Topology Hmwk 5 All problems are from Allen Hatcher Algebraic Topology (online) ch 1

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1 Topology Hmwk 5 All problems are from Allen Hatcher Algebraic Topology (online) ch Andrew Ma November 22, Claim: A nice space X has a unique universal abelian covering space X ab Proof. Given a space X, let H be the commuatator subgroup of π (X) and let X ab be it s corresponding covering space. Since H is a normal subgroup, by prop.39, for the group od deck transformations G( X ab ) = π (X)/H which is an abelian group. This is shows that X ab is an abelian cover cover. Next I will show that X ab is the unique cover space covering all other abelian cover spaces of X. Suppose H π (X) s.t. the cover corresponding to H is abelian. By prop.39 this implies that π (X)/H is abelian. Therefore H H and by the discussion under Thm.3.8. this implies that X ab covers H s corresponding cover space. By similar argument we also see that the cover space X ab is unique. Claim: The attached drawing is the universal abelian cover space for S S. Proof. For any loop in the space we can consider the region inside of the loop, which may be viewed as a graph made by a finite number of straight line segments. By a previous hmwk problem the fundamental group of this graph can be generated by loops the around each bounded square region contained in the loop. Each bounded square loop maps via the projection p to elements in the commutator subgroup (notice that the loop γ loop around a square γ is a conjugate of [a, b] for a, b < a, b >). Furthermore, this cover space is abelian because any deck transformation can be generated by a unit horizontal shift and a unit vertical shift and these generator clearly commute. Thus this space corresponds to the commutator subgroup and is the universal abelian cover. Claim: The attached drawing is the universal abelian cover of S S S.

2 Proof. This is analogous to the previous problem. I will hand wave an argument for it. Notice that every loop in this space can be factored (equivalent by path homotopies) into to a loop in the x y plane, a loop in the y z plane, and a loop in the x z plane. By repeating the argument for S S we get that each loop in a plane maps by p to the commutator of π (S S S ). Therefore the original loop maps via p to the commutator subgroup. Further the deck group is generated by a unit shift along the x-axis, a unit shift along the y-axis, and a unit shift along the z-axis. Since these generators commute the deck group is abelian. Therefore this space corresponds to the commutator subgroup of < a, b, c > and is the universal abelian cover space The Klein bottle can be covered by a Klein bottle Proof. Consider the attached drawing of three Klein bottles attached along a corresponding edge so that the orientations match correctly, call it K. This clearly covers one Klein bottle. To see that this is a non-normal cover, consider a loop by in the cover space which is a path labeled by a in the picture. Then consider the conjugate by b i.e. bab π (K). We can see that this element is not in the fundamental group of the cover since it does not correspond to any loop in the cover beginning at any base point (simply by brute force checking). Thus p (π ( K)) is not a normal subgroup of π (K) because it is not closed under conjugation. *An additional side note: Alicia explained that in any normal cover of a space, a loop should be a loop no matter what point of p (x) you start at. This is seen by saying p (π ( X)) is the stabilizer of x, and conjugation of this group will be the stabilizer of the element f f x where f π (X). By the normal assumption this should just be an element of the original subgroup.* I don t have a good answer for the problem of covering the Klein bottle by a Torus. I struggled to show that the cover is non-normal Claim: The action described in the problem statement is a cover space action Proof. The mapping can be described as a linear transformation ( ) 2 0 T = 0 2 This is clearly invertible (it has all non-zero eigenvalues) and it defines Z action on X by nx = T n x. Now to show that this is a cover space action, I will need to show that for 2

