THE FUNDAMENTAL GROUP AND THE BROUWER FIXED POINT THEOREM

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1 THE FUNDAMENTAL GROUP AND THE BROUWER FIXED POINT THEOREM NATHAN GILL Abstract. We introduce the concept of the fundamental group of a topological space and demonstrate its utility in classifying spaces by computing several examples. We also use the techniques developed in the paper to prove the Brouwer Fixed Point Theorem for the disk. We assume a knowledge of the basic elements of group theory and point-set topology. Contents. Motivation. Paths and Homotopies 3. The Fundamental Group 4 4. The Fundamental Group of S 5 5. The Fundamental Group of S n 7 6. The Brouwer Fixed Point Theorem For The Disk 8 Acknowledgments 9 References 9. Motivation One of topology s main concerns is determining whether or not two spaces are homeomorphic or homotopy equivalent (in this short paper we will not discuss the latter at length, but the reader should know that the techniques developed here are in fact more relevant to questions of homotopy equivalence than of homeomorphism). Often, obtaining an a rmative answer to a problem of this type is much more di cult than obtaining a negative one. Where the former involves constructing a homeomorphism (which can be very di cult for unfamiliar spaces), the latter only requires that we find a homeomorphism-invariant property that the two spaces do not share, e.g. connectedness, path connectedness, compactness, etc. These properties are su cient to di erentiate between spaces like R and R,but what about a more interesting pair, like the torus, S S, and the sphere, S? Intuition tells us that these spaces ought not be homeomorphic; the torus has a big gaping hole, while the sphere does not. Unfortunately, the familiar invariants of point-set topology prove inadequate here. We need a new invariant that can distinguish between two such spaces; the fundamental group, which we will develop in this paper, is one such invariant.

2 NATHAN GILL. Paths and Homotopies Recall from point-set topology the notion of a path in a topological space. Definition.. Let X be a topological space. A path in X is a continuous map f : I! X. The points f(0) and f() are called the initial and final points, respectively. If x 0 and x are the initial and final points, then f is a path from x 0 to x. If a path f has the same initial and final point x 0,thenf is called a loop based at x 0. We also have a natural way of combining two paths. Definition.. Suppose f is a path from x 0 to x and g is a path from x to x. We define the product of f and g, denoted f g, to be the path h defined by f(x) 0 apple x apple h(x) =. g(x ) apple x apple Note that in order to form the product of two paths, the starting point of the second path must be the ending point of the first path. To look at all possible paths in a space at once would be a daunting task. Instead, we take a more organized approach, in which we group paths into equivalence classes derived from the homotopy relation, whose definition follows. Definition.3. Let f : X! Y and f 0 : X! Y be continuous maps. We say that f and f 0 are homotopic, written f ' f 0, if there exists a continuous map F : X I! Y such that F (x, 0) = f(x) and F (x, ) = f 0 (x) for all x X. The map F is called a homotopy between f and f 0. Definition.4. Let f : I! X and f 0 : I! X be paths in X with the same initial and final points, x 0 and x respectively. We say that f and f 0 are path homotopic if there exists a homotopy F : I I! X satisfying the additional requirement that for all t I. F (0,t)=x 0 and F (,t)=x We will denote path homotopy in the same way that we denote homotopy. For the remainder of this paper, we will be concerned exclusively with path homotopy, so there will not be any confusion. As mentioned above, the notion of homotopy allows us to define a useful equivalence relation on the set of all paths in a particular space. Lemma.5. Path homotopy is an equivalence relation. Proof. Transitivity is the only property that is not entirely straightforward to check. If f ' g and g ' h, then the map F (x, t) 0 apple t apple H(x, t) =, G(x, t ) apple t apple where F is a path homotopy between f and g and G is a path homotopy between g and h, is a path homotopy between f and h.

