Topology Hmwk 3 All problems are from Allen Hatcher Algebraic Topology (online) ch 1

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1 Topology Hmwk 3 All problems are from Allen Hatcher Algebraic Topology (online) ch Andrew Ma December 23, 203 This assignment has been corrected post - grading...6 (a) Proof. Assume for a contradiction that there exists a retract. By prop..7 a retract implies that the homomorphism induced by the inclusion of A X is injective. This implies π ( A) π ( X ), or Z injects into the trivial group which is a contradiction. (b) Proof. Similarly apply prop..7 for a contradiction. If a retract existed then this would imply an injection from π (S D2 ) = π ( S ) π ( D 2 ) = Z to π (S S ) = Z Z, which is impossible. (c) Proof. Note that the space A is a deformation retract of a space, I ll call T the twisted torus (essentially a torus with a twist in it - I ll try to remember to attach a picture of this space). Then see that T also deformation retracts to something homeomorphic to S S therefore A is homotopically equivalent to S S so that the fundamental groups are isomorphic. That is π ( A) = π (S S ). Then see that π (S D2 ) = Z. The as in previous reasoning pi (S S ) does not inject into Z. (d) Proof. X clearly deformation retracts to a point so that π ( D2 D2 ) = 0. It is also clear S contains loops which aren t homotopic to a point. that π (S S ) 6 0 because S = Thus by prop..7 there can t exist a retract because π (S S ) 6 0.

2 (e) Proof. X has a clear deformation retract to a circle therefore π ( X ) = Z. = π ( S ) However π (S S ) = Z Z (this is from chapter.2 with Van Kampen s thm but even if you didn t have this you could probably show that the group π (S S ) isn t generated by just one loop - perhaps by showing it directly). At any rate there can t be a retract because π (S S ) does not inject into Z. (f) Proof. The Mobius band can deformation retract to a circle so π ( X ) = π ( S ) = Z. The boundary circle of the Mobius band is homeomorphic to S S so that π ( A) = π (S S ), by reasoning similar to that in problem (d) we see that pi ( A) 6 π ( X ) so that there cannot be a retract from X to A...7 Proof. Let the map rn : S S S be a map which fixes one copy of S but maps the other copy S to n loops in the first copy of S based at the wedge point. This map is will be a retraction to the first copy of S since this is a cts map and it fixes the first copy of S. Finally see that rn 6 rm if n 6= m. We can see this is true by restricting these maps to the copy of S which is sent to n loops or m loops respectively. There can be no homotopy between these maps and it is enough to show this since there is an obvious homotopy between the maps rn and rm when restricted to the copy of S which is fixed, then by cancellation (which we have shown for path homotopies, but I believe works in this scenario too), this induces a homotopy of rn and rm when restricted to the copy of S that gets mapped to loops. To see that no such induced homotopy can exists, note that any cts map from S to S is a loop in the space S. Without loss of generality we can assume that these loops are based at the wedge point (because S, which we are mapping to, is path connected). Then a homotopy of the restricted maps would be a path homotopy of an n loop to an m loop which can t exist if n 6= m. Thus each rn is nonhomotopic to another rm when n 6= m and therefore we get an infinite number of retracts..2.5 **Note to self : I believe that I could prove a recursive algorithm to show that given a set of generating loops any loop could be created from the generators, however I did not have time to complete this argument ** I got the solution to this problem from a post on math stack exchange. Here is a link for reference 2

3 because I am going to type up the general idea of this proof very briefly. Proof. In the graph X we can take a maximum spanning tree, M. If M = X then we are done because the fundamental group of a tree is a trivial group. If M 6= X then label the finite number of edges in X \ M as e,..., en. Consider a cover of X by sets M ei for i =... n. (I have yet to see why these sets are open - but I m running only a few minutes before class). I believe that we can make the M ei open by taking slightly larger open endpoints of this set. The triple intersection and pairwise intersections both contain M so they are path connected. We may now apply Van Kampen s thm. The pairwise intersection of each set M ei M e j for i 6= j is simply M and therefore is trivial. The normal group containing the inclusions of these trivial fundamental groups is therefore trivial. Therefore, the fundamental group of X is simply n π ( M ei ). All that is left is to compute the π ( M ei ) for each ei. This fundamental group can be generated by a loop starting from a fixed base point, x0 X, which travels once around the bounded region that has edge ei. This shows the claim..2.6 I don t know how to show the last part of this problem for dimension equal to 3. Claim: The inclusion X Y induces an isomorphism on π Proof. Without going into too much detail I believe that we can repeat the argument given in the proof of Prop.26 using n-cells instead of 2-cells. The only difference would be noticing that the π ( A B) = 0 because each open set Aα (used to apply Van Kampen s thm to A B) can deformation retract to a point due to the n-cell, being an n dimensional closed ball, with a point removed is still contractible (because n 3). Thus we get that the injection of fundamental groups is actually an isomorphism. **Note to self : In showing that the space has trivial fundamental group I don t why we would use CW complexes when you can produce a wedge sum and solve the problem in similar way to determining π of the wedge of circles** Claim: The complement of a discrete subspace of R n is simply connected for n 3 Proof. Let A be the discrete subspace in R n (for n 3). First I will show that R n \ A is path connected. Since A is discrete, for each pt a in A, there exists an open ball in R n s.t. this ball intersects with A contains only a, call this ball Bα. Then consider the collection of balls in R n. Note the fact that the boundaries of these n-dimensional balls are path connected (this is shown in Munkres example 5 on page 56 - Using a cts map from R n \ {0} Sn ). Now consider any two points of R n \ A. If these two points are not contained in any ball Bα, then draw a straight line between the two points and if this line intersects the 3