3 any point (x, y) there exists an open nbhd U of (x, y) s.t. T n U T m U = if n = m. By applying inverse this is equivalent to showing that T k U U =. So, given a point (x, y) X I will find a rectangular set V around (x, y) s.t. TV V =. First note that for the intersection to be trivial we need that if x < x < x 2 then x 2 < 2x. So first pick a x close enough to x s.t. x < x < 2x < x x < 2x. Then choose an x 2 s.t. x < x < x 2 < 2x < x x 2 < x 2 x < 2 (note that the existence of x 2 can be done by density of the rational numbers). In a similar fashion choose y < y < y 2 s.t. 2 y 2 < y. Then the rectangle V = (x, x 2 ) (y, y 2 ) has been constructed so that (x, y) V and TV V =. By an induction argument it can be shown that for any k N T k V V =. Finally, since balls are the basis of the standard topology, we may take a ball containing (x, y) but contained in V to be the nbhd U. X/Z is not a Hausdorff space Proof. I ll try to attach pictures for reference. First define sets A = {(x, y) : xy = c = 0, x > 0, y > 0} ([, 2] R) and B = {(x, y) : xy = c = 0, x > 0, y > 0} ( Rtimes[ 2, ]). Note that these sets map to the same set in the quotient. Take a sequence of points a k = (, 2 k ) A and b k = (, ) each for k N. Note that a 2 k k A and a k (, 0). Similarly b k B and b k (0, ). Convergence carries through a continuous map. Even though both sequences a k and b k are equal in the quotient, the points they converge to are not. Since convergence is non-unique the space must not be Hausdorff. To see that the quotient space is a like a union of four spaces homeomorphic to S R I will attach a picture. Claim: π (X/Z) = Z Z Z Z Proof. I ll attach pictures to go along with this explanation. I will attempt to compute this fundamental group by using Van Kampen s thm. Take the sets A = [ {(x, y) : xy = c = 0, x < 0, y > 0} and let A be the projection to the quotient space s.t. A is an infinitely long cylinder. Similarly for each of the other positive and negative portions of each axis define sets B, C, D, each of which are cylinders. Each of these sets is open because the complement is closed in the quotient space. The pairwise intersections of each will either be the empty set or a portion or a part of an infinitely long cylinder in R 3. In the latter case the intersection is path connected because we may take a path in X and project it through the quotient map to get a path in the quotient space. Finally all triple intersections are empty so that the triple intersections are trivially path connected. So we may apply Van Kampen s thm. Since the set A is an infinitely long cylinder π (A) =< a > = Z, where a is a loop once around the cylinder. Similarly π (B) =< b >, π (C) =< c >,... etc. To compute the normal group N in Van Kampen s theorem note that for any intersection of fundamental groups which is non-trivial, such as π (A B), a loop w π (A B) is on a 3

4 part of the an open ended-infintely long cylinder s.t. i AB (w) = i BA (w). Thus the group N will always be trivial and we have that π (X/Z) =< a, b, c, d > = Z Z Z Z Claim: If G is conjugate to G 2 the Y/G Y/G 2 Proof. Say G 2 = αg α for α Homeo(Y). Then since Y αy we have that in the projection Y/G 2 (which is equal to Y/αG α ) the element [αg α y] G is homeomorphic to [αg y] G. Next see that α : Y Y induces a continuous map α : Y/G Y/G 2 by α[g y] G = [αg y] G (which is given by the universal property of quotient maps). Similarly there is an induced map α : Y/G 2 Y/G. It is straightforward to check that α α = and α α =. Thus we have a homeomorphism from the Y/G to Y/G 2. Ch A 6 Proof. In the first problem I argued that the cover space corresponding to the commutator is a lattice and it s fundamental group is generated by loops around unit squares starting from a fixed basepoint, wlog say this basepoint is the origin. By prop.3 the map p a st is injective. Further, each generator for the fundamental group of the cover is in the commutator (as discussed in the first problem) so that p of the fundamental group do the cover, maps injectively into the commutator. By construction I think it is clear that the any element of the commutator corresponds to a loop in the cover based at the origin. Thus p is an isomorphism from the fundamental group of the cover space to the commutator. Since the loops around squares freely generate the fundamental group of the cover, then they correspond to the free generators of the commutator i.e. conjugates of [a, b]. Ch A 8 Claim: A finitely generated free group only has a finite number of subgroups of a given finite index Proof. Consider a group free on m generators, < a,..., a m >. This is equivalent to π ( m S ) where each copy of S corresponds to a generator a i. Let H π ( m S ) s.t. [π ( m S ) : H] = n. This group must have a corresponding cover of m S and by prop.32, it must be an n-sheeted cover. In a very hand-wavy way, consider the wedge point x 0 of m S and it s pre-image p (x). There are n points in the set p (x) and each point must have one in edge and one out edge for each generator a i. Since there are only finitely many ways to produce graphs with n points and mn edges (m edges per base point) there can only be finitely many subgroups of index n in π ( m S ). 4

5 Claim: In any finitely generated group there are only finitely many subgroups of a given index Proof. Any finitely generated group G of m generators can be seen as quotient F/N where F is a free group on m generators and N F a normal subgroup. For any subgroup H G of index n by the lattice isomorphism theorem there is a corresponding group HN F s.t. n = [G : H] = [F : HN]. Therefore if there existed an infinite number of subgroups of index n in G then there would be an infinite number of subgroups of index n in F. This would contradict the previous claim. Ch B.2 Proof. Let Y be any K(G, ). Then π (Y, y 0 ) = G. Let f : (X, x 0 ) (Y, y 0 ), as always this induces a homomorphism f : π (X, x 0 ) π (Y, y 0 ) = G. However, by assumptions f must be a trivial homomorphism i.e. f induces the trivial homomorphism. Additionally it is clear that the constant map sending all of X to y 0 induces the trivial homomorphism too. By prop..9.b these maps that induce the trivial homomorphism must be homotopic. Therefore f is nullhomotopic. 5