Solutions to homework problems

Solutions to homework problems November 25, 2015 Contents 1 Homework Assignment # 1 1 2 Homework assignment #2 6 3 Homework Assignment # 3 9 4 Homework Assignment # 4 14 5 Homework Assignment # 5 20 6 Homework Assignment # 6 25 7 Homework Assignment # 7 32 8 Homework Assignment # 8 37 9 Homework Assignment #9 41 10 Homework Assignment # 10 46 1 Homework Assignment # 1 1. Show that the ɛ-δ definition of continuity of a map f : X Y between metric spaces is equivalent to the definition of continuity in terms of open subsets. Proof. Suppose f is ɛ-δ-continuous (i.e., continuous in the sense of the ɛ-δ definition of continuity) and that U is an open subset of Y. We need to show that f 1 (U) is an open subset of X. The definition of an open subset U of a metric space implies that U can be written as a union of open balls B i, i I. Then f 1 (U) = f 1 ( i I B i ) = i I f 1 (B i ), and hence it suffices to show that the inverse image of a ball B in Y is an open subset of X. So let B = B ɛ (y 0 ) = {y Y d(y, y 0 ) < ɛ} and let x 0 be an element of f 1 (B). Then 1

the continuity of f in the ɛ-δ sense implies that there is some δ > 0 such that B δ (x 0 ) f 1 (B ɛ (y 0 )). This shows that f is continuous. Conversely, assume that f is continuous (as defined via open sets). To show that f is ɛ-δ-continuous let x X and ɛ > 0. Then f 1 (B ɛ (f(x)) is an open subset of X containing x. Hence there is some δ > 0 such that the ball B δ (x) is contained in f 1 (B ɛ (f(x))). In other words, f(b δ (x)) B ɛ (f(x)), which implies y X d(y, x) < δ d(f(y), f(x)) < ɛ 2. a) Suppose T, T are topologies on a set X which are generated by bases B and B, respectively. Show that the topologies agree if and only if the bases B, B are equivalent in the sense that every B B is the unions of subsets belonging to B and every B B is the union of subsets belonging to B. b) Show that the metric topology on R n determined by the Euclidean metric is equal to the metric topology determined by the l 1 -metric d 1 (x, y) = n i=1 x i y i. c) Show that the product topology on R m R n (with each factor equipped with the metric topology) agrees with the metric topology on R m+n = R m R n. Hint: Use the freedom provided by part (b) to choose whether to work with the Euclidean metric or the l 1 -metric to make this easier. Proof. Part (a). Suppose U T, i.e., U is an open subset of X w.r.t. the topology T. Then U is the union of subsets B i belonging to B. By assumption, each B i is the union of subsets belonging to B, and hence U is the union of subsets belonging to B and consequently U T. Analogously, U T implies U T, and hence the topologies T and T coincide. Part (b). We recall that the metric topology on a set X equipped with a metric d(x, y) is generated by the collection of subsets of X consisting of the open balls B ɛ (x) = {y X d(y, x) < ɛ} of any radius ɛ > 0 around any point x X. Specializing to X = R n, let us denote by B ɛ (x) (resp. B ɛ(x)) the open ball of radius ɛ around x R n for the Euclidean metric d (resp. the l 1 -metric d 1 ), and by B (resp. B ) the collection of all balls B ɛ (x) (resp. B ɛ(x)). By part (a) it suffices to show that 1. every ball B ɛ (x) is a union of balls B δ (y) B ɛ(x), and 2. every ball B ɛ(x) is a union of balls B δ (y) B ɛ(x). Comparing balls defined w.r.t. the Euclidean metric d with those w.r.t. the metric d 1 will require inequalities comparing the distances d(x, y) and d 1 (x, y) for x, y R n. We recall that d(x, y) = x y, and d 1 (x, y) = x y 1, where v := v1 2 + + vn 2 and v 1 := v 1 + + v n is the standard norm (resp. the 1-norm) of a vector v R n. These norms are related by the inequalities v v 1 v 1 n v. 2

The first inequality follows from The second from v 2 = v 1 2 + + v n 2 ( v 1 + + v n )( v 1 + + v n ) = v 2 1. v 1 = v 1 + + v n n max 1 i n v i = n max 1 i n v i 2 n v 1 2 + + v n 2 = n v. We note that these inequalities imply B ɛ(x) B ɛ (x) and B ɛ/n (x) B ɛ(x). In particular, for y B ɛ (x) and δ := ɛ d(y, x) > 0, we have B δ(y) B δ (y) B ɛ (x). (1.1) Here the second inclusion is a consequence of the triangle inequality for the metric d, since if z B δ (y), then d(z, x) d(z, y) + d(y, x) < δ + d(y, x) = ɛ, and hence z B ɛ (x). This proves (1). Statement (2) is proved similarly: assume y B ɛ(x) and set δ := ɛ d(y, x) > 0. Then B δ/n (y) B δ(y) B ɛ(x), where the second inclusion is a consequence of the triangle inequality for the metric d 1. Part (c). For (x, y) R m R n = R m+n, the relationship between x 1, y 1 and (x, y) 1 is given by (x, y) 1 = x 1 + + x m + y 1 + + y n = (x, y) 1, which is simpler than the corresponding relationship between the Euclidean norms of these vectors. This equality implies in particular B ɛ 1 (x) B ɛ 2 (y) B ɛ 1 +ɛ 2 ((x, y)) and B ɛ((x, y)) B ɛ(x) B ɛ(y). (1.2) This motivates us to compare the d 1 metric topology on R m+n with the product topology of R m and R n, with each factor equipped with the d 1 metric topology. A basis for the metric topology for R m+n is given by the balls B ɛ((x, y)) R m+n. A basis for the product topology is given by U V, where U R m, V R n are subsets which are unions of open balls in R m and R n, respectively. In particular, any such product U V is a union of subsets of the form B ɛ 1 (x) B ɛ 2 (y) which shows that these products of balls form a basis for the product topology. As in part (b), the inclusions (1.2) imply that every open ball B ɛ((x, y)) is a union of product of balls, and that every product of balls is a union of balls. By part (a) this implies that the topologies they generate agree. 3. Let X, Y 1, Y 2 be topological spaces. Then the projection maps p i : Y 1 Y 2 Y i are continuous and a map f : X Y 1 Y 2 is continuous if and only if the component maps f i = p i f are continuous for i = 1, 2. 3

