or else take their intersection. Now define

Transcription

1 Samuel Lee Algebraic Topology Homework #5 May 10, 2016 Problem 1: ( 1.3: #3). Let p : X X be a covering space with p 1 (x) finite and nonempty for all x X. Show that X is compact Hausdorff if and only if X is. Proof. First I will show that X Hausdorff implies that X is Hausdorff. Let x and ỹ be elements of X and define x = p( x) and y = p(ỹ). Then there are neighborhoods U x and U y of x and y satisfying the requirements of a covering space, namely that they are the homeomorphic image of disjoint open sets in X. Let U x be the connected component of p 1 (U x ) containing x and Ũy be the connected component of p 1 (U y ) containing ỹ. Now since X is Hausdorff, there are disjoint sets V x U x and V y U y seperating x and y (if the sets guaranteed by the Hausdorff condition are not subsets of U x and U y we can just take the intersections V x U x and V y U y to get disjoint sets). Moreover, each of these has a homeomorphic copy in the connected component of their respective preimage, i.e there are V x and Ṽy with p Ux ( V x ) V x and Ṽy with p Uy (Ṽy) V y. Suppose z V x Ṽy, then p(z) V x and p(z) V y which is a contradiction since they are disjoint. Hence V x and Ṽy are disjoint sets containing x and ỹ showing that X is Hausdorff. Now suppose that X is Hausdorff, x, y X and again let U x and U y be the neighborhoods that satisfy the conditions for p : X X to be a cover. By hypothesis, p 1 (x) = { x 1,..., x m } and p 1 (y) = {ỹ 1,..., ỹ m } are finite sets each contained in a neighborhood Ũ x i or Ũỹ j homeomorphic to U x or U y. By an easy argument following the assumption that X is Hausdorff, there exist disjoint neighborhoods à x 1,..., à x m, Ãỹ 1,..., Ãỹ m containing x 1,..., x m, ỹ 1,..., ỹ m, respectively. We can assume that à x i is contained in Ũ x i or else take their intersection. Now define m m A x = p(ã x i ) and A y = p(ãỹ i ). i=1 If z A x A y, then z p(ã x i ) p(ãỹ j ) for all combinations of i, j. Then z has lifts in each of the à xi and each of the Ãỹ j. But each of these is disjoint and so p 1 (z) = 2m which is a contradiction since p 1 (z) = m. Hence X is Hausdorff. Note: This is making the assumption that X and X are connected so that the number of sheets is constant but it seems easy enough to generalize by considering each connected component of X and then of each of the finitely many connected components of the covering space of that component, which would be a connected component of X For compactness, one direction is easy: If X is compact, then its image under a continuous map is compact so since p 1 (x) is assumed to be nonempty for all x, p is surjective and then p( X) = X is compact. Suppose that X is compact. We will use the following notions of compactness and nets: 1. A net is a function I X, where I is a directed set, whose image is denoted by {x i } i I. 2. If I is a directed set with i, j I, then there is a k I with k i and k j. This property can be extended to a finite collection, i.e if i 1,..., i m I then there is a k I with k i j for j = 1,..., m. i=1

2 3. A topological space X is compact if and only if every net has a convergent subnet. Let { x i } i I be a net in X. It s image {p( x i )} i I in X is then also a net and by compactness, has a convergent subnet {p( x j )} j J, where J is some subset of I, converging to a point x X. By hypothesis, x has finite preimage p 1 (x) = { x 1,..., x m } each of which is contained in a neighborhood Ũk homeomorphic to some fixed neighborhood U of x (U is the neighborhood guaranteed by the covering map p). For the sake of contradiction, assume that { x j } j J has no convergent subnet. Then for each k there exists a neighborhood Ṽ k of xk and an M k such that for j M k, x j Ṽk, or else x k would be an accumulation point. We can take Ṽk = Ṽ k Ũk, which also satisfies the previous condition, but additionally has a homeomorphic image in U under p Ṽk. Now let M I be such that M M k for each k = 1,..., m which is possible by notion 2 above. For any j M, x j Ṽk, but since Ṽk is homeomorphic to p(ṽk) U, this implies that p( x j ) p(ṽk. This contradicts the fact that p( x j ) x and so eventually stays in every neighborhood of x. Hence { x j } j J and therefore { x i } has a convergent subnet. By notion (theorem) 3 stated before, this implies that X is compact. Problem 2: ( 1.3: #4). Construct simply connected covering space of the space X R 3 that is the union of a sphere and a diameter. Do the same when X is the union of a sphere and a circle intersecting it in two points. Solution: The covering space of a sphere with a diameter is the infinite string of spheres connected by lines as shown. Each sphere in X maps homeomorphically to the sphere in X and similarly the lines in X all go to the diameter. The covering space for the space X of a sphere with an S 1 glued on in two places is the infinite grid of spheres shown below. Several mappings of p : X X are shown in pink.

