 1)(1) 1(1) (1, 0) for each t I. Thus 1 p 2, so [1] [2]. There are no exercises in this section. The sole purpose of the section is to calculate the fundamental group of a familiar figure and thus provide you with a concrete
example. In the next section, we generalize some of the theorems in this section and
give some exercises.
8.4 Covering Spaces
In this section we generalize Theorems 8.15 and 8.16 by replacing the function
p: → S1 defined by p(x) (cos 2x, sin 2x) with a function called a “covering
map” from an arbitrary topological space (E, ) into another topological space
(B, ). These two generalized theorems are used in Section 9.4.
Definition. Let (E, ) and (B, ) be topological spaces, and let p: E → be a
continuous surjection. An open subset U of B is evenly covered by p if p1(U)
can be written as the union of a pairwise disjoint collection {V: } of open sets
such that for each , p is a homeomorphism of V onto U. Each V is called a
slice of p1(U). 284 Chapter 8
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The Fundamental Group and Covering Spaces
Definition. Let (E, ) and (B, ) be topological spaces, and let p: E → B be a continuous surjection. If each member of B has a neighborhood that is evenly covered
by p, then p is a covering map and E is covering space of B. Note that the function p: → S1 in Section 8.3 is a covering map. Covering
spaces were introduced by Poincaré in 1883.
Definition. Let (E, ), (B, ), and (X, ) be topological spaces, let p: E → B be a
covering map, and let f: X → B be a continuous function. A lifting of f is a continuous function f: X → E such that p f f (see Figure 8.8). E
f'
X
f
p
B
Figure 8.8
Notice that Theorem 8.15 provides a lifting of a path in S1, where p is the specific function discussed in Section 8.3 rather than an arbitrary covering map.
THEOREM 8.19. Let (E, ) and (B, ) be topological spaces, let p: E → B be a
covering map, let e0 E, let b0 p(e0), and let : I → B be a path such that
(0) b0. Then there exists a unique lifting : I → E of f such that (0) e0.
Proof. Let {U: } be an open cover of B such that for each , U is
evenly covered by p. By Theorem 4.4, there exists t0, t1, ..., tn such that
0 t0 t1 ... tn 1 and for each i 1, 2, ..., n, ([ti1, ti]) U for some
. We define the lifting inductively.
Define (0) e0, and assume that (t) has been defined for all t [0, ti1],
where 1 i n. There exists such that ([ti1, ti]) U. Let {V:
} be the collection of slices of p1(U). Now (ti1) is a member of exactly
one member V of . For t [ti1, ti], define (t) (pV)1((t)). Since
pV: V → U is a homomorphism, is continuous on [ti1, ti] and hence on [0, ti].
Therefore, by induction, we can define on I .
It follows immediately from the definition of that p .
The uniqueness of is also proved inductively. Suppose is another lifting of such that (0) e0, and assume that (t) (t) for all t [0, ti1], where
1 i n. Let U be a member of the open cover of B such that ([ti1, ti]) U,
and let V be the member of chosen in the definition of . Since is a lifting of
, ([ti1, ti]) p1(U) V. Since is a collection of pairwise disjoint
open sets and ([ti1, ti]) is connected, ([ti1, ti]) is a subset of one member of .
ˇ
8.4 Covering Spaces 285
Since (ti1) (ti1) V, ([ti1, ti]) V. Therefore, for each t [ti1, ti],
(t) V p1((t)). But V p1((t)) {(t)}, and so (t) (t).
Therefore (t) (t) for all t [0, ti]. By induction, . THEOREM 8.20. Let (E, ) and (B, ) be topological spaces, let p: E → B be a
covering map, let e0 E, let b0 p(e0), and let F: I I → B be a continuous function such that F(0, 0) b0. Then there exists a lifting F: I I → E of F Such that
F(0, 0) e0. Moreover, if F is a path homotopy then F is also.
Proof. Define F(0, 0) e0. By Theorem 8.19, there exists a unique lifting F: {0} I → E of F{0}I such that F(0, 0) e0 and a unique lifting F: I {0} → E of FI{0}
such that F(0, 0) e0. Therefore, we assume that a lifting F of F is defined on
({0} I) (I {0}) and we extend it to I I.
Let {U: } be an open cover of B such that for each , U is evenly
covered by p. By Theorem 4.4, there exists s0, s1, ..., sm and t0, t1, ..., tn such that
0 s0 s1 ... sm 1, 0 t0 t1 ... tn 1, and for each i 1, 2, ..., m
and j 1, 2, ..., n, F([si1, si] [tj1, tj]) U for some . For each i 1,
2, ..., m and j 1, 2, ..., n, let Ui Jj [si1, si] [tj1, tj]. We define F inductively on the rectangles Ii Jj in the following order:
I1 J1, I2 J1, ..., Im J1, I1 J2, I2 J2, ..., Im J2, I1 J3, ..., Im Jn.