Proof. If U 1 is an open subset of Y 1 then p 1 1 (U 1 ) = U 1 Y 2 is an open subset of the Cartesian product Y 1 Y 2 and hence p 1 is continuous; similarly for p 2. Now assume that f : X Y 1 Y 2 is continuous. Then the i-th component map f i = p i f is continuous as a composition of the continuous maps p i and f. Conversely, let us assume that the component maps f 1 and f 2 are continuous. To show that f is continuous, it suffices to show that f 1 (B) is open for an open subset B Y 1 Y 2 belonging to the basis B for the product topology consisting of products B = U 1 U 2 of open sets U i Y i (since any open set is of the form U = B i for B i B and hence f 1 (U) = f 1 (B i ) which is an open subset of X, provided the subsets f 1 (B i ) are open). Now f 1 (U V ) = f1 1 (U) f2 1 (V ) is the intersection of open subsets of X (since f i is continuous and U i is an open subset of Y i ) and hence open. 4. Show that the map f : GL n (R) GL n (R), A A 1 is continuous (here GL n (R) R n2 is equipped with the metric topology). Proof. We recall from a lemma in class that f is continuous if and only if the composition GL n (R) f GL n (R) i R n2 is continuous, where i is the inclusion map. Moreover, this composition is continuous if and only if all component maps p ij i f are continuous (thinking of R n2 as the space of all n n-matrices, p ij for 1 i, j n is the projection map that takes a matrix M R n2 to its component M ij R). We conclude that it suffices to show that for 1 i, j n the map GL n (R) R, A (A 1 ) ij is continuous. We recall from linear algebra that the inverse of A can be calculated by the formula A 1 = Ct det(a), where det(a) is the determinant of A, and C t is the transpose of the n n matrix C whose entry C ij is ( 1) i+j times the (i, j)-minor of A (the determinant of the (n 1) (n 1) matrix that results from deleting row i and column j of A). This shows that each matrix entry (A 1 ) ij is of the form p(a), where p(a) and q(a) are q(a) polynomial functions of the matrix entries of A. In particular the functions p, q : GL n (R) R are continuous. Since q(a) = det(a) in non-zero for all A GL n (R), we conclude with a lemma from class that the function p(a) is continous. q(a) 5. Consider the following topological spaces The subspace T 1 := {v R 3 d(v, S) = r} R 3 equipped with the subspace topology, where S = {(x, y, 0) R 2 x 2 + y 2 = 1} and 0 < r < 1. The product space T 2 := S 1 S 1 equipped with the product topology. The quotient space T 3 := ([ 1, 1] [ 1, 1])/ equipped with the quotient topology, where the equivalence relation is generated by (s, 1) (s, 1) and ( 1, t) (1, t). 4

These three spaces are homeomorphic, but we don t yet have the tools to prove this in an efficient way. For this homework problem construct two continuous bijections between pairs of these spaces. Hint: An important part of this problem is to figure out which of these three spaces should be the (co)domains of the two maps you are looking for. Of course, any bijection has an inverse, but determining continuity is a different matter. Think about which type of topological space (subspace, product space, quotient space) you would prefer to have as domain (resp. codomain) for a map whose continuity you want to prove. Proof. We recall from class that checking continuity of a map f : X Y is easy if the domain X is a quotient space and the range Y is a subspace or a product space. This suggests to construct continuous bijections f : T 3 T 1 and g : T 3 T 2. We first discuss the simpler map g given by the formula g([s, t]) := ((cos πs, sin πs), (cos πt, sin πt)) S 1 S 1 R 2 R 2. (1.3) To argue that this is a well-defined bijection, we note that the map [ 1, 1] S 1, s (cos πs, sin πs) is surjective, and its only failure to be injective is due to the fact that it maps s = 1 and s = +1 to the same point in S 1. The map above when precomposed with the projection map [ 1, 1] [ 1, 1] T 3 is the product of two copies of this map. Hence it is surjective, and its only failure of injectivity comes from points (s, t) with s = ±1 or t = ±1. This shows that g is in fact a bijection. To prove continuity of g it suffices to prove continuity of the composition [ 1, 1] [ 1, 1] pr [ 1, 1] [ 1, 1]/{±1} g S 1 S 1 i R 2 R 2 = R 4, where pr is the projection map, the i is the inclusion map. This map is continuous, since all its component functions, given explicitely in equation (1.3) are continuous. Now we define the map f, first describing it geometrically. From this point of view, it will be obvious that f is bijective. Then we will derive an explicit formula for f, which will make it easy to argue that f is continuous. We note that there is a surjective map p: T 1 S which sends a point v R 3 to the unique point (x, y, 0) S which is closest to v. Similarly, there is a surjective map q : T 3 = ([ 1, 1] [ 1, 1])/ [ 1, 1]/{±1} given by [s, t] [s]. We want to construct the map f such that it fits into a commutative diagram T 3 = ([ 1, 1] [ 1, 1])/ f T 1, q [ 1, 1]/{±1} f 0 S p where f 0 is the homeomorphism we discussed in class that sends [s] to (cos πs, sin πs, 0) S R 3. We note that the fiber q 1 ([s]) for any fixed [s] [ 1, 1]/{±1} is a circle. Similarly, for any point (x, y, 0) of the circle S R 3, the fiber p 1 (x, y, 0) T 1 is a circle of radius r 5

in the plane spanned by the unit vectors (x, y, 0) and (0, 0, 1) with center (x, y, 0). We note that the map [ 1, 1]/{±1} R 3 [t] (x, y, 0) + r cos πt(x, y, 0) + r sin πt(0, 0, 1) = ((1 + r cos πt)x, (1 + r cos πt)y, r sin πt) is a bijection onto the circle p 1 (x, y, 0). In particular, defining f on each fiber q 1 ([s]) to be this map, we obtain a bijection between T 3 and T 1. To argue that f is continuous, we write down f : T 3 T 1 explicitly: f 0 sends a point [s] [ 1, 1]/{±1} to (x, y, 0) = (cos πs, sin πs, 0) and hence f([s, t]) =((1 + r cos πt)x, (1 + r cos πt)y, r sin πt) =((1 + r cos πt) cos πs, (1 + r cos πt) sin πs, r sin πt). (1.4) Then f is continuous if and only if the composition [ 1, 1] [ 1, 1] pr [ 1, 1] [ 1, 1]/{±1} f T 1 i R 3 is continuous. This is continuous since its component functions are, which can be read off from equation (1.4). 2 Homework assignment #2 1. Show that the quotient space D n /S n 1 is homeomorphic to the sphere S n. Proof. Generalizing the case n = 1 which we did in class, we define a map f : D n R R n = R n+1 by f(v) := (cos π v, v sin π v ) v sin πt We recall from Calculus that t is a continuous function on R \ {0} which extends to t a continuous function on all of R (by defining its value at t = 0 to be π). It follows that f is a continuous function since all its components are continuous. It s easy to check that f(v) belongs to S n R n+1 : f(0) = (0, 1) S n, and for v 0 we have f(v) 2 = cos 2 π v + sin 2 π v 2 = 1 We conclude that the map D n S n, v f(v) is continuous w.r.t. the subspace topology on S n since its composition with the inclusion map i: S n R n+1 is the map f. Abusing notation, we write again f : D n S n. We note that f maps any unit vector v D n to the point ( 1, 0) R n R. Hence f induces a well-defined map f : D n /S n 1 S n [v] f(v) This map is continuous since its composition with the projection map p: D n D n /S n 1 is the map f which is continuous. We won t present the details that f is bijective (if you can come up with a map which in fact is a bijection, I believe that you could prove that it is). 6