4 boundary of any ball Bα then use the path connectedness of the boundary to connect one point of the intersection to the opposite point of intersection instead of going straight through the ball. Doing this for each ball the line meets will result in creating a new path which still connects the two points in question. If the two points are each contained in balls. For each point take a straight line path from to the boundary of each respective ball and finally connect the boundaries of each ball by a straight line in a similar manner to the above. In fact using this same idea we can produce paths between pairs of points in any case i.e. one point is in a ball the other is not in a ball, both points are in the same ball, etc. Now I will argue that the space R n \ { A} has trivial fundamental group. Given that for each pt a A there is a nbhd of a which contains no other points of A, I ll call this nbhd Bα. We can deformation retract the space R n \ A to a collection of boundaries of balls (one for each Bα ) each of which is connected by a path as described in the previous paragraph to the origin. That is each is connected by a straight line to the origin which paths around any boundary of a sphere which it intersects. (Aside : The retract may be done by considering a smaller closed ball inside each of the open balls Bα ) *At first I thought the ability to perform this deformation retract was clear by thinking of examples in R3 of balls, however I realize now it may not be so clear, but it is too late in the night for me to turn back*. Assuming that this deformation retract can be done we can then consider these boundaries of balls as copies of en cells which have been attached to a collection of line segments (these line segments are the paths connecting the boundary of each ball to the origin. The fundamental group of this collection paths is trivial. If n > 3 then by the earlier part of 6 the attaching the en cells do not change the fundamental group. In the case of n = 3 we can say that... I guess I m not sure. Hunh?.2.8 We can be more specific in this problem. There is a more detailed answer at www3.nd.edu/ Inicolae/ProblemsHatcher.pdf. Proof. Consider the attachment of two torii seen as two labeled squares with the attached along a side. Say the edges are labeled a, b, c where the squares are attached along edge a. Then by applying Van Kampen s theorem for CW - complexes (this space is a CW - Complex) the resulting fundamental group is < a, b, c aba b, aca c >. 4

5 Proof. For this proof I will include a picture to illustrate the sets and deformation retracts which I will describe here. Consider the two torii as unit squares each with opposite edges associated. To have these torii associated by a circle in each would mean to intersect these unit squares much like the intersection of two perpendicular planes. Apply Van Kampen s thm to this set of intersected squares. Define an open set A which is the original space but with the opposite edges of one of the square removed, that is remove a circle from one of the torii. Define a similar set B but with the opposite edges of the other square removed. Each individual set, A or B, can deformation retract to one unit square which represents a torus. Thus π ( A) = π ( B ) = Z Z. The intersection, A B, can deformation retract to a line, which represents a circle in the quotient space. Under inclusions, the normal subgroup we get, I believe, will be Z, the fundamental group of the circle. By Van Kampen s thm this gets that the fundamental group of the (Z Z ) (Z Z ). space is Z.2.9 Claim: Mh does not retract to C Proof. Consider the 4g polygon which quotients to become Mh (as described on page 5 of chapter 0). The corresponding polygon for Mh would be the same 4g polygon but with an open circle removed from it s center. We can deformation retract this radially to get that π ( Mh ) is isomorphic to the fundamental group of the image of the perimeter of this polygon under the quotient map. That is π ( Mh ) =< a, b,..., a g, bg >, where this last group can be thought of as π ( 2g S ). Note that in this isomorphism the loop of C will correspond to the element a b a b... a g bg a g b g i.e. the loop that is the image of the perimeter of the 4g-gon under the quotient map, call this loop P. Now assume for a contradiction that there exists a retract from Mh C. This implies that π (C ) π ( Mh ). By the previous isomorphism this would imply that π ( P) π ( 2g S ). From this we can make an exact sequence 0 π ( P) π ( 2g S ) 0 We ll now apply a functor F which will be abelianization to this exact sequence. This gets 0 F (π ( P)) F (π ( 2g S )) 0 Since abelianization is a right exact functor this sequence ought to be exact, however, F (π ( P)) cannot inject into F (π ( 2g S )). Note that F (π ( P)) = π ( P) since π ( P) corresponds to pi (C ) = Z which is already abelian. Also in the abelianization of π ( 2g S ) the element P is one of the elements which gets quotiented out by the commutator. Therefore this cannot be an injection which gives a contradiction. Since a retract of Mg to C would imply a retract of Mh to C by a restriction of the map, it can t be that such a retract exists. Claim: Mg retracts onto C 5