I1 Jn
I2 Jn
…
Im Jn
I1 J2
I2 J2
…
Im J2
I1 J1
I2 J2
…
Im J1
Suppose 1 p m, 1 q n, and assume that F is defined on C ({0} I)
(I {0}) m
i1
q1(I
j1 i
Jj) p1
i1 (Ii
Jq). We define F on Ip Jq.
There exists such that F(Ip Jq) U. Let {V: } be the
collection of slices of p1(U). Now F is already defined on D C (Ip Jq).
Since D is connected, F(D) is connected. Since is a collection of pairwise disjoint
open sets, F(D) is a subset of one member V of . Now pV is a homeomorphism
of V onto U. Since F is a lifting of FC, ((pV) F)(x) F(x) for all x D. For
x Ip Iq, define F(x) (pV)1(F(x)). Then F is a lifting of F(C(Ip Jq)). Therefore, by induction we can define F on I I.
Now suppose F is a path homotopy. Then F(0, t) b0 for all t I. Since F is a
lifting of F, F(0, t) p1(b0) for all t I. Since {0} I is connected, F is continu-
286 Chapter 8
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The Fundamental Group and Covering Spaces
ous, and p1(b0) has the discrete topology as a subspace of E, F({0} I) is connected
and hence it must be a single point. Likewise, there exists b1 B such that
F(1, t) b1 for all t I, so F(1, t) p1(b1) for all t I, and hence F({1} I)
must be a single point. Therefore F is a path homotopy. THEOREM 8.21. Let (E, ) and (B, ) be topological spaces, let p: E → B be a
covering map, let e0 E, let b0 p(e0), let b1 B, let and be paths in B from b0
to b1 that are path homotopic, and let and be liftings of and respectively
such that (0) (0) e0. Then (1) (1) and p .
Proof. Let F: I I → B be a continuous function such that F(x, 0) (x) and
F(x, 1) (x) for all x I and F(0, t) b0 and F(1, t) b1 for all t I. By Theorem 8.20, there exists a lifting F: I I → E of F such that F(0, t) e0 for all t I
and F({1} I) is a set consisting of a single point, say e1. The continuous function
F(I{0}) is a lifting of F(I{0}) such that F(0, 0) e0. Since the lifting of paths is
unique (Theorem 8.19), F(x, 0) (x) for all x I. Likewise, F(I{1}) is a lifting
of F(I{1}) such that F(0, 1) e0. Again by Theorem 8.19, F(x, 1) (x) for all
x I. Therefore (1) (1) e1 and p . EXERCISES 8.4
1.
Let (E, ) and (B, ) be topological spaces, and let p: E → B be a covering
map. Show that p is open.
2.
Let (E1, 1), (E2, 2), (B1, 1), and (B2, 2) be topological spaces, and let
p1: E1 → B1 and p2: E2 → B2 be covering maps. Prove that the function
p: E1 E2 → B1 B2 defined by p(x, y) (p1(x), p2(y)) is a covering map.
3.
In this exercise, let p1 denote the function p that is used throughout Section
8.3, and define p: → S1 S1 by p(x, y) (p1(x), p1(y)). By Exercise
2, p is a covering map. Since the torus is homeomorphic to S1 S1, is
a covering space of the torus. Draw pictures in and describe verbally
the covering map p.
4.
Let (X, ) be a topogical space, let Y be a set, let be the discrete topology
on Y, and let 1: X Y → X be the projection map. Prove that 1 is a
covering map.
5.
Let (E, ) be a topological space, let (B, ) be a connected space, let n ,
and let p: E → B be a covering map such that for some b0 B, p1(b0) is a set
with n members. Prove that for each b B, p1(B) is a set with n members.
6.
Let (E, ) be a topological space, let (B, ) be a locally connected, connected space, let p: E → B be a covering map, and let C be a component of
E. Prove that pC: C → B is a covering map.
8.5 Applications and Additional Examples of Fundamental Groups 287
7.
Let (E, ) be a pathwise connected space, let (B, ) be a topological space,
let p: E → B be a covering map, and let b B.
(a) Prove that there is a surjection : 1(B, b) → p1(b).
(b) Prove that if (E, ) is simply connected, then is a bijection.
8.
The projective plane, which we have previously introduced, can also be defined
as the quotient space obtained from S2 by indentifying each point x of S2 with
its antipodal point x. Let p: S2 → P2 denote the natural map that maps each
point of S2 into the equivalence class that contains it. Prove that P2, defined in
this manner, is a surface and that the function p is a covering map.
