SOLUTIONS TO THE FINAL EXAM Short questions 1 point each) Give a brief definition for each of the following six concepts: 1) normal for topological spaces) 2) path connected 3) homeomorphism 4) covering space 5) deformation retraction 6) second countable State each of the following four theorems: 7) Urysohn s lemma 8) Tychonoff s theorem 9) the Seifert-van Kampen theorem 10) Brouwer s theorem about invariance of domain In each case, give an example with the desired properties; proofs are not required. 11) A sequentially compact topological space that is not compact 12) A connected topological space that is not path connected 13) A topological space that is not first countable 14) A regular topological space that is not normal 15) A locally path connected topological space with no universal covering space 16) A topological space whose fundamental group is not abelian Solution. Check the textbook for 1) 10). For 11), we can use the first uncountable ordinal S Ω. For 12), the topologist s sine curve will do. For 13), take the box topology on the product of countably many copies of R. For 14), we can use R l R l. For 15), the union of the countably many circles x, y) R 2 nx 1) 2 + ny) 2 = 1 } will do. For 16), use the figure eight space. Longer problems 8 points each) 1. Let A and B be two open subsets of a topological space X such that A B and A B are connected. Show that A and B must be connected. Solution. After replacing X by A B, we may assume without loss of generality that X is connected and that X = A B. By symmetry, we only have to prove that A is connected. Suppose that A = C D was a separation; because A is open, C and D are disjoint open subsets of X. By assumption, A B is connected, hence contained either in C or in D. If A B C, then B is disjoint from D, and B C) D is a separation of X; if A B D, then B is disjoint from C, and Date: December 16, 2014. 1
2 SOLUTIONS TO THE FINAL EXAM C B D) is a separation. In either case, we reach a contradiction since X is supposed to be connected. 2. Let X be a countable set with a topology. Show that if X is regular as a topological space, then it is also normal. Solution. The argument is almost the same as for Theorem 11.1 from class. We have to prove that every pair of disjoint closed sets A, B X can be separated by open sets. For any x A, the open set X \B contains x; because X is regular, there is a neighborhood of x whose closure does not intersect B. Since A is countable, this produces countably many open sets U 1, U 2,... with A U n and U n B =. By the same token, there are countably many open sets V 1, V 2,... with B V n and V n A =. To get disjoint open sets containing A and B, consider the collection of open sets U n = U n \ V k and V n = V n \ U k. Since every V k is disjoint from A, we have A U n = A U n, and therefore A U n = U; for the same reason, B is contained in V, which is the union of the V n. Any point x U V would have to belong to some U m and some V n. If m n, then x U m U m and x V n = V n \ U k, which is clearly a contradiction; if n m, we similarly reach a contradiction. This means that U and V are disjoint, and so X is normal. 3. Let p: E B be a covering space with E path connected and locally path connected. Let b 0 B and e 0 E be two base points such that pe 0 ) = b 0. a) Give a proof for the result that p : π 1 E, e 0 ) π 1 B, b 0 ) is injective. b) Let H 0 = p π1 E, e 0 ) ), and let H be an arbitrary subgroup of π 1 B, b 0 ) that is conjugate to H 0. Show that H = p π1 E, e) ) for some e p 1 b 0 ). Solution. For a), it is enough to show that ker p = e}, due to p being a group homomorphism. Suppose [α] π 1 E, e 0 ) is an element of ker p, meaning that p α is path homotopic to the constant path at b 0. By the homotopy lifting property, α must be path homotopic to the constant path e 0 ; hence [α] = e. For b), let [α] π 1 B, b 0 ) be some element with the property that H = [α] 1 H 0 [α] = [ᾱ] H 0 [α].