3 THE FUNDAMENTAL GROUP AND THE BROUWER FIXED POINT THEOREM 3 Definition.6. Let f and g be two paths in a space X with f() = g(0), and let [f] and [g] denote the path homotopy classes of f and g. We define an operation on these classes by [f] [g] =[f g]. We must of course check that Lemma.7. The operation is well-defined. Proof. Let f, f 0, g, and g 0 be paths in X such that f() = g(0) and f 0 () = g 0 (0), and suppose f ' f 0 and g ' g 0. We will show that f g ' f 0 g 0. Let F and G be homotopies between f and f 0 and g and g 0, respectively. Define a function H : I I! X by F (s, t) 0 apple s apple H(s, t) = The map H is well-defined, since G(s,t). apple s apple F (,t)=f() = g(0) = G(0,t), and is continuous by the pasting lemma of point-set topology. We check that H is the desired path homotopy. Let s [0, ]. Then we have Similarly, for s [, ], H(s, 0) = F (s, 0) = f(s) =(f g)(s). H(s, 0) = G(s, 0) = g(s ) = (f g)(s). A similar argument shows that H(s, ) = (f 0 g 0 )(s) for all s I. Furthermore, H(0,t)=F (0,t)=f(0) = (f g)(0) = (f 0 g 0 )(0), and similarly for H(,t). We conclude that H is indeed a path homotopy between f g and f 0 g 0. Lemma.8. The operation has the following properties: If h : X! Y is a continuous map, and f and g are paths in X, then h (f g) =(h f) (h g). If h : X! Y is a continuous map, and F is a path homotopy between the paths f and g in X, then h F is a path homotopy between the paths h f and h g in Y. Existence of right and left identities. In particular, if f is a path in a space X from x 0 to x, then the constant path at x 0 is the left identity, and the constant path at x is the right identity. Existence of inverses. In particular, if f is a path in a space X from x 0 to x, then its inverse, or reverse, is the path f(s) =f( s). This path satisfies [f] [ f] =[e x0 ] and [ f] [f] =[e x ], where e x0 and e x denote the constant paths at x 0 and x. Associativity. These properties make the set of all path homotopy classes in a topological space into a groupoid, an algebraic structure with a partial function defined for only certain pairs of elements (as opposed to a binary operation that is defined for all pairs), but that otherwise satisfies the usual group axioms.

4 4 NATHAN GILL 3. The Fundamental Group As a group is the most basic and familiar algebraic structure, we would like to turn the groupoid associated with a topological space into a proper group. To do this, we consider only loops. Clearly, if f and g are loops based at x 0,thenf g is always defined and is itself a loop based at x 0. In addition, the left and right identities mentioned above coincide, as they are both equal to the constant path at x 0. This observation suggests the following definition. Definition 3.. Let X be a topological space, and let x 0 X. Endowed with the operation, the set of all path homotopy classes of loops based at x 0 forms a group, called the fundamental group of X relative to the base point x 0, and is denoted (X, x 0 ). Notice that the fundamental group is defined using a particular (and arbitrary) point from a space. Since we aim to use it as an invariant, our new construction will not be of much use if it varies from point to point. We should therefore hope that many familiar spaces do not have this undesirable property. This turns out to be the case. Lemma 3.. In a path connected space, the fundamental group is independent of base point up to isomorphism. Proof. Let x 0 and x be two points in a path connected space X, and let be a path between them. Define a map ˆ : (X, x 0 )! (X, x )by ˆ ([f]) = [ ] [f] [ ]. This map is well-defined since is well-defined, as we have already shown, and the path f is a loop based at x. First we check that ˆ is a homomorphism. Let [f], [g] (X, x 0 ). We have ˆ ([f] [g]) = [ ] ([f] [g]) [ ] =[ ] [f] ([ ] [ ]) [g] [ ] =([ ] [f] [ ]) ([ ] [g] [ ]) = ˆ ([f]) ˆ ([g]), as desired. We show that ˆ is in fact an isomorphism by exhibiting an inverse function. Consider the map ˆ : (X, x )! (X, x 0 ), defined by ˆ ([f]) = [ ] [f] [ ]. For any [f] (X, x 0 ), we have (ˆ ˆ )([f]) = ˆ ([ ] [f] [ ]) =[ ] ([ ] [f] [ ]) [ ] =[f]. A similar argument shows that (ˆ ˆ ) is the identity map of (X, x ). We therefore conclude that ˆ is an isomorphism. All of the spaces that we deal with in this paper are path connected, so we do not need to worry much about this point.