So f : D n /S n 1 S n is a continuous bijection, and hence it is a homeomorphism, since its range is Hausdorff (since S n is a metric space), and its domain is compact (D n R n is closed and bouneded and hence compact by the Heine-Borel Theorem; it follows that the quotient space D n /S n 1 is compact). 2. Use the Heine-Borel Theorem to decide which of the topological groups GL n (R), O(n), SO(n) are compact. Provide proofs for your statements. Hint: A strategy often useful for proving that a subset C of R n is closed is to show that C is of the form f 1 (C ) for some closed subset C R k (often C consists of just one point) and some continuous map f : R n R k. Proof. The general linear group GL n (R) M n n (R) = R n2 is not compact, since it is not bounded. To see this, consider the sequence of diagonal matrices i 1 A i =.... 1 Considered as a vector in R n2, the norm squared of A i is given by A i 2 = i 2 + 1 2 + + 1 2 = i 2 + n 1 which shows that GL n (R) is an unbounded subset of R n2. We claim that O(n) and SO(n) are compact. We begin by showing that they are bounded subspaces of R n2. So let A O(n) be a matrix with column vectors v 1,..., v n R n. The assumption that A belongs to O(n) means that the v i s form an orthogonal basis of R n. Hence A 2 = v 1 2 + + v n 2 = n, which shows that O(n) and hence also SO(n) O(n) are bounded subsets of R n2. To show that O(n) is a closed subset of R n2 we note that a matrix A with column vectors v i R n belongs to O(n) if and only if the v i s are orthogonal, i.e., if v i, v j = δ i,j for all 1 i, j n, where v, w denotes the scalar product of vectors v, w, and the delta symbol δ i,j is defined by declaring δ i,i = 1, and δ i,j = 0 for i j. More elegantly, this can be rephrased by saying A O(n) if and only if A t A = I, where A t is the transpose of A, and I is the identity matrix. In particular, if we define f : M n n M n n by A A t A, then O(n) = f 1 (I). Since f is continuous (since each component is), and the one-element subset {I} M n n = R n2 is closed, it follows that O(n) is a closed subset of R n2. To show that SO(n) is a closed subset, we note that SO(n) = O(n) SL n (R), where SL n (R) is the special linear group consisting of all n n matrices with determinant 1. In other words, SL n (R) = det 1 (1), where det: M n n R is the map that sends a matrix to its determinant. Since the determinant is a polynomial function of the entries of the matrix, this is a continuous function, and hence SL n (R) = det 1 (1) is a closed subset of R n2. This implies that the intersection SO(n) = O(n) SL n (R) is closed as well. 7

4. Let X be a topological space which is the union of two subspaces X 1 and X 2. Let f : X Y be a (not necessarily continuous) map whose restriction to X 1 and X 2 is continuous. (a) Show f is continuous if X 1 and X 2 are open subsets of X. (b) Show f is continuous if X 1 and X 2 are closed subsets of X. (c) Give an example showing that in general f is not continuous. Remark. This result is needed to verify that various constructions (e.g., concatenations of paths) in fact lead to continuous maps. In a typical situation, we have continuous maps f 1 : X 1 Y and f 2 : X 2 Y which agree on X 1 X 2 and hence there is a well-defined map { f 1 (x) x X 1 f : X Y given by f(x) = f 2 (x) x X 2 The above result then helps to show that this map is continuous. Proof. To prove part (a) let U Y be open. Then U i := f 1 i (U) is an open subset of X i, i.e., it is of the form U i = V i X i, where V i X is open. If follows that U i is an open subset of X, and hence f 1 (U) = f 1 1 (U) f 1 2 (U) = U 1 U 2 is an open subset of X, proving the continuity of f. To prove part (b) we note that f : X Y is continuous if and only if f 1 (U) is a closed subset of X for every closed subset U Y. Then repeating the previous sentences with open replaced by closed provides a proof of part (b). For part (c) consider the map f : X = R Y = R given by f(t) = 0 if t (, 0), and f(t) = 1 if t [0, ). The restrictions of f to (, 0) resp. [0, ) are constant and hence continuous, but f is not. 5. Which of the topological groups GL n (R), O(n), SO(n) are connected? Hint: Use without proof the fact that every element in SO(n) is represented by a matrix of block diagonal form whose diagonal blocks are the 1 1 matrix with entry +1 and/or 2 2 rotational matrices ( ) cos φ sin φ R =. sin φ cos φ Here block diagonal means that all other entries are zero. Proof. The determinant function det: M n n R is a continuous function. Since the determinant for matrices in GL n (R) or O(n) is non-zero, its restriction to G = GL n (R), O(n) provides a continuous map f : G R \ {0}. It follows that G = f 1 ((, 0)) f 1 ((0, )) is a decomposition of G into a disjoint union of open subsets. Both of these are not empty, since the determinant of the identity matrix is 1, and the determinant of the diagonal matrix with diagonal entries ( 1, 1,..., 1) is 1. This shows that the spaces GL n (R) and O(n) are both not connected. To prove that SO(n) is connected it suffices to show that SO(n) is path connected, i.e., any two elements A, B SO(n) can be connected by a path. We will show this by 8

constructing for every A SO(n) a path γ A : [0, 1] SO(n) with γ A (0) = I (the identity matrix) and γ A (1) = A. This implies that any two points A, B SO(n) can be connected by a path γ AB, obtained by first taking a path from A to I, by running the path γ A backwards, and then taking the path γ B from I to B. In formulas, the path γ AB is given by { γ A (1 2t) for t [0, 1 2 γ AB (t) = ] γ B (2t 1) for t [ 1, 1] 2 This is in fact a path (i.e., a continuous map) by the previous problem. Let us first construct γ A for n = 2, using the fact that every A SO(2) is a rotation, given by a matrix of the form ( ) cos θ sin θ A =. sin θ cos θ We define ( ) cos tθ sin tθ γ A : [0, 1] SO(2) by γ A (t) := sin tθ cos tθ which is continuous since it components are continuous functions of t, and has the desired property γ A (0) = I and γ A (1) = A. For a general dimension n, we use the fact that for any A SO(n) there is a choice of basis for R n such that the matrix representing the isometry A is of block diagonal form as described in the hint. Replacing for each block R i the rotation angle θ i (which may be different for the different blocks) by tθ i we obtain a path γ A : [0, 1] SO(n) from I to A, generalizing our construction in the n = 2 case. 3 Homework Assignment # 3 1. Show that the connected sum RP 2 #RP 2 of two copies of the real projective plane is homeomorphic to the Klein bottle K. Proof. Consider the topological spaces given by identifying egdes with the same label for the disjoint union of polygons given by the following pictures. b b c a a a c c a c a a c b c b b b b b The left picture is our standard construction of the Klein bottle K. The second and third picture shows the quotient obtained from two disjoint triangles by identifying edges with the same label. The second space is homeomorphic to the first since identifying the two edges labeled c in the second picture, we obtain the first. The third space is homeomorphic to the second, since the top right triangle of picture 2 is just redrawn as the left triangle of 9