8.5 Applications and Additional Examples of
Fundamental Groups
We know that the fundamental group of a contractible space is the trivial group (see
Exercise 8 of Section 8.1 and Exercises 2 and 3 Section 8.2) and that the fundamental group of the circle is isomorphic to the group of integers. In this section, we give
some applications and some theorems that are useful in determining the fundamental group of a space. We begin by showing that S1 is not a retract (see Exercise 5 of
Section 8.2) of B2 {(x, y) : x2 y2 1}.
EXAMPLE 7. We show that B2 is contractible. Define H: B2 I → B2 by
H((x, y), t) (1 t)(x, y). Then H is continuous, H((x, y), 0) (x, y) and
H((0, 0), t) (0, 0) for each t I.
THEOREM 8.22. S1 is not a retract of B2. The diagram in Figure 8.9 is helpful in following the proof.
1(B 2)
j
1(S 1)
*
r*
id
1(S 1)
Figure 8.9
Proof. Suppose S1 is a retract of B2. Then there is a continuous function r: B2 → S1
such that r(x) x for each x S1. Define j: S1 → B2 by j(x) x for each x S1.
Then r j: S1 → S1 is the identity function, and hence ( j) * : 1(S1, (1, 0)) →
1(S1, (1, 0)) is the identity homomorphism. By Theorem 8.12, (rj) * r * j * .Thus
288 Chapter 8
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r * maps 1(B2, (1, 0)) onto 1(S1, (1, 0)). This is a contradiction, because 1(B2,
(1, 0)) is the trivial group and 1(S1, (1, 0)) is isomorphic to the group of integers. Recall that a topological space (X, ) has the fixed-point property if for each
continuous function f: X → X there exists x X such that f(x) x. The following
theorem is due to the Dutch mathematician L.E.J. Brouwer (1881–1966).
THEOREM 8.23. (Brouwer Fixed-Point Theorem). B2 has the fixed-point property.
Proof. Suppose there is a continuous function f : B2 → B2 such that f(x) x for any
x B2. For each x B2, let Lx denote the line segment that begins at f(x) and passes
through x, and let yx Lx S1 { f(x)}. (Note that yx is uniquely determined—see
Figure 8.10.)
Lx
yx
x
f (x)
Figure 8.10
It is a straightforward but tedious exercise (see Exercise 4) to show that the function
r : B2 → S1 defined by r(x) yx for each x B2 is continuous. Since r(x) x for
each x S1, r is a retraction of B2 onto S1. This contradicts Theorem 8.22. Now we introduce a property of subspaces that is stronger than retract.
Definition. A subspace A of a topological space (X, ) is a deformation retract of
X provided there is a continuous function H: X I → X such that H(x, 0) x and
H(x, 1) A for all x X and H(a, t) a for all a A and t I. The homotopy H
is called a deformation retraction. ˇ
EXAMPLE 8. Let X {(x, y) : 1 x2 y2 4}. Define H: X x
I → X by H(x, t) (1 t)x t x
for each (x, t) X I. Then H is conx
tinuous, H(x, 0) x and H(x, 1) x
S1 for all x X, and H(x, t) x for all
1
x S and t I. Therefore H is a deformation retraction, and hence S1 is a deformation retract of X.
Using Example 8 and Theorem 8.18, the following theorem tells us that the fundamental group of the space X in Example 8 is isomorphic to the group of integers.
8.5 Applications and Additional Examples of Fundamental Groups 289
THEOREM 8.24. If A is a deformation retract of a topological space (X, ) and
a A, then 1(X, a0) is isomorphic to 1(A, a0).
Proof. Let H: X I → X be a continuous function such that H(x, 0) x and
H(x, 1) A for all x X and H(a, t) a for all a A and t I. Define h: X → A
by h(x) H(x, 1) for each x X, and let h * : 1(X, a0) → 1(A, a0) be the homomorphism, given by Theorem 8.10, induced by h. This means that h * is defined by
h * ([]) [h ] for all [] 1(X, a0).
Now we show that h * is one-to-one. Suppose [], [] 1(X, a) and h * ([]) h * ([]). Then [h ] [h ]. Since A X, h may be considered as a path in X.
We complete the proof that [] [] by showing that p h in X. Define F: I I → X by F(x, t) H((x), t) for all (x, t) I I. Then F(x, 0) H((x), 0) (x) and F(x, 1) H((x), 1) (h )(x) for all x I, and F(0, t) H((0), t) H(a0, t) a0 and F(1, t) H((1), t) H(a0, t) a0 for all t I. Since [] is an
arbitrary member of 1(X, a0), this also shows that p h in X. Therefore, in
1(X, a0), we have [] [h ] [h ] [], and so h * is one-to-one.
Now let [] 1(A, 0). Then : I → A and since A X, we may also consider to be a path in X. Let []X denote the member of 1(X, a0) that contains .