5 THE FUNDAMENTAL GROUP AND THE BROUWER FIXED POINT THEOREM 5 The following theorem demonstrates the utility of the fundamental group in classifying spaces. Theorem 3.3. Suppose h :(X, x 0 )! (Y,y 0 ) is a homeomorphism of the space X with the space Y. Then (X, x 0 ) and (Y,y 0 ) are isomorphic. Proof. Define a map h : (X, x 0 )! (Y,y 0 )by h ([f]) = [h The map h is a homomorphism because h ([f] [g]) = [h (f g)] = [(h f) (h g)] = h ([f]) h ([g]), and is an isomorphism because (h ) : (Y,y 0 )! (X, x 0 ), defined in the obvious way, is its inverse map. This theorem tells us that two spaces with di erent fundamental groups cannot be homeomorphic. f]. 4. The Fundamental Group of S Before we can compute the fundamental goup of S, we must introduce some preliminary notions. Definition 4.. Let p : E! B be a continuous surjective map. The map p is said to be a covering map, and E a covering space of B, if for each b B, there exists an open set U B such that b U and p (U) can be expressed as the union of a collection disjoint open sets {V }, called slices, each of which is mapped homeomorphically to U by p. The open set U is said to be evenly covered by p. The most crucial tool in this section comes in the following definition and lemma. Definition 4.. If p : E! B is a map and f : X! B is a continuous map, then a map f : X! E is called a lifting of f if p f = f. Lemma 4.3. Let p : E! B be a covering map with p(e 0 )=b 0.Anypathf : I! B beginning at b 0 has a unique lifting to a path f in E beginning at e 0. Proof of Existence. Since p is a covering map, each point of B has a neighborhood, call it U b that is evenly covered by p. Now the collection {p (U b )} bb is an open cover (since p is by definition surjective) of the compact space I. Therefore, using the Lebesgue number lemma of point-set topology, we form a partition {a 0,a,...,a n } of I with the property that for all i, wehave[a i,a i+ ] p (U b ) for some b B. Thenp([a i,a i+ ]) U b, so the image of each subdivision lies entirely within one of the sets U b. Now we construct the lifting. Let f(0) = e 0. Now suppose that f is defined for all x [0,a i ]. Let {V } denote the slices of the evenly-covered set U, whereu satisfies f([a i,a i+ ]) U. Since f(a i ) p (U), we must have f(a i ) V for some. Now we define, for x [a i,a i+ ], f(x) =(p V ) (f(x)). Since p V is a homeomorphism, (p V ) is continuous. If we continue in this fashion along the entire partition, the resulting function is continuous by the pasting lemma, and is a path in E beginning at e 0, with the property that p f = f.

6 6 NATHAN GILL Proof of Uniqueness. Suppose that ḡ is another lifting of the path f beginning at e 0. We will show that ḡ = f. Partition I using the same partition employed in the construction of f. Clearly, ḡ(0) = f(0) = e0. Suppose that ḡ(x) = f(x) for all x [0,a i ]. Now consider [a i,a i+ ]. Let U B be the open set such that f([a i,a i+ ]) U from the previous paragraph. For x [a i,a i+ ], we have ḡ(x) p (f(x)) p (U). Since ḡ is continuous, the set ḡ([a i,a i+ ]) is connected, and ḡ([a i,a i+ ]) p (U). Therefore, since the collection {V } (the slices of p (U)) consists of disjoint open sets, the set ḡ([a i,a i+ ]) lies entirely within one of the slices V. But ḡ(a i )= f(a i ) V for some particular slice V,soḡ([a i,a i+ ]) lies entirely within V. The image of x under ḡ is thus simply the point of V that maps to f(x) underp, whichisprecisely f(x). Lemma 4.4. Let p : E! B be a covering map with p(e 0 )=b 0. Any map f : I I! B with F (0, 0) = b 0 has a unique lifting to a continuous map F : I I! E with F (0, 0) = e 0.IfF is a path homotopy between two paths f and g, then F is a path homotopy between f and ḡ. Proof. The proof is similar to that of the previous theorem. We take su ciently small rectangles in I I in place of subintervals in I and again construct the lifting inductively using these pieces. We are now in a position to compute the fundamental group of S. Theorem 4.5. The fundamental group of S is isomorphic to the additive group of integers. Proof. Consider the map p : R! S given by p(x) = (cos( x), sin( x)), which is in fact a covering map. Let b 0 =(, 0). Define a map : (S,b 0 )! Z by ([f]) = f(), where f is the lifting of f induced by p beginning at 0. The map is well-defined because if f ' g, then f ' ḡ by the previous theorem, and hence f() = ḡ(). To show that is a homomorphism, let [f], [g] (S,b 0 ). We compute ([f] [g]). Let ([f]) = m and ([g]) = n. Consider the path h : I! R given by f(x) 0 apple x apple h(x) =. m +ḡ(x ) apple s apple This path is the lifting of f g beginning at 0, since f(x) 0 apple s apple (p h)(x) =, g(x ) apple s apple which is simply f g. Wethenhave ([f] [g]) = h() = m + n = ([f]) + ([g]). To complete the proof, we show that is a bijection. Let n Z. SinceR is path connected, there is a path f in R from 0 to n. Thenp f (S,b 0 ), and since f is the lifting of p f beginning at 0, we have ([p f]) = f() = n. Therefore, is surjective. Now suppose that ([f]) = ([g]). Then f and ḡ are paths in R with the same initial and final points. Since R is simply connected, there is a homotopy H between f and ḡ. Butthenp H is a homotopy between p f = f and p ḡ = g, and hence [f] =[g]. We conclude that is injective, and hence an isomorphism.