picture 3 (flipped over such that the a-edge of that triangle lines up with the a-edge of the second triangle). The fourth space is homeomorphic to the third since it is obtained from the third space by identifying the two edges labeled a in the third picture. Reading off the word bbcc by starting at the top left corner the last picture and going around counterclockwise, the topological space given by that picture is Σ(bbcc). From class we know the homeomorphisms Σ(bbcc) Σ(bb)#Σ(cc) RP 2 #RP 2. Composing all the homeomorphisms mentioned above we obtain the desired homeomorphism between the Klein bottle K and the connected sum RP 2 #RP 2. 2. a) Show that the surface of genus g and the connected sum Σ g = T #... #T }{{} g RP 2 #... #RP 2 }{{} k are both homeomorphic to the topological space Σ(W ) associated to a word W. What is the word W in these two cases? b) Calculate the Euler characteristic of these manifolds. Proof. Part (a) From class we know that T Σ(aba 1 b 1 ), RP 2 Σ(cc), and Σ(W )#Σ(W ) if W, W are words from disjoint alphabets. This implies and Σ g = T #... #T Σ(a 1 b 1 a 1 1 b 1 1 )#... #Σ(a g b g a 1 g Σ(a 1 b 1 a 1 1 b 1 1... a g b g a 1 g b 1 g ) b 1 g ) (3.1) RP 2 #... #RP 2 Σ(a 1 a 1 )#... #Σ(a k a k ) Σ(a 1 a 1... a k a k ). (3.2) Part (b). The homeomorphism 3.1 implies that Σ g is the quotient space of a polygon with 4g edges (the length of the word a 1 b 1 a 1 1 b 1 1... a g b g a 1 g b 1 g ) and edge identification encoded by this word. After gluing, this polygon will provide a pattern of polygons on Σ g with one face and 2g edges (of the original 4g edges of the polygon, the pairs of edges with the same label are glued, resulting in 2g edges after gluing). Each of the original 4g vertices is indentified with any other by gluing, resulting in one vertex in the quotient space (this is easy to check for the first four vertices corresponding to the first torus, and then arguing inductively). It follows that χ(σ g ) = #V #E + #F = 1 2g + 1 = 2 2g. 10

The argument for RP 2 #... #RP 2 is completely analogous. The homeomorphism 3.2 implies that this surface is the quotient space of a 2k-gon with edge identifications encoded in the word a 1 a 1... a k a k. Using the resulting pattern of polygons on RP 2 #... #RP 2 to calculate its Euler characteristic, we obtain χ(rp 2 #... #RP }{{} 2 ) = #V #E + #F = 1 k + 1 = 2 k. k 3. Important examples of quotient spaces are orbit spaces of a group G acting on a topological space X. We recall that a (left) of a group G on a set X is given by a map G X X typically written as (g, x) gx, such that g 1 (g 2 x) = (g 1 g 2 )x for g 1, g 2 G, x X (associativity) and ex = x for e the identity element element of G, x X (identity element property). Given a G-action on a topological space X the orbit space denoted X/G is the quotient space X/ where two elements x, y X are declared equivalent if and only if there is some g G with gx = y. In particular, the equivalence class of x is the subset {gx g G}, which is called the orbit of x, and hence X/G is the space of orbits. Although we haven t used this terminology, we ve already encountered orbit spaces, namely Here the actions are given by RP n = S n /{±1} and CP n = S 2n+1 /S 1. {±1} S n S n (t, (v 0,..., v n )) (tv 0,..., tv n ) and S 1 S 2n+1 S 2n+1 (z, (z 0,..., z n )) (zz 0,..., zz n ), where z S 1 C and (z 0,..., z n ) S 2n+1 C n+1. (a) Consider the action Z 2 R 2 R 2, (m, n), (x, y) (x + m, y + n). Show that the quotient space R 2 /Z 2 is homeomorphic to the torus, described as the quotient space b a a b Use without proof the fact that this orbit space is Hausdorff (this will come up later this semester). (b) Consider the action G R 2 R 2 where G is the subgroup of the group of isometries of the metric space R 2 generated by the isometries g, h: R 2 R 2 defined by g(x, y) = (x + 1, y) and h(x, y) = ( x, y + 1) 11

Show that the quotient space R 2 /G is homeomorphic to the Klein bottle, described as the quotient space of the square [0, 1] [0, 1] with edge identifications b a a b Again, use without proof the fact that this orbit space is Hausdorff. Hint: Show that every orbit can be represented by a point (x, y) [0, 1] [0, 1]. To do this, it might be helpful to argue that the composition ghgh 1 is the identity and to use this to show that every element of G can be uniquely written in the form g m h n with m, n Z. Proof. Part (a). We recall that the torus T is the quotient (I I)/ where I = [0, 1] and the equivalence relation is generated by (x, 0) (x, 1) and (0, y) (1, y). Consider the map f : T = (I I)/ R 2 /Z 2 [(x, y)] Z 2 (x, y), where we write Z 2 (x, y) for the orbit of the Z 2 -action through (x, y). Next we will argue that f is well-defined, surjective, injective and continuous. This implies that f is a homeomorphism since its domain is compact (as the image of the closed bounded subset I I R 2 under the projection map I I (I I)/, and we use the fact that R 2 /Z 2 is Hausdorff. The map f is well-defined since the points (x, 0), (x, 1) I I belong to the same orbit of the Z 2 -action on R 2, since (0, 1)(x, 0) = (x, 1). Similarly, (0, y) and (1, y) belong to the same Z 2 -orbit. For every (x, y) R 2 the element (x [x], y [y]) belongs to I I, where [x] Z is the largest integer smaller or equal to x R. Since ([x], [y]) Z 2 applied to (x [x], y [y]) gives (x, y), this shows that (x [x], y [y]) and (x, y) belong to the same orbit and hence the map f is surjective. Two elements (x, y), (x, y ) R 2 belong to the same Z 2 -orbit if and only if x x Z and y y Z. Hence the only points (x, y) I I mapping to the same orbit are of the form (0, y), (1, y) or (x, 0), (x, 1). Since these points are identified in I I/, the map f is injective. To prove continuity of f consider the commutative diagram I I i R 2. p (I I)/ f R 2 /Z 2 The inclusion map i and the projection maps p, q are continuous, and so the composition q i is continuous. By commutativity of the diagram it follows that f p = q i is continuous, and hence f is continuous. 12 q