Since h(a) H(a, 1) a for all a A, (h )(x) (x) for all x I. Therefore
h * ([]X) [h ] [], and so h * maps 1(X, x0) onto 1(A, a0).
Therefore h * is an isomorphism. By Example 8, S1 is a deformation retract of X {(x, y) : 1 x2 4}. Therefore, by Theorems 8.24 and 8.18, the fundamental group of X is isomorphic to the group of integers.
y2
EXAMPLE 9. Let p (p1, p2) , and let S {(x, y) R: (x p1)2 (y p2)2 1}. In Exercise 1, you are asked to show that S is a deformation retract of
{p}. Thus it follows that the fundamental group of – {p} is isomorphic
to the group of integers.
Since the torus is homeomorphic to S1 S1, we can determine the fundamental
group of the torus by proving a theorem about the fundamental group of the Cartesian
product of two spaces. The definition of the direct product of two groups is given in
Appendix I.
THEOREM 8.25. Let (X, ) and (Y, ) be topological spaces and let x0 X and
y0 Y . Then 1(X Y, (x0, y0)) is isomorphic to 1(X, x0) 1(Y, y0).
Proof. Let p: X Y → X and q: X Y → Y be the projection maps (since 1 is
used as part of the symbol to denote the fundamental group of a space, we now use
different symbols to denote the projection maps, but p and q are the functions that
we have previously denoted by 1 and 2), and let p * : 1(X Y, (x0, y0)) →
1(X, x0) and q * : 1(X Y, (x0, y0)) → 1(Y, y0) denote the induced homeomorphisms. Define : 1(X Y, (x0, y0)) → 1(X, x0) 1(Y, y0) by ([]) ([p ],
290 Chapter 8
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The Fundamental Group and Covering Spaces
[q ]). Since [p ] p * ([]) and [q ] q * ([]), by Exercise 12 of Appendix I,
is a homomorphism.
We use Exercise 13 of Appendix I in order to prove that is one-toone. Let [] 1(X Y, (x0, y0)) such that ([]) is the identity element of
1(X, x0) 1(Y, y0). This means that [p ] is the identity element of 1(X, x0)
and [q f ] is the identity element of 1(Y, y0). Then p p ex0 and q p ey0,
where ex0 and ey0 are the constant paths defined by ex0(t) x0 and ey0(t) y0 for
all t I. Let F: I I → X and G: I I → Y be the homotopies such that
F(x, 0) (p )(x) and F(x, 1) x0 for all x I, F(0, t) F(1, t) x0 for all
t I, G(x, 0) (q )(x) and G(x, 1) y0 for all x I, and G(0, t) G(1, t) y0 for all t I. Define H: I I → X Y by H(x, t) (F(x, t), G(x, t)) for all
(x, t) I I. Then H(x, 0) (F(x, 0), G(x, 0)) ((p )(x), (q )(x)) (x)
and H(x, 1) (F(x, 1), G(x, 1)) (x0, y0) for all x X, and H(0, t) (F(0, t),
G(0, t)) (x0, y0) and H(1, t) (F(1, t), G(1, t)) (x0, y0) for all t I. Therefore
p e, where e is the constant path defined by e(t) (x0, y0) for all t I. Therefore by Exercise 13 of Appendix I, is one-to-one.
Now we show that maps 1(X Y, (x0, y0)) onto 1(X, x0) 1(Y, y0). Let
([], []) 1(X, x0) 1(Y, y0). Define : I → X Y by (t) ((t), (t)) for all
t I. Then [] 1(X Y, (x0, y0)), and ([]) ([p ], [q ]) ([], []),
and so is an isomorphism. If A is a subset of a set X, the inclusion map i: A → X is the function defined by
i(a) a for each a A. A homomorphism from a group G into a group H is called
the zero homomorphism if for each g G, (g) is the identity element of H. The following theorem is a special case of a famous theorem called the Van Kampen Theorem.
THEOREM 8.26. Let (X, ) be a topological space, let U and V be open subsets
of X such that X U V and U V is pathwise connected. Let x0 U V, and
suppose the inclusion maps i: U → X and j: V → X induce zero homomorphisms
i * : 1(U, x0) → 1(X, x0) and j * : 1(V, x0) → 1(X, x0). Then 1(X, x0) is the
trivial group.