SOLUTIONS TO THE FINAL EXAM 3 By the path lifting property, α can be uniquely lifted to a path γ : I E with γ0) = e 0. Setting e = γ1), we have π 1 E, e) = [ γ] π 1 E, e 0 ) [γ], and therefore p π1 E, e) ) = [ᾱ] H 0 [α] = H. 4. Let G be a group with group operation and unit element e. We say that G is a topological group if G has a topology that makes the two functions m: G G G, mg, h) = g h i: G G, ig) = g 1 continuous. Show that if the point e G is closed, then G is Hausdorff. Solution. By one of the homework problems, the Hausdorff property for G is equivalent to saying that the diagonal = g, h) G G g = h } is closed in G G. Since g = h iff g h 1 = e iff m g, ih) ) = e, we have = id i) 1 m 1 e) ) ; this is a closed subset of G G because both functions id i and m are continuous and e G is closed. 5. Let T 2 = T #T be the connected sum of two copies of the torus T = S 1 S 1. Show that T 2 is not a covering space of T. Solution. The fundamental group of the torus is the abelian group Z Z. If T 2 was a covering space of T, then the fundamental group of T 2 would be isomorphic to a subgroup of Z Z, and therefore also abelian; but according to a result from class, the fundamental group of T 2 is not abelian. 6. Let X be the topological space obtained by taking two copies of the real projective plane P 2 and identifying a single point p in one copy with a single point q in the other copy. a) Determine the fundamental group of X. b) Draw a picture of the universal covering space of X. Solution. For a), we first note that π 1 P 2, p) = Z 2. Let x X denote the common image of the points p and q. Let U be the first copy of P 2 union a small neighborhood of q in the second copy; let V be the second copy of P 2 union a small neighborhood of p in the first copy. Then U and V are path connected and U V is contractible; by the Seifert-van Kampen theorem, π 1 X, x) = Z 2 Z 2 is the free product of two groups of order 2. For b), recall that the universal covering space of P 2 is just the sphere S 2. The universal covering space of X is an infinite string of spheres, something like this:
4 SOLUTIONS TO THE FINAL EXAM Each sphere maps in the usual way to one of the two copies of P 2 ; as we move from one sphere to the next, we alternate between the two copies. It is not hard to see that this gives a simply connected covering space of X. 7. Let X = a, b, c, d}, with the topology induced by the basis } a}, c}, a, b, c}, c, d, a}. Show that X is path connected but not simply connected. Solution. Picture the space as something like a necklace with four beads, then it should be clear why it is path connected but not simply connected. There is a path from the point a to the point b, given by a if 0 t < 1, α: [0, 1] X, αt) = b if t = 1; this is continuous because every open set containing b also contains a. Similarly, b if t = 0, β : [0, 1] X, βt) = c if 0 < t 1 is a path from the point b to the point c, and c if 0 t < 1, γ : [0, 1] X, γt) = d if t = 1 is a path from the point c to the point d. To show that X is not simply connected, it is enough to produce a nontrivial path connected covering space. Let Y = a, b, c, d, a, b, c, d }, with topology induced by the basis } a }, a }, c }, c }, a, b, c }, a, b, c }, c, d, a }, c, d, a } ; for the same reason as above, Y is path connected. The function p: Y X with pa ) = pa ) = a, pb ) = pb ) = b, pc ) = pc ) = c, pd ) = pd ) = d is surjective and continuous; this can be checked by looking at the preimages of the four basic open sets in X. Now the preimage of the open set a, b, c} is the disjoint union of the two open sets a, b, c } and a, b, c }, both of which are homeomorphic to a, b, c}; similarly, the preimage of c, d, a} is the disjoint union of the two open sets c, d, a } and c, d, a }. Therefore every point in X has an open neighborhood that is evenly covered by p, and so p is a covering space. Bonus problem. Let A be a subspace of R 2 that is homeomorphic to the closed unit interval. Show that A is a deformation retract of R 2. Solution. We first construct a retraction r : R 2 A. Let h: A [0, 1] be a homeomorphism between A and the closed unit interval. By the Tietze extension theorem, we can extend h to a continuous function f : R 2 [0, 1]; now r : R 2 A, rx) = h 1 fx) ) is continuous and satisfies rx) = x for every x A. To show that A is a deformation retract of R 2, we need a homotopy H : R 2 [0, 1] R 2
SOLUTIONS TO THE FINAL EXAM 5 such that Hx, 0) = x and Hx, 1) = rx) for all x R 2, and Hx, t) = x for all x A. We can construct such a homotopy by another application of the Tietze extension theorem. On the closed subset X = R 2 0, 1} ) A [0, 1] ) R 2 [0, 1], we already have a well-defined continuous function F : X R 2 x if t = 0 or x A,, F x, t) = rx) if t = 1. Let p 1 and p 2 denote the projections from R 2 to the first and second coordinate. By applying the Tietze extension theorem to F 1 = p 1 F, we obtain a continuous extension H 1 : R 2 [0, 1] R; similarly, F 2 = p 2 F has a continuous extension H 2 : R 2 [0, 1] R. Together, they determine a continuous function H : R 2 [0, 1] R 2 that extends F, and is therefore a deformation retraction of R 2 onto A.