Chapter 8 The and Covering Spaces

In the first seven chapters we have dealt with point-set . This chapter pro- vides an introduction to . 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 Appen- dix I). This is done in such a way that if (X, ) and (Y, ) are homeomorphic, then → GX and GY are isomorphic. Also, a f : X Y “induces” a homo- → morphism f : GX GY, and, if f is a , 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 that distinguishes between two nonhomeomorphic spaces. The group that we study is called the funda- mental group.

8.1 of Paths In Section 3.2 we defined paths and the path product of two paths. This section stud- ies an equivalence relation defined on a certain collection of paths in a . Review Theorems 2.15 and 2.27, because these results about continuous func- tions will be used repeatedly in this chapter and the next. The closed [0, 1] will be denoted by I.

Definition. Let (X, ) and (Y, ) be topological spaces and let f, g: X → Y be con- tinuous functions. Then f is homotopic to g, denoted by f g, if there is a continu- ous 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 ■ The Fundamental Group and Covering Spaces

In the preceding definition, we are assuming that I has the determined by the usual topology on . Thus X I has the product topology deter- mined 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 con- tinuous 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 func- tion 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)) andg (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 , 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)

H | (( 3/2, 1/2), (1, 0)) (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 func- tions 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

F(x, 2t), for all x X and 0 t 1 H(x, t) 2 { 1 G(x, 2t 1), for all x X and 2 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 andH(1, t) x1 for all t I . The function H is called a path homotopy between f and g. 268 Chapter 8 ■ 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 I 3 I

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 “cross- ing the hole” in the space. → If (X, ) is a topological space and x0 X, let (X, x0) { : I X: is con- tinuous 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 269

(I) (I) (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

F(2x, t), if 0 x 1 and 0 t 1 H(x, t) 2 { 1 G(2x 1, t), if 2 x 1 and 0 t 1.

Then H is continuous,

F(2x, 0), if 0 x 1 (2x), if 0 x 1 H(x, 0) 2 1 2 ( )(x) 1 1 1 * 1 {ˇ { G(2x 1, 0) if 2 x 1 1(2x 1), if 2 x 1 270 Chapter 8 ■ The Fundamental Group and Covering Spaces and

F(2x, 1), if 0 x 1 (2x), 0 x 1 H(x, 1) 2 2 2 ( )(x) 1 1 2 * 2 {ˇ { G(2x 1, 1) if 2 x 1 2(2x 1), 2 x 1 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,

(4x), if 0 x 1 1 4 ( * )(2x), if 0 x 2 1 1 [( ) ](x) (4x 1), if x * * {(2x 1), if 1 x 1 4 2 2 {(2x 1), if 1 x 1 2 and

1 (2x), if 0 x 2 1 3 [ * ( * )](x) (4x 2), if 2 x 4 {(4x 3), if 3 x 1. 4

Next we draw a picture to illustrate how we arrive at the definition of the desired path homotopy H .

1 3 (0, 1) ( 2 , 1) ( 4 , 1) (1, 1)

L

t

(0, 0)1 1 (1, 0) ( 4 , 0) ( 2 , 0) 8.1 Homotopy of Paths 271

If t I, then the intersection of the horizontal line L and the line segment joining 1 1 (4, 0) and (2, 1) has coordinates ((1 t)4, t), and the intersection of L and the line 1 3 segment joining (2, 0) and (4, 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 1 t , if 0 t 1 and 0 x (1 t) 4 1 t 2 t H(x, t) (4x 1 t), if 0 t 1 and x 4 4 4x 2 t 2 t , if 0 t 1 and x 1. { ( 2 t ) 4

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 defini- tion of the path homotopy H to show that * e p .