Proof. Let [] 1(X, x0). By Theorem 4.4, there exist t0, t1, ..., tn such that
t0 0, tn 1, and, for each k 1, 2, ..., n, tk1 tk and ([tk1, tk]) is a subset of
one of U or V. Among all such subdivisions t0, t1, ..., tn of I, choose one with the
smallest number of members. We prove by contradiction that for each
k 1, 2, ..., n 1, f(tk) U V. (Note that if (I) is a subset of one of U or V,
then n 1, and we have nothing to prove.) Suppose k {1, 2, ..., n 1} and
(tk) U. Then neither ([tk1, tk]) nor ([tk, tk1]) is a subset of U, so ([tk1, tk]) ([tk, ti1]) V. Thus we can discard tk and obtain a subdivision with a smaller
number of members that still have the desired properties. Therefore for each
k 1, 2, ..., n 1, (tk) U. The same argument shows that (tk) V for each
k 1, 2, ..., n 1.
Since (t0) (tn) x0 U V, (tk) U V for each k 0, 1, ..., n.
8.5 Applications and Additional Examples of Fundamental Groups 291
For each k 1, 2, ..., n, define k: I → X by k(x) ((1 x)tk1 xtk)
for each x I. We show that k is path homotopic (in X) to a path that lies entirely
in U.
If k(I) U, define Hk: I I → X by Hk(x, t) k(x) for each (x, t) I I.
Suppose k(I) U. Then, since k(I) ([tk1, tk]), k(I) V. Since U V is
pathwise connected, there exist paths , in U V such that
(0) (0) x0, (1) k(0), and (1) k(1)( see Figure 8.11).
k (0) 5 (tk21)
k
x0
k (1) 5 (tk)
Figure 8.11
Then the path product ( * k) * (recall that is the path defined by
(x) (1 x) for each x I) is a loop in V at x0. Since the inclusion map
j: V → X induces the zero homomorphism, the loop ( * k * ) is path homotopic in
X to the constant loop e: I → X defined by e(x) x0 for each x I. Therefore, by
Exercise 2, k is path homotopic in X to the path * . Let Fk: I I → X be this path
homotopy. Then Fk(x, 0) k(x) and Fk(x, 1) ( * )(x) for each x I and
Fk(0, t) k(0) and Fk(1, t) k(1) for each t I. Define F: I I → X
by F(x, t) Fk((x tk1)(tk tk1), t) for each x [tk1, tk] and t I.
Since Fk1(1, t) k1(1) k(0) Fk(0, t) for each k 1, 2, ..., n and t I, F is
well-defined and continuous. Let x I. Then there exists k such that x [tk1, tk]. So
F(x, 0) Fk((x tk1)(tk tk1), 0) k((x tk1)(tk tk1))
((1 (x tk1)(tk tk1))tk1 ((x tk1)(tk tk1)tk))
(x),
and
F(x, 1) Fk((x tk1)(tk tk1), 1) ( * )((x tk1)(tk tk1)) U.
Also, F(0, t) F1(0, t) 1(0) (0) and F(1, t) Fn(1, t) n(1) (1) for
each t I. Finally, define a path by (x) F(x, 1) for each x I. Then is a
path in U and is path homotopic to . Since i * : 1(U, x0) → 1(X, x0) is the zero
homomorphism, is path homotopic in X to the constant loop e. Therefore p e,
and so 1(X, x0) is the trivial group. As a consequence of Theorem 8.26, we establish the important fact that if
2
n 1, Sn {(x1, x2, ..., xn1) n1: n1
i1 xi 1} is simply connected. In the
292 Chapter 8
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The Fundamental Group and Covering Spaces
proof, we use that if X {(x1, x2, ..., xn, xn1) n1: xn1 0}, then X is homemorphic to n.
THEOREM 8.27. For n 1, Sn is simply connected.
Proof. Let p and q be the members of Sn given by p (0, 0, ..., 0, 1) and
q (0, 0, ..., 0, 1), and let X {(x1, x2, ..., xn, xn1) n1: xn1 0}. First
we show that Sn {p} is homeomorphic to n. Define f : Sn {p} → X by
f(x1, x2, ..., xn, xn1) (1(1 xn1))(x1, x2, ..., xn, 0). (We can describe f geometrically as follows: For each x Sn {p}, f(x) is the point of intersection of the line
determined by p and x with X.) It is clear that f is a one-to-one continuous function
that maps Sn {p} onto X. For each (y1, y2, ..., yn, 0) X, let t 2(1 y21 … y2n), and define g: X → Sn {p} by g(y1, y2, ..., yn, 0) (ty1, ty2, ..., tyn, 1 t). Again
it is clear that g is a one-to-one continuous function. In Exercise 3, you are asked
to show that g is an inverse of f. This completes the proof that Sn {p} is homeomorphic to n.
Since the function h: Sn {q} → Sn {p} defined by h(x1, x2, ..., xn, xn1) (x1, x2, ..., xn, xn1) is clearly a homeomorphism, Sn {q} is also homeomorphic
to n.