3 Problem 3: ( 1.3: #8). Let X and Ỹ be simply connected covering spaces of the path connected, locally path connected spaces X and Y. Show that if X Y then X Ỹ (Hint: ( 0: #11)) Proof. Denote by p : X X and q : Ỹ Y the covering maps of X and Y and suppose that X Y. Then there is a map f : X Y and g : Y X such that f g 1 Y and g f 1 X. Note that f p is a map from X Y and g q is a map from Ỹ to X. Since the fundamental groups π 1 ( X, x 0 ) and π 1 (Ỹ, ỹ 0) are both trivial (since X 0 and Ỹ0 are simply connected) we have that (g q) (π 1 (Ỹ, ỹ 0)) = {0} = p (π 1 ( X, x 0 )) and then an application of Prop gives a lift g : Ỹ X of g q and a lift f : X Ỹ of f p. These are illustrated by the following diagram: X X p The claim is that f and g satisfy the requirements for X and Ỹ being homotopy equivalent... g f g f Problem 4: ( 1.3: #10). Find all the connected 2 sheeted and 3 sheeted covering spaces of S 1 S 1, up to isomorphism of covering spaces without basepoints. Ỹ Y q

4 Solution: I will give a brief outline of why the following diagrams represent all 2 and 3 sheeted covering spaces of S 1 S 1. Diagrams 1, 2 and 3 exhaust all possible the 2-vertex, 4-valent graphs. To see this pick one of the vertices (preimages of the point of gluing in S 1 S 1 ), it can either loop to itself, producing the S 1 (S 1 S 1 ) or it will not, producing the other space as in 3. Diagrams 1 and 2, however, represent distinct covering spaces, for if we picked the intersection of the left and middle circles in each as the basepoint, then π 1 ( 1 ) has the element a whereas π 1 ( 2 ) does not (respectively, with the previously stated basepoints). Any other reasonable map from 3 to S 1 S 1 would just change how the diagram has been labeled, and would be homeomorphically equivalent. It is different than 1 and 2 since it s fundamental group (with basepoint being either of the two vertices)contains both a 2 and b 2, which is not the case in either 1 or 2. For the 3-sheeted cover I claim that 4-10 exhaust all possibilities. We can make a similar argument to before that, minus the labelings, these represent all potential graphs having 3 vertices each with valence 4 as follows. Pick one vertex v 1. If v 1 has a loop then the remaining edges can either both go to the same vertex or to the remaining two vertices. If they both go to one vertex v 2 then the remaining edges incident to v 2 must go to the last vertex v 3 which must then also have a loop. This looks like 4. If the two remaining edges of v 1 are split between v 2 and v 3, then we can either have loops at v 2 and v 3, resulting in 7 and 8, or no loops at v 2 and v 3 yielding the graph of 9 and 10 (it is clear why these are the only possiblities in this scenario). If there are no loops at any vertices, the only possible graph is that of 5 and 6.

5 The map represented by the labeling in 4 is the only one up to isomorphism on this graph. To see this we can just notice that interchanging the a and b s gives the same covering space, again up to isomorphism. 5 and 6 are distinct, since there is a choice of basepoint x 4 in 5 such that π 1 ( 5, x 5 ) contains the element ab, whereas there is no point x 6 such that π 1 ( 6, x 6 ) has that element (the analogous loop would be ab 1 which is distinct from ab). Similarly for pairs 7, 8 and In 9, 10 we can find elements of the fundamental group of the first that are not in the latter. 7, consider b π 1 ( 7, x 7 ) ( x 7 is any of the vertices), which is not in π 1 ( 8, x 8 ) for any point x 8. In 9, consider b π 1 ( 9, x 9 ) ( x 9 is the bottom vertex), which is not in π 1 ( 10, x 10 ) for any point x 10

6 Problem 5: ( 1.3: #11). Construct finite graphs X 1 and X 2 having a common finite sheeted covering space X 1 = X 2, but such that there is no space having both X 1 and X 2 as covering spaces. Solution: X 1, as drawn below, does not cover any space other than itself. If it did, then it would have to be a 2-sheeted covering, with each of the three-valent points being mapped to a common vertex. The only graphs with exactly one vertex with three edges would have to have one (or more) vertex with one edge, for which there would be no corresponding point in X 1 mapping to it. Since X 2 does not cover X 1, there can be no space that they both cover. The space X 1 = X 2, shown below, covers each of X 1 and X 2 via the assignment of edges as indicated on each diagram. Problem 6: ( 1.3: #12). Let a and b be the generators of G = π 1 (S 1 S 1 ) corresponding to the two S 1 summands. Draw a picture of the covering space of S 1 S 1 corresponding to the normal subgroup generated by a 2, b 2 and (ab) 4, and prove that this covering space is indeed the correct one. Solution: The normal subgroup H generated by a 2, b 2 and (ab) 4 ) is the group generated by all conjugates by elements in G. Consider the covering space below with the covering map p : X X being indicated by the labeling. The fundamental group π 1 ( X, x 0 ) certainly contains a 2, b 2 and