(0, 1) (1, 1)

L

e t

(0, 0) 1 (1, 0) ( 2 , 0) 272 Chapter 8 ■ The Fundamental Group and Covering Spaces

1 For each t I, the intersection of the line L with the line segment joining (2, 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

2x 1 t , if 0 x (1 t) 2 H(x, t) 1 t x , if x 1. { 0 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) 1 (1, 0) ( 2 , 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 1 t (2x), if 0 x 2 1 t 1 t H(x, t) (1 t), if x 2 2 1 t (2x 1), if x 1. { 2 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 alge- braic 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 . Poincaré published 30 books and over 500 papers on mathe- matics, and he is regarded as the founder of . 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 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 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 . If f, g: X → Y are continuous functions, then f is homotopic to g . 274 Chapter 8 ■ 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 equiva- lence relation on (X, x0) . → 2. Let (X, ˇ ) 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 homo- topic 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. (b) Prove that every contractible space is pathwise connected. 9. Prove Theorem 8.7. 10. Prove Theorem 8.8. n 11. Let X be a convex subset of and let x0 X . Prove that X is contractible to x0. 12. Let (G, , ) be a (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 iso- morphic 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 well- defined, 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 ■ 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 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 con- tinuous 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 compo- nent 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 topo- logical 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 X x → Y y *: 1( , 0) 1( , 0).

→ 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 suffi- * * cient 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 t y h h h (1, ) 0. Therefore p , and so * is well defined. h X x Now we show that * is a homomorphism. Let [ ], [ ] 1( , 0) . Then h h h h h h h *([ ] [ ]) *([ * ]) [ ( * )] and *([ ]) *([ ]) [ ] [ ] [(h ) * (h )]. Since

(h )(2x), if 0 x 1 (h ( ))(x) 2 ((h ) (h ))(x), * { 1 * (h )(2x 1), if 2 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 y Y h k X x → Y y h k x h k 0 , and let , : ( , 0) ( , 0) be maps such that rel 0 .Then * * . → 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 X x h h k k G I I → Y [ ] 1( , 0). Then *([ ]) [ ] and *([ ]) [ ] . Define : 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 ] . h k Thus * *.

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 k h k h maps. Then ( )* * * .

Proof. X x k h k h k h Let [ ] 1( , 0). Then ( )*([ ]) [( ) ] and ( * *)[ ]) k h k h *([ ]) [ ( )]. Since the composition of functions is associative (see Exer- k h k h k h k h cise 21 in Appendix C), ( ) ( ) . Therefore ( )* * * . 278 Chapter 8 ■ 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 iso- morphic 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 i x f g Y y → Y y g f X rel 0 . By Theorem 8.11, ( )*: 1( , 0) 1( , 0) and ( )*: X x → X x f g 1( , 0) 1( , 0) are the identity homomorphisms. By Theorem 8.12, ( )* f g g f g f f X x → Y y * * and ( )* * * . By Theorem C.2, *: 1( , 0) 1( , 0) 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 1 1 1 that x0 h (y0) . Since h h iX and h h 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 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 279

4. A topological space (X, ) is 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 func- tion r is called a 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) h h such that ( 1)* ( 0)* .

8.3 The Fundamental Group of the Circle Our goal in this section is to show that the fundamental group of the circle is isomor- phic 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 1 cos 2 x and sin 2 x are both positive, so p (U) n(n, n 4) . Moreover, for 1 each n , p 1 is a one-to-one function from the closed interval [n, n ] onto [n, n 4] 4 1 1 U. Thus, since [n, n ] is compact, p 1 is a homeomorphism of [n, n ] onto 4 [n, n 4] 4 1 U. Therefore p 1 is a homeomorphism of (n, n ) onto U. Throughout the (n, n 4) 4 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.

→ 1 THEOREM 8.15. Let : I S 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 1 such that (Ut) is a proper subset of S . Since I is compact, the open {Ut: t I} of I has a finite subcover {U1, U2, ..., Um} . If Ui I for some i, then (I) is a connected 1. proper subset of S If Ui I for any i, choose Ui so that 0 Ui . There exists t I such that t0, [0, t] Ui, and t Ui Uj for some j i . If Ui Uj I, 1 ([0, t] ) and ([t, 1]) are connected proper subsets of S . If Ui Uj I, choose t I such that tt, [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. Sup- 1 pose 0 i n and Pi1 is true. Let Vi1 denote the component of p ( [ti1, ti])) → that contains i1(ti1), and let pi1 p V . Then pi1: Vi1 ([ti1, ti]) is a → i 1 homeomorphism. Define i: [0, ti] by

(t), if 0 t t (t) i1 i1 i { 1 (pi1) ( (t)), if ti1 t ti .

