MAT 530: Topology&Geometry, I Fall 2005

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1 MAT 530: Topology&Geometry, I Fall 2005 Problem Set 11 Solution to Problem p433, #2 Suppose U,V X are open, X =U V, U, V, and U V are path-connected, x 0 U V, and i 1 π 1 U,x 0 j 1 π 1 U V,x 0 i 2 π 1 V,x 0 j 2 π 1 X,x 0 Figure 1: Van Kampen s Theorem Setting are the homomorphisms induced by inclusions. Suppose in addition that i 2 is surjective. Let M π 1 X,x 0 be the least normal subgroup containing i 1 ker i 2. a Show that j 1 induces a surjective homomorphism b Show that h is an isomorphism. h: π 1 U,x 0 / M π 1 X,x 0. a By the weak version of van Kampen s Theorem, the homomorphism is surjective. Since j 1 j 2 : π 1 U,x 0 π 1 V,x 0 π 1 X,x 0 j 1 i 1 = j 2 i 2 : π 1 U V,x 0 π 1 X,x 0, i.e. the diagram in Figure?? is commutative, and i 2 is surjective in this case, Imj 2 = Im j 2 i 2 = Imj 1 i 1 Imi 1 π 1 X,x 0. Thus, the homomorphism j 1 is surjective. In addition, since the diagram in Figure?? is commutative ker i 2 ker i 2 j 2 = ker i 1 j 1 = i 1 ker i 2 ker j 1. Since ker j 1 is a normal subgroup of π 1 U,x 0 and contains i 1 ker i 2, it must contain M as well. Thus, j 1 induces a homomorphism h: π 1 U,x 0 / M π 1 X,x 0. Since j 1 is surjective, so is h. b Define homomorphisms φ 1 : π 1 U,x 0 π 1 U,x 0 /M and φ 2 : π 1 V,x 0 π 1 U,x 0 /M by φ 1 α = αm and φ 2 i2 α = φ 1 i1 α. Since i 2 is surjective and i 1 ker i 2 M, φ 2 is well-defined. It is immediate that φ 1 i 1 = φ 2 i 2 : π 1 U V,x 0 π 1 U,x 0 /M,

2 i.e. the diagram i 1 π 1 U,x 0 φ 1 π 1 U V,x 0 i 2 j 1 j 2 π 1 X,x 0 ϕ π 1 U,x 0 /M π 1 V,x 0 φ 2 Figure 2: Amalgated Product Setting of solid lines is commutative. Since Figure?? is an amalgated product by van Kampen s Theorem, there exists a unique homomorphism In particular, ϕ: π 1 X,x 0 π 1 U,x 0 /M s.t. φ 1 = ϕ j 1 and φ 2 = ϕ j 2. ϕ hαm = ϕ j 1 α = φ 1 α = αm αm π 1 U,x 0 /M = ϕ h = id π1 U,x 0 /M. Thus, h is an injective homomorphism. On the other hand, it is surjective by part a. We conclude that h is an isomorphism and its inverse is ϕ. Solution to Problem p438, #5 Let S n R 2 be the circle with center at n,0 and of radius n. Let Y be the subspace of R 2 consisting of the circles S n, with n Z +. Denote the common point of the circles by p. S 1 S 2 S 3 Figure 3: Some Circles S n a Show that Y is not homeomorphic to either a countably infinite wedge X of circles or the infinite earring Z of Example 1 on p436. b Show that π 1 Y,p is a free abelian group with {[f n ]} as a system of free generators, where f n is a loop representing a generator of π 1 S n,p. a Since Y is an unbounded in the standard metric subset of R 2, Y is not compact and is second countable. Since Z is closed and bounded with respect to the standard metric on R 2, Z is compact. On the other hand, if q X is the point common to all of the circles in X, X {q} is homeomorphic to Z + S 1 {q}. Since X {q} is not second countable, neither is X. Thus, Y is homeomorphic to neither X not Z. Remark: In fact, X does not have a countable basis at p. So, X is not even first countable. b For each n Z +, let i n : π 1 S n,p π 1 Y,p