Part (b). The proof is analogous to that of part (a). Explicitly the Klein bottle K is the quotient (I I)/ where the equivalence relation is generated by (x, 0) (1 x, 1) and (0, y) (1, y). Consider the map f : T = (I I)/ R 2 /Z 2 [(x, y)] G(x, y), where we write G(x, y) for the orbit of the G-action through (x, y). We will prove this in the same steps as in (a), but first we prove the following result. Lemma 3.3. The G-orbit G(x, y) of a point (x, y) R 2 is given explicitly by G(x, y) = {(m + ( 1) n x, n + y) m, n Z}. To prove this result we first verify that the relation ghgh 1 = 1 holds in G: (x, y) h 1 ( x, 1 + y) g (1 x, 1 + y) h ( (1 x), 1 + ( 1 + y)) = ( 1 + x, y) g (x, y) A general element of G is a word made from the letters g, g 1, h, h 1. proved implies hgh 1 = g 1 which implies h ±1 g ±1 = g 1 h ±1. The relation just In particular, we can move all factors g ±1 to the left of the factors h ±1, which shows that every element of G can be written in the form g m h n for m, n Z. Since this proves the lemma. g m h n (x, y) = g m (( 1) n x, n + y) = (m + ( 1) n x, n + y), The map f is well-defined since by the lemma the points (x, 0), (x, 1) I I (resp. the points (0, y) and (1, y)) belong to the same G-orbit. The map f is surjective since by the lemma for the orbit through any point (x, y) R 2, there is a unique point (x, y ) [0, 1) [0, 1) in the same orbit as (x, y). For every (x, y) R 2 the element (x [x], y [y]) belongs to I I, where [x] Z is the largest integer smaller or equal to x R. Since ([x], [y]) Z 2 applied to (x [x], y [y]) gives (x, y), this shows that (x [x], y [y]) and (x, y) belong to the same orbit and hence the map f is surjective. Two elements (x, y), (x, y ) R 2 belong to the same Z 2 -orbit if and only if x x Z and y y Z. Hence the only points (x, y) I I mapping to the same orbit are of the form (0, y), (1, y) or (x, 0), (x, 1). Since these points are identified in I I/, the map f is injective. To prove continuity of f consider the commutative diagram I I i R 2. p (I I)/ f R 2 /Z 2 The inclusion map i and the projection maps p, q are continuous, and so the composition q i is continuous. By commutativity of the diagram it follows that f p = q i is continuous, and hence f is continuous. q 13

4. a) Show that a subspace of a Hausdorff space is Hausdorff and that a product of Hausdorff spaces is Hausdorff. b) Show that a subspace of a second countable space is second countable and that a product of second countable spaces is second countable. Proof. Part (a). Suppose A X is a subspace of a Hausdorff space X. To show that A is Hausdorff, let x, y A be distinct points. Since X is Hausdorff, there are disjoint open subsets U, V X with x U and y V. Then U A and V A are disjoint open neighborhoods of x resp. y in A, proving that A is Hausdorff. Suppose X, Y are Hausdorff spaces. To show that X Y is Hausdorff, let (x, y), (x, y ) X Y be distinct points. Without loss of generality we may assume x x. This allows us to find disjoint open neighborhoods U, V X of x resp. x. Then U Y and V Y are disjoint open neighborhoods of (x, y) resp. (x, y ), which proves that X Y is Hausdorff. Part (b). Let A be a subspace of a second countable space X. Let {U i } i I be a countable basis for X. We claim that the countable collection {U i A} i I is a basis for A, thus proving that A is second countable. To prove the claim, we recall that every open subset of A is of the form U A for some open subset U. Now U can be expressed as a union of U i s for i belonging to some subset J I. It follows that U A = ( i J U i ) A = i J(U i A), proving that {U i A} i I is a basis for the topology of A. Let X, Y be second countable spaces with countable bases {U i } i I and {V j } j J, respectively. We claim that the countable collection {U i V j } (i,j) I J is a basis for X Y, thus showing that X Y is second countable. To prove the claim, we need to show that every open neighborhood W of (x, y) X Y contains a neighborhood of (x, y) of the form U i V j. By definition of the product topology, W contains a neighborhood of the form U V, where U is an open neighborhood of x X and V is an open neighborhood of y Y. Since the U i s form a basis for the topology of X, this implies that there is some U i with x U i and U i U. Similarly, we find some V j with y V j V. Then which finishes the proof. (x, y) U i V j U V W 4 Homework Assignment # 4 1. Let ω n : (I, I) (S 1, 1) be the based loop defined by ω n (s) = e 2πins. Show that the map Φ: Z π 1 (S 1, 1) given by n [ω n ] is a group homomorphism. Proof. We note that Φ(m + n) = [ω m+n ], and Φ(m)Φ(n) = [ω m ][ω n ] = [ω m ω n ] 14

and so we need to construct a homotopy relative endpoints between ω m+n and the concatenation ω m ω n (note that they are not the same path unless m = n: both wind around the circle m + n times, but ω m+n goes around the circle at a constant speed, while ω m ω n winds around the circle n times on the interval [0, 1/2], while its winds around m times on [1/2, 1]). To produce the desired homotopy we note that ω m+n = p α 1 and ω m ω n = p α 2, where p: R S 1 is defined by p(s) = e 2πis, and the paths α i : [0, 1] R are given by { 2ns 0 s 1/2 α 1 (s) = (m + n)s α 2 (s) = 2ms m + n 1/2 s 1 We note that paths α 1 and α 2 have the same start/end point and hence they are homotopic via the linear homotopy H t connecting them. Then p H t is the desired homotopy between the paths ω m+n = p α 1 and ω m ω n = p α 2. 2. Let γ : I X be a path in a topological space X, and let γ : I X be the path γ run backwards, that is, γ(s) = γ(1 s). Then there are the following homotopies γ γ c γ(0) γ γ c γ(1) γ c γ(0) γ c γ(1) γ γ, (1) where c x for x X denotes the constant path at x. (a) Prove the first and the third homotopy. (b) Use the homotopies (1) to finish our proof that π 1 (X, x 0 ) is a group. In other words, show that the constant map c x0 gives an identity element for π 1 (X, x 0 ), and that [ γ] is the inverse for [γ] π 1 (X, x 0 ). (c) Let β be a path from x 0 to x 1. Show that the map Φ β : π 1 (X, x 0 ) π 1 (X, x 1 ) [γ] [β γ β] is an isomorphism of groups. Proof. Part (a). The strategy of constructing these homotopies is analoguous to the strategy used in problem 1. We note that γ γ = γ α 1 and c γ(0) = γ α 2, where α 1, α 2 : [0, 1] [0, 1] are the paths in [0, 1] given by { 2s 0 s 1 2 α 1 (s) = 1 2(1 s) s 1 and α 2 (s) = 0. 2 Let H t be the linear homotopy between α 1 and α 2. This is a homotopy of paths since the initial and terminal points of the paths α 1 and α 2 agree. This implies that γ H t is the desired homotopy between γ α 1 = γ γ and γ α 1 = c γ(0). Similarly, γ c γ(0) = γ β 1 and γ c γ(0) = γ β 2, 15