Now let U Sn {p} and V Sn {q}.Then U and V are open subsets of Sn
and Sn U V. By Example 4, n is simply connected, and therefore the inclusion
maps i: U → Sn and j: V → Sn induce zero homomorphisms. In order to use Theorem 8.26 to conclude that Sn is simply connected, it is sufficient to show that U V
is pathwise connected. Now U V Sn {p, q}, and the homeomorphism f
(defined in the first paragraph of this proof) restricted to Sn {p, q} is a homeomorphism of Sn {p, q} onto X {(0, 0, ..., 0)}, and X {(0, 0, ..., 0)} is homeomorphic
to n {(0, 0, ..., 0)}. We show that n {(0, 0, ..., 0)} is pathwise connected. Let
x n {(0, 0, ..., 0)}. If x (y, 0, 0, ..., 0), where y 0, then x and (1, 0, 0, ..., 0)
can be joined by a straight-line path. If x (y, 0, 0, ..., 0), where y 0, then x and
(0, 1, 0, 0, ..., 0)
can
be
joined
by
a
straight-line
path,
and
(0, 1, 0, 0, ..., 0) and (1, 0, 0, ..., 0) can be joined by a straight-line path. In either
case, x and (1, 0, 0, ..., 0) can be joined by a path. Therefore n {(0, 0, ..., 0)} is
pathwise connected. In the proof of Theorem 8.27, we showed that if n 2 then n {(0, 0, ..., 0)}
is simply connected. In Example 9, we showed that 2 {(0, 0)} is not simply
connected. Therefore by Theorem 8.14, if n 2, then n {(0, 0, ..., 0)} is not
homeomorphic to 2 (0, 0)}.
EXERCISES 8.5
1.
Let p (p1, p2) , and let S {(x, y) : (x p1)2 (y p2)2 1}. Prove that S is a deformation retract of {p}.
8.6 Knots 293
2.
Let (X, ) be a topological space, let x0 X, and let , , and be paths in
X such that (0) x0, (1) (0), (0) x0, and (1) (1), and let
H: I I → X be a continuous function such that H(x, 0) (( * ) *
)(x) and H(x, 1) x0 for each x I and H(0, t) H(1, t) x0 for each
t I. Prove that there is a continuous function F: I I → X such that
F(x, 0) (x) and F(x, 1) ( * )(x) for each x I and F(0, t) (0)
and F(1, t) (1) for each t I.
3.
Let X {(x1, x2, ..., xn, xn1) n1: xn1 0} and let p (0, 0, ..., 0, 1) Sn. Define f : Sn {p} → X by f(x1, x2, ..., xn1) (1(1 xn1))(x1, x2, ...,
xn, 0), and define g: X → Sn {p} by g(y1, y2, ..., yn, 0) (ty1, ty2, ..., tyn,
1 t), where t 2(1 y21 y22 ... y2n). Prove that g is an inverse of f .
4.
Show that the function r: B2 → S1 defined in the proof of Theorem 8.23 is
continuous.
8.6 Knots
We conclude this chapter with a brief discussion of knot theory. No proofs are given in
the text. This section is somewhat unique in that it is not rigorous. Instead, it is
designed to appeal to the geometric and intuitive side of topology. We refer to the onepoint compactification of a topological space, which is defined in Section 6.6. In keeping with the spirit of this section, one can consider the intuitive nature of the one-point
compactification rather than approach it from a rigorous viewpoint. A reference is provided at the end of the section for those who wish to continue the study of knot theory.
An invariant of knot theory was first considered by Karl Friedrick Gauss
(1777–1855) in 1833. Early work on the subject was done by Max Dehn
(1878–1952) in 1910, J.W. Alexander in 1924, E. Artin in 1925, E.R. van Kampen
in 1928, and others. The first comprehensive book on knot theory was written by
K. Reidemeister in 1932.
The overhand knot and the figure-eight knot (see Figure 8.12) are familiar to
almost everyone. A little experimenting with a piece of string should convince you
that one of these knots cannot be transformed into the other without untying one of
them. One goal of knot theory is to prove mathematically that this cannot be done.
Thus we must have a mathematical definition of a knot and of when two knots are
considered to be the same. The latter definition must prevent untying. We do this by
Overhand knot
Figure-eight knot
Figure 8.12
294 Chapter 8
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The Fundamental Group and Covering Spaces
getting rid of the ends; that is, we splice the ends together. The overhand knot with
the ends spliced together is often called the trefoil or cloverleaf knot. The figureeight knot with the ends spliced together is often called the four-knot or Listing’s
knot. These are shown in Figure 8.13.
Trefoil or cloverleaf knot
Figure-eight or Listing’s knot
Figure 8.13
Definition. A subset K of 3 is a knot if there is a homeomorphism that maps S1
onto K.
Three additional knots are shown in Figure 8.14.