Then i is continuous, i(0) x0, and p i [0, t ] . We must show that i is → i unique. Suppose i: [0, ti ] is a continuous function such that i (0) x0 and p t t t t P t i [0, ti ]. If 0 i1, then i( ) i( ) since i1 is true. Suppose i1 1 t ti. Since (p i) (t) (t) ([ti1, ti]), i(t) p ( ([ti1, ti])) . Therefore 1 i(t) Vi1 since i is continuous and Vi1 is a component of p ( ([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.

→ 1 Proof. For each x X, define x: I S 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 neigh- borhood 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 : t I} of I has a finite subcover {N , N ..., N } . Let M n M . Then M is a t t1 t2 tn i1 ti neighborhood of x0 , and, for each i 1, 2, ....., n, H(M Nti) is contained in a 1 proper connected subset of S . 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 func- → tion 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 1 Pj1 is true. Let Uj1 be a connected proper subset of S that contains H(M Ij1), 1 let Vj1 denote the component of p (Uj1) that contains Gj1(M {tj1}) , and → letpj1 p V . Then pj1: Vj1 Uj1 is a homeomorphism. Define Gj: M → j 1 [0, tj] by

G (x, t), if 0 t t G (x, t) j1 j1 j { 1 (pj1) (H(x, t)), 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) 1 Uj1, Gj(x, t) p (Uj1). Therefore, since Gj is continuous and Vj1 is a compo- 1 nent of p (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 ■ 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).

1 EXAMPLE 6. Define : I → S 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) ˇ2x for each x I. Therefore deg() (1) 2 .

1 THEOREM 8.17. Let 1 and 2 be loops in S at (1, 0) such that 1 p 2 . Then deg( 1) deg( 2).

→ 1 Proof. Since 1 p 2, there is a continuous function H: I I S 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 dis- crete 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.

1 THEOREM 8.18. 1(S , (1, 0)) is isomorphic to the group of integers.

1 → Proof. Define : 1(S , (1, 0)) by ([ ]) deg( ) . By Theorem 8.17, is 1 well-defined. Let [ 1], [ 2] 1(S , (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 defi- → nition, deg( 1) 1(1) and deg( 2) 2(1). Define : I by (2x), if 0 x 1 (x) 1 2 { 1 1(1) 2(2x1), if 2 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 )(2x), if 0 x 1 ( p )(x) 1 2 { 1 (p 2)(2x 1), if 2 x 1 (2x), if 0 x 1 1 2 ( )(x). { 1 1 * 2 2(2x 1), if 2 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. 1 Now we show that maps 1(S , (1, 0)) onto . Let z , and define a path → → 1 1: I by (t) zt for each t I. Then (0) 0 and (1) z, so p : I S 1 1 is a loop in S at (1, 0). Therefore [p ] 1(S , (1, 0)), and, by definition, deg(p ) (1) z. Therefore ([p ]) deg(p ) z . 1 Finally, we show that is one-to-one. Let [ 1], [ 2] 1(S , (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) t 2(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) → 1 2(1) for each t I . Therefore p F: I I S 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 calcu- late 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 1 p: → S 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 ■ The Fundamental Group and Covering Spaces

Definition. Let (E, ) and (B, ) be topological spaces, and let p: E → B be a con- tinuous surjection. If each member of B has a neighborhood that is evenly covered by p, then p is a covering map and E is 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 con- tinuous function f: X → E such that p ff (see Figure 8.8).