7 THE FUNDAMENTAL GROUP AND THE BROUWER FIXED POINT THEOREM 7 What about the torus? Since the fundamental group of S is Z, we might guess that the fundamental group of S S is Z Z. It is straightforward to show that the product of two covering maps is a covering map of the product of the codomains. Using this fact, the map p p is a covering map of S S, and we can use reasoning similar to that in the previous proof to show that the fundamental group of S S is indeed Z Z. 5. The Fundamental Group of S n As another example, we will compute the fundamental group of a sphere in n dimensions. In particular, we will see that the -sphere and the torus are not homeomorphic. Theorem 5.. Suppose X = U [ V, where U and V are open sets of X. Suppose that U \V is path connected, and that x 0 U \V. Let i and j be the inclusion maps of U and V. Then the images of the induced homomorphisms i and j (defined as in the proof of Theorem 3.3) generate (X, x 0 ). Proof. Let [f] (X, x 0 ). Using the Lebesgue number lemma, form a partition {a 0,a,...,a n } of I so that the image of each segment of the partition under f lies entirely in either U or V. Now from this partition, we remove all points a i such that f(a i ) / U \ V. This process yields a new partition {b 0,b,...,b m }, which in fact has the same property that the images of its segmenets lie entirely in either U or V.Toseethis,let[a i,a i ] and [a i,a i+ ] be two segments of the original partition. Assume without loss of generality that f([a i,a i ]) U, and suppose we remove the point a i. Since f(a i ) U, wemusthavef(a i ) / V. But then f([a i,a i+ ]) cannot be entirely contained in V, and thus must be entirely contained in U. We then have f([a i,a i+ ]) U. Let i be a path in U \ V from x 0 to f(a i ), and let F i denote the composition of the positive linear map of I onto [a i,a i ]withf (intuitively, we can visualize this as the portion of the path f between f(a i ) and f(a i ), but the formalism is needed so that we can perform the product operation in the next step). Then the product f i = i F i i is a loop based at x 0 that lies entirely in either U or V. Therefore, the element [f i ] (X, x 0 ) is in the image of either i or j.since [f] =[f ] [f n ], we see that (X, x 0 ) is indeed generated by the images of i and j. Corollary 5.. Suppose X = U [ V, where U and V are open sets of X, and U \ V is nonempty and path connected. If U and V are simply connected, then X is simply connected. Proof. Since U and V are path connected, and U \ V is nonempty, the space X is path connected as well, since given two points x,x X, we can form a path from x to a point x 0 U \ V, and then take the product with a path from x 0 to x to form a path from x to x. By the preceding theorem, the images of the inclusion maps of U and V generate (X, x 0 ). But because U and V have trivial fundamental goups, the image of the homomorphisms induced by both inclusion maps is simply the identity loop based