3 be the homomorphism induced by the inclusion S n,p Y,p. We will show that the homomorphism i n : π 1 S n,p π 1 Y,p n Z + n Z + is an isomorphism. First, let r N : Y Y N S n be the retraction obtained by collapsing the circles S n with n>n to the point p. By the existence of such a retraction, the homomorphism j N : π 1 Y N,p π 1 Y,p induced by the inclusion Y N,p Y,p is injective. If n N, let i N,n : π 1 S n,p π 1 Y N,p be the homomorphism induced by the inclusion S n,p Y N,p. By Theorem 71.1, the homomorphism i N,n : π 1 S n,p π 1 Y N,p is an isomorphism. Thus, the homomorphism i n = j N i N,n : π 1 S n,p π 1 Y,p is injective, and so is i n. It remains to show that every element [α] of π 1 Y,p lies in the image of the homomorphism j N for some N Z +. Let α: I, {0,1} Y,p be a loop in Y based at p. Since αi is compact, αi is bounded and thus for some N Z +. Let αi Y N Y +1 H : Y N I Y N {2n,0} be a deformation retraction of YN onto Y N, i.e. a homotopy from id Y N to r N Y N such that Hx,t=x for all x Y n. Such a homotopy is obtained by retracting the open upper and lower semicircles of S n, with n>n, to p. Then, H {α id I } is a path homotopy from the loop α in Y to the loop r N α in Y N. In particular, [α] = [ r N α ] [ π 1 Y,p and rn α ] Im j N, as needed.

4 Solution to Problem p441, #3 Suppose G is a group, h G, and N is the least normal subgroup of G containing h. Show that if π 1 X G for some compact path-connected normal topological space X, then π 1 Y G for some compact path-connected normal topological space Y. Let p: I S 1, ps=e 2πis, be the usual quotient map. Choose a representative α: I, {0,1} X,x 0 for h π 1 X,x 0. Since α0 = α1, α induces a continuous map f : S 1 X such that α = f p, i.e. the diagram commutes. Let I, {0,1} p S 1 f,1 X,x 0 α Figure 4: A Commutative Diagram X α = X B 2 /, x fx x S 1 B 2. Let q : X B 2 X α be the quotient map. If X is compact, then so are X B 2 and thus X α. Since X is path-connected and B 2 are path-connected, so are qx and qb 2. Since X α = qx qb 2 and qx qb 2, it follows that X α is path-connected. It is shown in the next paragraphs that X α is normal. Finally, by Figure??, f π 1 S 1,1 π 1 X,x 0 is generated by h=[α]. Since the map is a homeomorphism, Theorem 72.1 implies that q B 2 S 1 : B2 S 1 X α π 1 Xα,qx 0 π 1 X,α / N. We now show that X α is normal, i.e. X α is T1 one-point sets are closed and disjoint closed sets can be separated by continuous functions. We begin by showing that the map q is closed. If A X is closed, then q 1 qa = q 1 qa X q 1 qa B 2 = q 1 X qa q 1 B qa = A f 1 A, 2 since q A is injective and qx qb 2 S 1 =. Since f is continuous, f 1 A is closed in S 1. Since S 1 is closed in B 2, it follows that f 1 A is closed in B 2 and thus q 1 qa is closed in X B 2. Since q is a quotient map, qa is then closed in X α. On the other hand, if A B 2, then q 1 qa = q 1 qa X q 1 qa B 2 = q 1 X qa q 1 B qa = fa S 1 A. 2