Stevedore’s knot
True lover’s knot
Square knot
Figure 8.14
8.6 Knots 295
Notice that any two knots are homeomorphic. Thus the question of when two
knots are considered to be the same is a characteristic of the way in which the two
knots are embedded in 3. Thus knot theory is a portion of 3-dimensional topology.
Definition. Two knots K1 and K2 are equivalent if there is a homeomorphism
h: 3 → 3 such that h(K1) K2. It is easy to show that knot equivalence is an equivalence relation. Notice that
knot equivalence does not say anything about “sliding” (as is the case with a homotopy) K1 until it lies on top of K2. The difference in these ideas is illustrated by the
fact that a reflection about a plane is a homeomorphism of 3 onto 3 that maps a
knot into its mirror image, but we cannot “slide” the trefoil knot into its mirror image
(see Figure 8.15).
Trefoil knot
Mirror image of trefoil knot
Figure 8.15
Definition. A polygonal knot is a knot that is the union of a finite number of closed
straight-line segments called edges. The endpoints of the straight-line segments are
called vertices. A knot is tame if it is equivalent to a polygonal knot, and otherwise
it is wild. An example of a wild knot (one that is obtained by tying an infinite number of
knots one after the other) is shown in Figure 8.16.
Wild knot
Figure 8.16
In order to picture knots and work with them effectively, we project them into a
plane.
296 Chapter 8
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The Fundamental Group and Covering Spaces
Definition. Let K be a knot in 3 and define p: 3 → 3 by p((x, y, z)) (x, y, 0).
A point x p(K) is called a multiple point if p1(x) K consists of more than one
point. The order of x p(K) is the cardinality of p1(x) K . A double point in
p(K) is a point of order 2, and a triple point in p(K) is a point of order 3. In general, p(K) can contain any number and kinds of multiple points, including
multiple points of infinite order. In the remainder of this section, p denotes the projection in the preceding definition.
Definition. A polygonal knot K is in regular position if: (1) there are only a finite
number of multiple points and they are double points, and (2) no double point is the
image of a vertex of K . In Exercise 1, you are asked to give an example of a polygonal knot in regular
position.
Definition. Let K be a polygonal knot in regular position. Then each double point
in p(K) is the image of two points of K . The one whose z-coordinate is larger is
called an overcrossing and the one whose z-coordinate is smaller is called an
undercrossing. In Exercise 2, you are asked to indicate the overcrossings and undercrossings of
the knot you have given in Exercise 1.
A rigorous proof of the following theorem is somewhat tedious, and thus we omit
it. In Exercise 3 and 4, you are asked to illustrate this theorem.
THEOREM 8.28. If K is a polygonal knot, then there is an arbitrarily small rotation of 3 onto 3 that maps K into a polygonal knot in regular position. It is, of course, an immediate consequence of Theorem 8.28 that every polygonal knot is equivalent to a polygonal knot in regular position.
Definition. A homeomorphism h: 3 → 3 is isotopic to the identity if there is a
homotopy H: 3 I → 3 such that H(x, 0) h(x) and H(x, 1) x for each
x 3 and, for each t I, H(3 {t}): 3 {t} → 3 is a homeomorphism. The
homotopy H is called an isotopy. Suppose K1 and K2 are knots and h: 3 → 3 is a homeomorphism that is isotopic to the identity under an isotopy H. Moreover, suppose that h(K1) K2. For
each t I, let ht H(3 {t}). Then {ht(K1): t I} is a “continuous” family of knots
that move from K2 to K1 as t moves from 0 to 1. (Obviously, we are not being precise
here. Instead we are trying to provide a “picture” of how one knot is moved onto
another under an isotopy).
We can extend the definition of a triangulation of a compact surface (given in
Section 6.4) to a compact 3-manifold.
ˇ
8.6 Knots 297
Definition. A triangulation of a compact 3-manifold (X, ) consists of a finite collection {T1, T2, ..., Tn} of closed subsets of X that cover X and a collection
{i : Ti → Ti : i 1, 2, ..., n} of homeomorphisms, where each Ti is a tetrahedron in
the plane. Each Ti is also called a tetrahedron. The members of Ti that are images of
the vertices of Ti are called vertices, the subsets of Ti that are images of the edges
of Ti are called edges, and the subsets of Ti that are images of the 2-faces of Ti are
called 2-faces. In addition, for each pair Ti and Tj of distinct tetrahedra
Ti Tj , Ti Tj is a vertex, Ti Tj is an edge, or Ti Tj is a 2-face. This is not the usual definition of a triangulation of a manifold, but for our purposes this definition will suffice.