8 8 NATHAN GILL at x 0.So (X, x 0 ) is trivial. Since it is also path connected, the space X is simply connected. Before we prove the next theorem, note that S n {p}, wherep =(0, 0,...,0, ), is homeomorphic to R n. Intuitively, in three dimensions, we can imagine poking a hole in a sphere and stretching it out to cover the entire plane. The map f : S n {p}!r n given by f(x,...,x n+ )= (x,...,x n ) x n+ provides the explicit homeomorphism. Theorem 5.3. For n, the n-sphere S n is simply connected. Proof. Consider the sets S n {p} and S n {q}, wherep =(0,...,0, ) and q = (0,...,0, ). Both sets are open, and (S n {p}) [ (S n {q}) =S n. In addition, since they are homeomorhpic to R n, each subspace is simply connected. Restricting the homeomorphism defined above to (S n {p}) \ (S n {q}) =S n ({p, q}), we see that S n {p, q} is homeomorphic to R n 0, which is path connected for n. The intersection (S n {p}) \ (S n {q}) is then path connected, and so S n is simply connected by the preceding corollary. Corollary 5.4. The sphere S and the torus S S are not homeomorphic. Proof. Since it has a trivial fundamental group, the sphere S cannot be homeomorphic to the torus by Theorem 3.3. At this point, one might ask why it took so much e ort to prove that S is simply connected. A perhaps more obvious (but flawed) argument, which better reflects our intuition, proceeds as follows. Let f be a loop in S. Choose a point x S such that x/ f(i). Let g be the homeomorphism between S {x} and R mentioned above. Then g f is a loop in R, and since R is simply connected, there exists a path homotopy H between g f and a constant path. Then g H is a homotopy in S {x}, and therefore S,betweenf and a constant path. Where is the flaw? The construction is beyond the scope of this paper, but there does in fact exist a loop : I! I I that is surjective. In addition, there exists a continuous surjective map : I I! S (imagine stretching out the interior of I I and collapsing the boundary to a single point). Then isasurjective loop in S. In this case, we cannot pick a point x S that does not lie in the image of the loop in question. 6. The Brouwer Fixed Point Theorem For The Disk As a final application of the fundamental group, we will prove one case of the famous Brouwer Fixed Theorem. The theorem does in fact hold in higher dimensions, but to show this requires more advanced tools. We begin with a new definition. Definition 6.. A retraction of a space X onto a subspace A is a continuous map r : X! A such that r A = i A. The subspace A is called a retract of X.

9 THE FUNDAMENTAL GROUP AND THE BROUWER FIXED POINT THEOREM 9 Theorem 6.. There is no retraction of the disk B onto S. Proof. We proceed by contradiction. Suppose that there existed a retraction r : B! S. Then r j, where j is the inclusion map of S into B, would be the identity map of S. The homomorphism r j would then be the identity homomorphism of (S,b 0 ). This in turn would force j : (S,b 0 )! (B,b 0 ) to be injective. But this cannot be the case, since the fundamental group of B is trivial and has only one element, while that of S is not trivial and has more than one element. Theorem 6.3 (The Brouwer Fixed Point Theorem for the Disk). If f : B! B is a continuous map, then there is a point x B such that f(x) =x. Proof. We proceed by contradiction. Suppose that there existed a continuous map f : B! B such that for all x B,wehavef(x) 6= x. Define a function : B! S by sending x to the point of intersection of S and the ray extending from f(x) and passing through x. Recall from elementary geometry that two distinct points determine a line. Since x and f(x) are always distinct, the map is welldefined. A rigorous demonstration of the continuity of would require a substantial amount of messy algebra, so we will omit one. Intuitively, however, the continuity of f guarantees that by taking the points x and x to be su ciently close, the slopes of the rays passing through those points, and therefore (x ) and (x ), can be made arbitrarily close. Notice now that for x S,wehave (x) =x. Therefore, the restriction S is simply the identity map of S. The map is then a retraction of B onto S, which is a contradiction of the preceding theorem. We conclude that any continuous map from B to itself must have a fixed point. Acknowledgments. It is a pleasure to thank my mentor, Mary He, for her guidance in exploring this topic. Were it not for her helpful explanations of the material and reading and exercise suggestions, this paper would not have been possible. I would also like to thank Peter May and all the members of the University of Chicago Mathematics department who gave up their time and e ort to put on this REU. References [] James Munkres. Topology. Pearson [] David S. Dummit and Richard M. Foote. Abstract Algebra. Wiley [3] Allan Hatcher. Algebraic Topology. Cambridge University Press. 00.