where β 1, β 2 : [0, 1] [0, 1] are the paths given by { 0 0 s 1 2 β 1 (s) = 1 2(s 1) s 1 2 and β 2 (s) = 0. As before, the linear homotopy G t between these maps is a path homotopy since β 1 (0) = β 2 (0) and β 1 (1) = β 2 (1). It follows that γ G t is a homotopy between γ c γ(0) and γ as desired. Part (b). Let γ be a loop in X, based at x 0 X. Then the third and fourth homotopy in part (a) implies the equalities in π 1 (X, x 0 ) [γ][c x0 ] = [γ] [c x0 ][γ] = [γ]. This shows that [c x0 ] is the identity element for the multiplication in π 1 (X, x 0 ). Similarly, the first and second homotopy from part (a) imply the equalities [ γ][γ] = [c x0 ] [γ][ γ] = [c x0 ] in π 1 (X, x 0 ). These equalities imply that [ γ] is the inverse of [γ]. In particular, π 1 (X, x 0 ) is a group. Part (c). First we show that Φ β : π 1 (X, x 0 ) π 1 (X, x 1 ) is a homomorphism. So let γ, δ be loops in X based at x 0. Then Φ β ([γ][δ]) = Φ β ([γ δ]) = [β γ δ β] = [β γ β β δ β] = [β γ β][β δ β] = Φ β ([γ])φ β ([δ]) Here we avoid using parantheses for the concatenation operation; this is OK by the fact that the paths resulting from different ways of putting in parantheses leads to homotopic paths. The first and second equality is just the definition of multiplication in π 1 (X, x 0 ) resp. the definition of Φ β. The third equation holds due to the homotopy β β c x0 (an application of the second homotopy of part (a)), and the fact that we can insert at any point an appropriate constant path into an iterated concatenation without changing the homotopy class of the path (follows from either the third or the fourth homotopy of part (a)). The last two equations again hold by construction of multiplication in the fundamental group resp. the map Φ β. It remains to show that Φ β is an isomorphism. We claim that its inverse is given by Φ β : π 1 (X, x 1 ) π 1 (X, x 0 ). Φ βφ β (γ) = Φ β([β γ β]) = [ β β γ β β] = [c x1 γ c x0 ] = [γ] Here the first two equations is just by construction of Φ β resp. Φ β, the third equation holds thanks to the first two homotopies of part (a), and the last equation holds due to the last two homotopies of part (a). 3. A subspace A X of a topological space X is called a retract of X if there is a continuous map r : X A whose restriction to A is the identity. 16

(a) Show that S 1 is not a retract of D 2. Hint: Show that the assumption that there is a continuous map r : D 2 S 1 which restricts to the identity on S 1 leads to a contradiction by contemplating the induced map r of fundamental groups. (b) Brouwer s Fixed Point Theorem states that every continuous map f : D n D n has a fixed point, i.e., a point x with f(x) = x. Prove this for n = 2. Hint: show that if f has no fixed point, then a retraction map r : D 2 S 1 can be constructed out of f. Proof. We prove part (a) be contradiction. Suppose that r : D 2 S 1 is a retraction of D 2 onto S 1. This means that r makes the following diagram commutative: S 1 i D 2 r id S 1 Choosing a basepoint x 0 S 1 and applying the fundamental group functor gives the commutative diagram i π 1 (S 1, x 0 ) π 1 (D 2, x 0 ) r id =id π 1 (S 1, x 0 ) This gives the desired contradiction since according to the diagram the identity map on π 1 (S 1, x 0 ) = Z factors through the trivial group π 1 (D 2, x 0 ). To prove part (b) we assume that f : D n D n does not have a fixed point, i.e., that for all x D n, f(x) x. Then the line through the points f(x) and x intersects the sphere S n 1 in two points. Let r(x) S n 1 be the intersection point closer to x. It is clear from the construction that r : D n S n 1 has the property r(x) = x for x S n 1, and so it remains to show the continuity of r to derive the desired contradiction from part (a). To show that r is continuous, we note that r(x) can be written in the form r(x) = x + α(x)(x f(x)), where α(x) R is the unique non-negative solution of the quadratic equation x + α(x)(x f(x)) 2 = 1 (4.1) (the solutions of this equation correspond to the two intersections of the line through x and f(x) with S n 1 ; hence it is clear geometrically, that there are two solution for every x, and that exactly one solution is non-negative). Writing out the quadratic equation (4.1) explicitly as x f(x) 2 α 2 (x) + 2 x, x f(x) α(x) + x 2 1 = 0 shows that its coefficients are continuous functions of x, and hence the quadratic formula shows that its non-negative solution α(x) is continuous function of x. We conclude that r(x) is a continuous function of x since its components are linear combinations of products of continuos functions. 17

4. We recall that the torus T and the Klein bottle K can both be described as quotient spaces of the square by identifying edges with the same label as shown in the following pictures. b b T = a a K = a a b b Each edge of the square can be interpreted as a path in the square, which projects to a based loop in the quotient space T resp. K. Here the base point x 0 of the quotient space is the projection of each of the four vertices of the square. Abusing language, let us denote by a resp. b the elements in the fundamental group of T (resp. K) represented by these loops. Prove the following identities: aba 1 b 1 = 1 π 1 (T, x 0 ) abab 1 = 1 π 1 (K, x 0 ). Proof. We begin with the torus T = ([0, 1] [0, 1])/ with equivalence relation generated by (0, y) (1, y) and (x, 0) (x, 1). For i = 1, 2, let α i, β i : I [0, 1] [0, 1] be the paths in the square [0, 1] [0, 1] given by its edges, with the direction provided by the arrow on each edge, as shown in the picture x 0 β 1 [0, 1] [0, 1] = α 1 α 2 β 2 It should be emphasized that this is a precise description by adopting the convention that all of these paths are the linear paths connecting their endpoints. Going counterclockwise around the boundary of the square, we obtain the loop γ := α 1 β 1 ᾱ 2 β 2, based at the top left corner point x 0 (strictly speaking we should put parentheses here, since we are talking about paths, but we don t want to burden the notation). Let p: I I T be the projection map. We note that the projection p α i of the paths α i is a loop in T based at the point x 0 := p( x 0 ). In fact, p α 1 and p α 2 are the same loop in T which we denote by α, and [α] π 1 (T, x 0 ) is the generator called a. Similarly, projecting the paths β i we obtain a loop β in T based at x 0 with [β] = b π 1 (T, x 0 ). Hence aba 1 b 1 = [α][β][ᾱ][ β] = [α β ᾱ β] = [p (α 1 β 1 ᾱ 2 β 2 )] = [p γ]. Since γ is a loop based at x 0 in the convex subset [0, 1] 2 R 2, the linear homotopy H t between this concatenation and the constant loop c x0 is a homotopy relative endpoints (we 18