E

f' p

X B f

Figure 8.8

Notice that Theorem 8.15 provides a lifting of a path in S1 , where p is the spe- cific 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: ˇ } 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: 1 } be the collection of slices of p (U) . Now (ti1) is a member of exactly 1 one member V of . For t [ti1, ti], define (t) (p V) ( (t)) . Since → p V: 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 1 , ([ti1, ti]) p (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 . 8.4 Covering Spaces 285

Since (ti1) (ti1) V, ([ti1, ti]) V . Therefore, for each t [ti1, ti], 1 1 (t) V p ((t)). But V p ((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 func- → tion 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 F I{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 induc- tively 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)

m q1 p1 (I {0}) i1 j1 (Ii Jj) i1 (Ii Jq). We define F on Ip Jq .

There exists such that F(Ip Jq) U . Let {V: } be the 1 collection of slices of p (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 p V is a homeomorphism of V onto U . Since F is a lifting of F C, ((p V) F)(x) F(x) for all x D . For 1 x Ip Iq, define F(x) (p V) (F(x)) . Then F is a lifting of F (C (Ip Jq)) . There- fore, 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 1 lifting of F, F(0, t) p (b0) for all t I . Since {0} I is connected, F is continu- 286 Chapter 8 ■ The Fundamental Group and Covering Spaces

1 ous, and p (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 1 F(1, t) b1 for all t I , so F(1, t) p (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 Theo- → rem 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 → 1 1 8.3, and define p: S S 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 , → 1 and let p: E B be a covering map such that for some b0 B, p (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, con- nected space, let p: E → B be a covering map, and let C be a component of → E. Prove that p C: 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 . → 1 (a) Prove that there is a surjection : 1(B, b) p (b) . (b) Prove that if (E, ) is simply connected, then is a . 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 fundamen- tal 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 fundamen- tal 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.

2 1(B )

j * r*

id 1 1 1(S ) 1(S )

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 . r j S1 → S1 j S1 → Then : is the identity function, and hence ( )*: 1( , (1, 0)) 1 S rj r j 1( , (1, 0)) is the identity homomorphism. By Theorem 8.12, ( )* * * .Thus 288 Chapter 8 ■ The Fundamental Group and Covering Spaces r B2 S1 B2 * maps 1( , (1, 0)) onto 1( , (1, 0)) . This is a contradiction, because 1( , 1 (1, 0)) is the trivial group and 1(S , (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 2 2 x B . For each x B , let Lx denote the line segment that begins at f(x) and passes 1 through x, and let yx Lx S { 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 2 → 1 2 r : B S defined by r(x) yx for each x B 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 ˇI 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.

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 con- x 1 tinuous, H(x, 0) x and H(x, 1) x S for all x X, and H(x, t) x for all x S1 and t I. Therefore H is a deformation retraction, and hence S1 is a defor- mation retract of X. Using Example 8 and Theorem 8.18, the following theorem tells us that the fun- damental 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 h x H x x X h X a → A a by ( ) ( , 1) for each , and let *: 1( , 0) 1( , 0) be the homo- h. h morphism, given by Theorem 8.10, induced by This means that * is defined by h h X a *([ ]) [ ] for all [ ] 1( , 0) . h X a h Now we show that * is one-to-one. Suppose [ ], [ ] 1( , ) and *([ ]) 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 con- sider 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 h h X x A a *([ ]X) [ ] [ ], and so * maps 1( , 0) onto 1( , 0) . h Therefore * is an isomorphism. By Example 8, S1 is a deformation retract of X {(x, y) : 1 x2 y2 4}. Therefore, by Theorems 8.24 and 8.18, the fundamental group of X is iso- morphic to the group of integers.

2 EXAMPLE 9. Let p (p1, p2) , and let S {(x, y): R (x p1) 2 (y p2) 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 homeomor- * → phisms. Define : 1(X Y, (x0, y0)) 1(X, x0) 1(Y, y0) by ([ ]) ([p ], 290 Chapter 8 ■ The Fundamental Group and Covering Spaces

q p p q q [ ]). Since [ ] *([ ]) and [ ] *([ ]) , by Exercise 12 of Appendix I, is a homomorphism. We use Exercise 13 of Appendix I in order to prove that is one-to- one. 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 allt I . Let F: I I → X and G: I I → Y be the 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 . There- fore 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 fol- lowing 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 U x → X x j V x → X x X x *: 1( , 0) 1( , 0) and *: 1( , 0) 1( , 0). Then 1( , 0) 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