5 Since A is closed in B 2 and S 1 is compact, A S 1 is closed in S 1 and thus compact. It follows that fa S 1 is a compact subset of X. Since X is Hausdorff, fa S 1 is a closed subset of X. Thus, q 1 qa is closed in X B 2 and qa is closed in X α. We conclude that the quotient map is closed and the space X α is Hausdorff. It remains to show that closed subsets of X α can be separated by continuous functions. First note that the map q X : X qx X α is continuous, bijective, and closed. Thus, it is a homeomorphism. Since X is normal, so is qx. Suppose that A,B X α are disjoint closed subsets. Then, A qx and B qx are disjoint closed subsets of qx. Since qx is normal, by Urysohn Lemma there exists a continuous function g X : qx [0,1] s.t. g X A qx = {0} and gx B qx = {1}. Then, is continuous function such that g X q: S 1 [0,1] g X q q 1 A S 1 = {0} and g X q q 1 B S 1 = {1}. Define g: S 1 q 1 A B 2 q 1 B B 2 [0,1] by g X qx, if x S 1 ; gx = 0, if x q A B 2 ; 1, if x q 1 B B 2. These definitions agree on the overlap and define a continuous function on each of the three closed sets. By the pasting lemma, g is continuous. Since B 2 is normal and S 1 q 1 A B 2 q 1 B B 2 B 2 is closed, by Tietze s Extension Theorem g extends to a continuous function g X qx, if x S 1 ; h B 2 : B 2 [0,1], i.e. h B 2x = gx = 0, if x q 1 A B 2 ; 1, if x q 1 B B 2. Let h X =g X q. Then, the function is continuous and h X h B 2 : X B 2 [0,1] h X fx = g X qfx = gx qx = h B2 x x S 1 B 2, h X h B 2 q 1 A = g X A qx gb 2 q 1 A B 2 = {0}, and h X h B 2 q 1 B = g X B qx gb 2 q 1 B B 2 = {1}. By the first property, h X h B 2 induces a map h: X α [0,1] such that h X h B 2 =h q, i.e. the diagram

6 X B 2 q h X h B 2 h X α [0,1] Figure 5: Construction of Separating Map commutes. The function h is continuous, because q is a quotient map. By the other two properties, ha = h X h B 2 q 1 A = {0} and hb = h X h B 2 q 1 B = {1}, as needed. Solution to Problem p445, #2 Show that for every finitely presentable group G, there exists a compact Hausdorff path-connected space X such that π 1 X G. Suppose G = α 1,...,α n r 1,...,r n, i.e. G = Z[α1 ]... Z[α n ] / Nr 1,...,r m, where Nr 1,...,r m is the smallest normal subgroup of Z[α 1 ]... Z[α n ] containing For each k=0,...,m, let {r 1,...,r m } Z[α 1 ]... Z[α n ]. H k = Nr 1,...,r k, G k = G/H k, h k = r k H k 1 G k 1 if k > 0. We note that the smallest normal subgroup N k of G k 1 containing h k is H k H k 1 h H k hh k 1 G k 1. Thus, G k G k 1 /N k. Let X 0 be the wedge of n circles. Let p be the point common to all of the circles. By Theorem 71.1, π 1 X,p Z[α 1 ]... Z[α n ] = G 0, where α i is the homotopy class of a loop going around the ith circle once. The space X 0 is compact Hausdorff and path-connected. Suppose k Z +, k n, and there exists a compact Hausdorff pathconnected space X n 1 such that π 1 X k 1 G k 1. Then, by Problem p441, #3, there exists a compact Hausdorff path-connected space X k such that π 1 X k G k 1 /N k G k. After applying this construction m times, we obtain a compact Hausdorff path-connected space X X m such that π 1 X = π 1 X m G m G. Remark: In brief, in order to obtain a compact Hausdorff path-connected space whose fundamental group is G we begin with the wedge of n circles and then make the elements r 1,...,r m null-homotopic by attaching m disks B 2. The jth disk is attached by wrapping its boundary, S 1, along a representative for r j, which can be taken to be a path going around some of the circles, possibly multiple times.