want to emphasize that the fact that γ is a loop is essential here; by contrast, the linear homotopy between the path α 1 and c x0 is not a homotopy relative endpoints). It follows that p H t : I T is a homotopy between p γ and the constant loop c x0, which by (??) implies the relation aba 1 b 1 = 1 in the fundamental group of the torus. The argument in the case of the Klein bottle K is completely analogous. As before we have paths α i, β i for i = 1, 2 in [0, 1] 2 which project to loops α, β based at x 0 = p( x 0 ) K under the projection map p: [0, 1] 2 ([0, 1] 2 / ) = K. Here is the picture: x 0 β 1 [0, 1] [0, 1] = α 1 α 2 β 2 The only difference to the previous situation is that we have the loop δ := α 1 β 1 α 2 β 2 in [0, 1] 2, which is homotopic to the constant loop c x0. The image of the loop δ under p is the loop α β α β which represents the element abab 1 in the fundemental group of the Klein bottle. This proves the relation abab 1 = 1. 5. Let Z 2 act on R 2 by translation (as in problem # 3 of the last set). Show that the projection p: R 2 R 2 /Z 2 is a covering map. Proof. We note that for (x, y) R 2 the subset p 1 (p(x, y)) R 2 is the orbit through (x, y) (this is true in general for the projection map p: X X/G to the orbit space for a G-action on a topological space X). More explicitly: p 1 (p(x, y)) = {(x + m, y + n) R 2 m, n Z}. In particular, if U R 2 is an open subset, then p 1 (p(u)) = (m + n) + U (4.2) (m,n) Z 2 is an open subset of R 2, since it is the union of the open subsets (m, n) + U given by the image of U under the homeomorphism R 2 R 2, (x, y) (x + m, y + n). It follows that p(u) is an open subset of R 2 /Z 2. In other words, p is an open map, which means that p takes open subsets of its domain to open subsets of its range. We claim that the open subset p(u) R 2 /Z 2 is evenly covered, provided the intersection ((m, n)+u) U is empty for (m, n) (0, 0). To prove the claim, assume that ((m, n)+u) U = for (m, n) (0, 0). This implies that the the sets (m, n) + U in the decomposition (4.2) of p 1 (p(u)) are pairwise disjoint, since if (x, y) ((m, n) + U) ((m, n ) + U) for (m, n) (m, n ), then (x m, y n ) ((m m, n n ) + U) U, contradicting our assumption that this intersection is empty. This implies that the restriction of p to (m, n)+u 19

is a continuous bijection between (m, n) + U and p(u). Moreover, this map is open (as the restriction of the open map p), and hence it is a homeomorphism. This finishes the proof of the claim above. To finish the proof we note that for any (x, y) R 2 the open ball U = B 1/2 (x, y) of radius 1/2 around (x, y) does not intersect any translated ball (m, n) + U. In particular p(u) is an evenly covered open subset. Since every point of R 2 /Z 2 is contained in an open subset of this form, this proves that p: R 2 R 2 /Z 2 is a covering map. 5 Homework Assignment # 5 1. Let f : S 1 S 1 be defined by f(z) = z n for some n Z. Calculate the induced homomorphism f : π 1 (S 1, 1) π 1 (S 1, 1). Clarification: We ve proved that the fundamental group π 1 (S 1, 1) is isomorphic to Z. In particular, any endomorphism of π 1 (S 1, 1) is given by multiplication by some integer k Z. Calculating f means determining that integer k for the endomorphism f End(π 1 (S 1, 1)). Proof. We recall that the homotopy class [ω 1 ] π 1 (S 1, 1) of the based loop ω 1 : I S 1, defined by ω 1 (s) = e 2πis, is a generator of π 1 (S 1, 1). So it suffices to calculate f [ω 1 ] = [f ω 1 ] π 1 (S 1, 1): f ω 1 (s) = f(e 2πis ) = (e 2πis ) n = (e 2πins ) = ω n (s). Hence f ([ω 1 ]) = [ω n ] = n[ω 1 ] π 1 (S 1, 1), where the second equation was proved in problem #1 of assignment # 4. This shows that the endomorphism f is given by multiplication by n Z. 2. Let p: X X be a covering space such that for each x X the fiber p 1 (x) is countable. (a) Show that if X is a manifold of dimension n, then so is X. Hint: to show that X has a countable basis, first show that X has a countable basis consisting of evenly covered open subsets. (b) If X is compact manifold of dimension 2 and p is a d-fold covering map, what is the Euler characteristic of X in terms of the Euler characteristic of X? Here a d-fold covering means that each fiber p 1 (x) consists of d N points. (c) Suppose that p: X X is a d-fold covering, d N where X is a surface of genus g and X is a surface of genus g. Give a formula expressing g in terms of g and d. Proof. Part(a). Given a point x X, let x = p( x) X, and let U X be an open neighborhood of x and φ: U V be a homeomorphisms between U and an open subset V R n. Replacing U by a smaller neighborhood of x if necessary, we may assume that U is 20