Let T be an edge, a triangle, or a tetrahedron. Consider two orderings of the vertices of T to be equivalent if they differ by an even permutation. Then there are
exactly two equivalence classes, each of which is called an orientation of T. If T is a
triangle or a tetrahedron that has been assigned an orientation by ordering its vertices in some way, say v0, v1,..., vk, and S is the face of T obtained by deleting vi, then
the vertices of S are automatically ordered. If i is even, the orientation of S given by
this ordering is called the orientation induced by T. If i is odd, the other orientation of S is called the orientation induced by T. For example, if T is a triangle with
vertices v0, v1, and v2, the orientation induced by T of the edges whose vertices are v1
and v2 is the one given by the ordering v1, v2, whereas the orientation induced by T
of the edge whose vertices are v0 and v2 is the one given by the ordering v2, v0.
Definition. Let (X, ) be a compact 2- (or 3-) manifold, and let T be a triangulation of X. We say that T is orientable if it is possible to orient the triangles (or tetrahedra) of T in such a way that two triangles with a common edge (or two tetrahedra
with a common 2-face) always induce opposite orientations on their common edge
(or 2-face). EXAMPLE 10. Let (X, ) be a compact 2-manifold with triangulation T shown
in Figure 8.17(a). Then the orientation shown in Figure 8.17 (b) shows that T is
orientable.
(a)
(b)
Figure 8.17
In Example 18 of Chapter 6, we showed that the one-point compactification of
is homeomorphic to S1. It is also true that the one-point compactification of 3 is
298 Chapter 8
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The Fundamental Group and Covering Spaces
homeomorphic to S3 (see Exercise 5). By Theorem 6.39, the one-point compactifications of two homeomorphic topological spaces are homeomorphic. We proved this
theorem by showing that if (X1, 1) and (X2, 2) are topological spaces, f : X1 → X2
is a homeomorphism, and, for each i 1, 2, (Yi, i) is the one-point compactification of (Xi, i), then there is a homeomorphism g: Y1 → Y2 that is an extension of f .
It is clear, from the proof, that the homeomorphism g is unique. It follows from this
discussion that each homeomorphism h: 3 → 3 can be extended in a unique
way to a homeomorphism hˆ: S3 → S3. We say that h is orientation preversing if hˆ
preserves the orientation of S3. Otherwise we say that h is orientation reversing
(see Exercise 9).
Suppose K1 and K2 are equivalent knots. Then there is a homeomorphism
h: 3 → 3 such that h(K1) K2. Thus h(3 K1): 3 K1 → 3 K2 is a homeomorphism, and so equivalent knots have homeomorphic complements. Therefore
one possible way of distinguishing between knots is to consider the fundamental
groups of their complements.
Definition. If K is a knot, the fundamental group 1(3 K) is called the knot
group of K . We do not attempt to calculate knot groups, since it is a rather tedious process.
Instead we describe, in a rather vague intuitive way, the knot group of a polygonal
knot K in regular position. We break up the knot into overcrossings and undercrossings that alternate as we go around the knot. This is illustrated in Figure 8.18 for
the trefoil knot and the square knot. In this figure, the heavier lines represent
overcrossings.
Trefoil knot
Square knot
Figure 8.18
The bottom line is that there is associated with each overcrossing a generator
of 1(3 K). Thus if K1 denotes the trefoil knot and K2 the square knot, then
1(3 K1) has a presentation that consists of 3 generators and 1(3 K2) has a
presentation that consists of 7 generators. In general, if K has n overcrossings, then
there is a presentation of 1(3 K) that consists of n generators. Now the number
of undercrossings is the same as the number of overcrossings, and there is associated
with each undercrossing a relation. The last undercrossing, however, yields a relation
that is a consequence of the others, so 1(3 K) has a presentation that consists
of n generators and n 1 relations. So there is a presentation of 1(3 K1) that
8.6 Knots 299
consists of 3 generators and 2 relations and a presentation of 1(3 K2) that consists of 7 generators and 6 relations.
EXERCISES 8.6
1.
Give an example of a polygonal knot in regular position.
2.
Indicate the overcrossings and undercrossings of the knot you have given in
Exercise 1.
3.
Give an example of a polygonal knot that is not in regular position.
4.
Give an arbitrarily small rotation of 3 onto 3 that maps the knot you have
given in Exercise 3 into a polygonal knot in regular position.
5.
Prove that the one-point compactification of 3 is homeomorphic to S3.
6.
Give a triangulation of S3.
7.
Show that the triangulation of the torus given in Figure 6.9 is orientable.
8.
In Exercise 13 of Section 6.4 you were asked to give a triangulation of the
Klein bottle. Experiment and guess whether this triangulation is orientable.
9.
Give a example of an orientation-preserving homeomorphism h: 3 → 3
and an orientation-reversing homeomorphism k: 3 → 3.
References
Armstrong, M.A., Basic Topology, Springer-Verlag, New York, 1983.