(1)5(t ) k k

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 byF(x, t) Fk((x tk1)(tk tk1), t) for each x [tk1, tk] and t I . SinceFk1(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 n n1 n1 2 n 1, S {(x , x , ..., x ) : 1} is simply connected. In the 1 2 n1 i1 xi 292 Chapter 8 ■ The Fundamental Group and Covering Spaces

n1 proof, we use that if X {(x1, x2, ..., xn, xn1) : xn1 0}, then X is home- morphic 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 n1 q (0, 0, ..., 0, 1), and let X {(x1, x2, ..., xn, xn1) : 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 geometri- cally 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 n 2 that maps S {p} onto X. For each (y1, y2, ..., yn, 0) X, let t 2(1 y1 … 2 → n yn), and define g: X S {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 homeo- morphic to n . n → n Since the function h: S {q} S {p} defined by h(x1, x2, ..., xn, xn1) n (x1, x2, ..., xn, xn1) is clearly a homeomorphism, S {q} is also homeomorphic ton . 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 Theo- rem 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 homeomor- phism 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

2 1. Let p (p1, p2) , and let S {(x, y) : (x p1) 2 (y p2) 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. n1 3. Let X {(x1, x2, ..., xn, xn1) : xn1 0} and let p (0, 0, ..., 0, 1) n n → S . Define f : S {p} X by f(x1, x2, ..., xn1) (1(1 xn1))(x1, x2, ..., → n xn, 0), and define g: X S {p} by g(y1, y2, ..., yn, 0) (ty1, ty2, ..., tyn, 2 2 2 1 t), where t 2(1 y1 y2 ... yn) . 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 one- point compactification of a topological space, which is defined in Section 6.6. In keep- ing 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 pro- vided at the end of the section for those who wish to continue the study of knot theory. An 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 ■ 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 figure- eight 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 3 → 3 h: 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 homo- topy) 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 ■ 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 pro- jection 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 rota- tion 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 polygo- nal 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 3 3 → 3 x and, for each t I, H (3 {t}): {t} is a homeomorphism. The homotopy H is called an isotopy. 3 → 3 Suppose K1 and K2 are knots and h: is a homeomorphism that is iso- topic to the identity under an isotopy H . Moreover, suppose that h(K1) K2 . For each t I, let ht H (3 {t})ˇ . 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-. 8.6 Knots 297

Definition. A triangulation of a compact 3-manifold (X, ) consists of a finite col- lection {T1, T2, ..., Tn} of closed subsets of X that cover X and a collection → { i : Ti Ti : i 1, 2, ..., n} of , 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 ofTi 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 pur- poses this definition will suffice. Let T be an edge, a triangle, or a tetrahedron. Consider two orderings of the ver- tices of T to be equivalent if they differ by an even . 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 ver- tices 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 orienta- tion 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 triangula- tion of X . We say that T is orientable if it is possible to orient the triangles (or tetra- hedra) 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 ■ The Fundamental Group and Covering Spaces homeomorphic to S3 (see Exercise 5). By Theorem 6.39, the one-point compactifica- tions 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 compactifica- → tion 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 3 3 3 3 → 3 → h: such that h(K1) K2 . Thus h ( K1): K1 K2 is a homeo- morphism, and so equivalent knots have homeomorphic complements. Therefore one possible way of distinguishing between knots is to consider the fundamental groups of their complements.

3 Definition. If K is a knot, the fundamental group 1( 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 undercross- ings 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 3 of1( K) . Thus if K1 denotes the trefoil knot and K2 the square knot, then 3 3 1( K1) has a presentation that consists of 3 generators and 1( K2) has a presentation that consists of 7 generators. In general, if K has n overcrossings, then 3 there is a presentation of 1( 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 3 that is a consequence of the others, so 1( K) has a presentation that consists 3 of n generators and n 1 relations. So there is a presentation of 1( K1) that 8.6 Knots 299

3 consists of 3 generators and 2 relations and a presentation of 1( K2) that con- sists 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 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 . 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.