evenly covered. Then p 1 (U) is the disjoint union of open subsets U α such that p Uα : U α U is a homeomorphism. Then x is contained in some U α, and the composition U α p Uα U φ V R n provides a homeomorphism between U α and V. This proves that X is locally homeomorphic to R n. To show that X is Hausdorff, let x, ỹ X with x ỹ. If p( x) p(ỹ), then we can find disjoint open neighborhoods U x, U y X of x := p( x) resp. y := p(ỹ), and then p 1 (U x ), p 1 (U y ) are disjoint open neighborhoods of x resp. ỹ. If x = y, let U be an evenly covered neighborhood of x, and let p 1 (U) = α A U α as above. If x belongs to U α, then ỹ belongs to U β for β α, and hence U α, U β are disjoint open neighborhoods of x and y. To show that X is second countable, let B be a countable basis for the topology of X, and let B be the subcollection of those open sets U X belonging to B which are evenly covered. We claim that this is a basis for the topology on X. To prove this, let x X and let V be an open neighborhood of x. Intersecting with an evenly covered open set containing x, we may assume that V is evenly covered. Since B is a basis, there is some U B with x U and U V. Since V is evenly covered, so is U, and hence U belongs to B. This proves that B is a basis for the topology on X; as a subcollection of the countable collection B it is countable. Now we construct a countable basis B of X as follows. For each U B, pick a point x U U. By construction U is evenly covered, and hence p 1 (U) is a disjoint union of open subsets of X, such that the restriction of p to each of these subsets is a homeomorphism. In particular, each of these subsets contains exactly one of the points x p 1 (x U ), allowing us to index these sets by the elements of p 1 (x U ): we will write Ũ x for the subset containing the point x p 1 (x U ). With this notation, we have p 1 (U) = Ũ x, x p 1 (x U ) and p Ũ x : Ũ x U is a homeomorphism. Let B be the following collection of subsets of X: B := {Ũ x U B, x p 1 (x U )}. This is a countable collection, since B is countable, and each fiber p 1 (x U ) is countable. We claim that B is a basis for X. To show this, it suffices to prove that for every ỹ X and open neighborhood Ṽ X there is some Ũ x B with ỹ Ũ x and Ũ x Ṽ. Let y := p(ỹ), let U B an open set containing y, and let U x B be the subset of p 1 (U) that contains the point ỹ. The open neighborhood U x of ỹ might not yet be what we are looking for; it might be too big in the sense that it might not be a subset of Ṽ. In order to fix this, consider U x Ṽ, a smaller open neighborhood of ỹ that via p maps homeomorphically to an open neighborhood of y. Then there is smaller open neighborhood U of y belonging to the basis B, and hence for some x p 1 (x U ) the open subset Ũ x is an open subset belonging to B, which contains the point ỹ and is contained in Ṽ. This shows that B is indeed a basis for the topology of X. 21

Part (b). Let P be a pattern of polygons on X. By passing to a refinement of P is necessary we can assume that each face is contained in an evenly covered open subset of X. Then the preimages of vertices, edges and faces of P will form a pattern of polygons P on X. Since the preimage of each vertex (resp. edge resp. face) of P consists of d vertices (resp. edges resp. faces) of P. In particular, if we denote by V (resp. E resp. F ) the number of vertices (resp. edges resp. faces) of P and by Ṽ (resp. Ẽ resp. F ) the corresponding number for P, we have χ( X) = Ṽ Ẽ + F = dv de + df = dχ(σ). Part (c). We recall that the Euler characteristic of a surface Σ g of genus g is given by χ(σ g ) = 2 2g. This implies It follows that g = 1 2 (2 χ(σ g)). g = 1 2 (2 χ( X)) = 1 2 (2 dχ(x)) = 1 (2 d(2 2g)) = g d + 1. 2 3. Show that assigning to a pointed space (X, x 0 ) its fundamental group π 1 (X, x 0 ), and to each basepoint preserving map f : (X, x 0 ) (Y, y 0 ) the induced map f : π 1 (X, x 0 ) π 1 (Y, y 0 ) defines a functor π 1 : Top Gp from the category Top of pointed topological spaces to the category Gp of groups. Proof. To prove that π is a functor, we need to show that it is compatible with compositions and that it maps identity morphisms in Top to identity morphisms of Gp. To show that it is compatible with compositions, suppose f : (X, x 0 ) (Y, y 0 ) and g : (Y, y 0 ) (Z, z 0 ) are morphisms in Top. We need to show that (g f) = g f as group homomorphisms from π 1 (X, x 0 ) to π 1 (Z, z 0 ). So let γ : I X be a based loop in (X, x 0 ) representing the element [γ] π 1 (X, x 0 ). Then (g f )([γ]) = g (f ([γ])) = g ([f γ]) = [g (f γ)] (g f) ([γ]) = [(g f) γ]. The associativity of maps implies that the based loops g (f γ) and (g f) γ in (Z, z 0 ) agree and proves compatibility with composition. To show that identity morphisms are preserved, let id X be the identity morphism of the pointed space (X, x 0 ). Then (id X ) ([γ] = [id X γ] = [γ], which shows that the induced homomorphism (id X ) : π 1 (X, x 0 ) π 1 (X, x 0 ) is the identity as required. 22

4. Given a set S, let F [S] be the free group generated by S. The elements of F [S] are equivalence classes of words whose letters consist of the symbols s and s 1 for s S (including the empty word). If a word contains letters s and s 1 adjacent to each other, the word obtained by deleting those two letters is declared equivalent to the original word, and the equivalence relation is generated by these relations. The multiplication is given by concatenation of words, and the identity element is provided by the empty word. (a) Show that there is a functor F : Set Gp which on objects sends a set S to the free group F [S]. (b) Show that for any group G the map Gp(F [S], G) Set(S, G) given by f f S is a bijection. Here we identify S with a subset of F [S] by regarding s S as a one letter word. Proof. Part (a). We need to construct the functor F on morphisms, and show that our construction is compatible with composition and identity morphisms. To construct F on morphisms, we need to associate to each map of sets f : S T a group homomorphism F (f): F (S) F (T ). Every element of F (S) is given by a word s ɛ 1 1... s ɛ k k with s i S and ɛ i {±1}. We define F (f) by F (f)(s ɛ 1 1... s ɛ k k ) := f(s 1 ) ɛ 1... f(s k ) ɛ k. (5.1) We note that this is well defined. If s i+1 = s i and ɛ i+1 = ɛ i, the word s ɛ 1 1... s ɛ k k is equivalent to the shorter word obtained by removing the segment s ɛ i i sɛ i+1 i+1 from it. Their images under F (f) are again equivalent, since the shorter one is obtained by removing the segment f(s i ) ɛ i f(s i+1 ) ɛ i+1 from the longer word. Next we need to show that our map F (f): F (S) F (T ) is indeed a group homomorphism. It is clear that it maps the identity element of F (S), given by the empty word, to the identity element of F (T ). It is also clear that it maps a product a b in F (S) to the product F (f)(a) F (f)(b), since the product is given by concatenation of words. The formula (5.1) shows that if f : S S is the identity map of S, then F (f) is the identity map of F (S). Finally we need to show that F is compatible with compositions. So let f : S T and g : T U be maps between sets. Then (F (g) F (f))(s ɛ 1 1... s ɛ k k ) = F (g)(f (f)(s ɛ 1 1... s ɛ k k )) = F (g)(f(s 1 ) ɛ 1... f(s k ) ɛ k ) = g(f(s 1 )) ɛ 1... g(f(s k )) ɛ k = ((g f)(s 1 )) ɛ 1... ((g f)(s k )) ɛ k = F (g f)(s ɛ 1 1... s ɛ k k ) 23