7 (ab) 4, as well as all conjugates which correspond to change of basepoint homomorphisms from π 1 ( X, x 0 ) to π 1 ( X, x k ) with x k being any of the other vertices in the graph. Hence H π 1 ( X, x 0 ). Let h be an element of π 1 ( X, x 0 ). Then h is, up to homotopy, represented by a word in a and b. If a α b β is the abelianization of h, then α and β are both congruent to 0(mod2) which follows from that going around around the circle requires four a s and going partially around requires doing one a in the clockwise direction for every a traveled in the counterclockwise direction (and similiarly for b). However, this is an equivalent condition to being the product of conjugates of a 2, b 2 and (ab) 4 which implies that H π 1 ( X, x 0 ). Problem 7: ( 1.3: #14). Find all the connected covering spaces of RP 2 RP 2. Solution: Up to basepoint changing isomorphisms, there is a bijective correspondence between covering spaces of a space X and supgroups of π 1 (X). Thus we need to find a covering space corresponding to each subgroup H of of which there are several types: 1. The trivial subgroup. π 1 (RP 2 RP 2 ) = a a 2 b b 2 = a, b a 2 = b 2 = id 2. Subgroups of the form a, (ba) k a 2, b 2 and b, (ab) k a 2, b 2 for each k. These are isomorphic to Z 2Z since, for example a, (ba) k a 2, b 2 = a, (ba) k 1 b a 2, b 2 which is generated by two elements each of order two. 3. Copies of Z of the form (ab) k for each integer k. 4. There are two Z 2Z s of the form a a 2 and b b 2. Recall that the universal cover of RP 2 is S 2. Each of the connected covering spaces are going to be unions of RP 2 s and S 2 s. I will write RPa 2 for the RP 2 corresponding to the a loop and similarly for the RP 2 corresponding to the b loop.

8 When H is trivial, the corresponding covering space X must be simply connected. An infinite string of pairs of spheres covers RP 2 RP 2 in this case. The covering map p takes adjacent spheres to the distinct copies of RP 2 in the wedge product and p restricted to one sphere is just the covering map S 2 RP 2. Thus the preimage of any point x RP 2 RP 2 is a set of antipodal points in every other sphere in the string. At the point of gluing, the preimage is the set of points where adjacent spheres intersect in X. For the second type, without loss of generality where H = a, (ba) k a 2, there are two cases. If k is even then the covering space will be a finite string of spheres with an RPa 2 on one end and an RPb 2 on the other. If k is odd, then it will be a finite string of spheres with an RP a 2 on each end: For the third type of subgroup, H = (ab) k, nontrivial loops in X are of the form (ab) mk. The covering space in this case is a ring of k pairs of spheres, each pair is associated with two universal covers of RPa 2 and RPb 2. Traversing all the way around the ring once will generate a nontrivial loop (ab) k.

9 In the last case, the only non-simply connected part of the covering space must generate the elements of Z 2Z. Since this is exactly the fundamental group of RP 2, one would think that there will be a copy of RP 2 in the covering space and this is exactly right. Here we get RP 2 with a string of sphere pairs. Problem 8. Consider the realization of the genus 2 surface S on page 5, with its fundamental group generated by a, b, c, d as shown where x 0 is the basepoint. What is the cover corresponding to the subgroup H = a, b of π 1 (H, x 0 )? What is the cover corresponding to the subgroup J = b, c. Solution: The loops a and b encapsulate the information for one genus and c and d the other. For H, π 1 (H, x 0 ) = a, b indicates that the information for the first genus is still being recorded. The covering space therefore looks something like a torus, but the lifts of loops c and d lie on the infinte skirt. Note that the c and d lifts are paths, and at the beginning/end point of each of these lifts of the basepoint there are 8 paths, corresponding to lifts of a, b, c and d. In the octagon construction, these correspond to the single identified vertex. So the skirt part of the cover is tiled by these octagons.

10 For J = b, c the covering space must make trivial a and d. To form this space we can fatten up neighborhoods of a circle wedge, representing b and c, but then instead of forming the surface by wrapping the other loops around, we send them to infinite cylinders. Topologically this is a pair of pants and just like before the cylinder surfaces are tiled by octagons.