6 Proof. Consider the 4g-gon which quotients to Mg as described on page 5 of chapter 0. Note that the circle C corresponds to one edge in the perimeter of this 4g-gon. We can then iteratively contract corresponding edges in the 4g-gon. For example if the perimeter is labeled a b a b... a g bg a g b g then we can first contract the edges a and a to a point and continue with the other edges until we are left with the edge bg whic, wlog, corresponds to the circle C. Then applying the universal property of quotient maps, this retraction of the 4g-gon to an edge induces a retraction from Mg to the circle C..2.0 Proof. I ll prove this with a series of deformation retracts for which I will attach a set of drawings. Let X be the space of D2 I without α β. In a similar manner to the deformation retract of R2 to S S we can deformation retract space X to a space made of two hollow cylinders (without tops and bottoms) linked like horseshoes. 2. We can next deformation retract one of these horseshoe tubes to get the space which is one tube with a copy of S attached at one point 3. Next deformation retract the other tube into another copy of S so that the space we have left is the wedge S S. Finally note that after the series of deformation retracts, the loop γ becomes a path once around each of the circles in S S. Therefore if γ was nullhomotopic in X then it would be nullhomotopic in the image of the series of deformation retracts, that implies the path once around each circle in S S is nullhomotopic, which is a clear contradiction..2. Compute a presentation for the fundamental group of the mapping torus of S S Proof. First I will compute the mapping torus of the wedge S S following the hint. Let X = S S and consider the CW complex X S. Say the space X S is generated by the loops a, b, γ, where γ is the loop of the mapping torus, a.k.a. the newly wedge copy of S. We can form the mapping torus by attaching 2 2-cells. The first 2-cell should be around the path aγ f ( a)γ and the second 2-cell should be around the path bγ f (b)γ. Now we may apply proposition.26 to compute a presentation of the mapping torus. We know π ( X S ) =< a, b, γ >, the fundamental group of the mapping torus is < a, b, γ aγ f ( a)γ, bγ f (b)γ >. 6

7 **Note to self : it helped me to think about the case of the mapping torus of a circle or a disc first** Compute a presentation of the fundamental group of the mapping torus of S S Proof. I got this argument from Alicia (another student in the class). Start with the wedge of two circles which is created by the loops a, b. Attach a copy of S, called γ which will be a part of the mapping torus. Then attach a 2-cell to create a torus from the wedge sum. Attach along the path ab a b. Next attach 2 2-cells, each along a respective generator as it is swept through the mapping torus. Similar to before we get the words aγ f ( a)γ and bγ f (b)γ. By prop.26 we can compute a presentation for the fundamental group as < a, b, γ ab a b, aγ f ( a)γ, bγ f (b)γ >...20 Proof. The fudamental group of X can be shown to be the same as n π (Cn ) by a similar argument to what was done to show that the infinite wedge sum is n Z. Take the open set An to be the circle Cn union a small open nbhd around the origin intersected with X. So small s.t. the nbhd doesn t contain any Ci for any i. Then any pairwise intersection will be path connected and can deformation retract to the origin. Similarly for any triple intersection. It then follows from the Van Kampen thm that π ( X ) = n π (Cn ). This is the same as the fundamental group of the infinite wedge sum because an individual Cn is homeomorphic to S, which implies that the fundamental groups are isomorphic. I ll show that the infinite wedge sum of circles is homotopically equivalent to X by showing that both are CW-complexes with attaching maps which are homotopic (It is also possible to show that both spaces can be made of deformation retracts of R +, R with holes at along the x axis at each positive integer x values). Consider the set which is the collection of circles of radius n. Call each circle Cn and the set X0. For each Cn we can attach this set to the origin by an attaching map f n which is to simply attach Cn by one point of Cn. There is another attaching map gn which is to map an arc of Cn to the origin so that the result is a copy of S attached to the origin. Note that both maps f n and gn are homotopic. Then we may combine the maps f n, gn to get attaching maps F, G respectively for X0 (note here F C = f n, similarly for G). n Both F and G will be homotopic simply by doing each homotopy between f n and gn for all n, simultaneously. Since the attaching maps of a CW complex are homotopic then the new spaces created will be homotopically equivalent, thus X and the infinite wedge sum of circles are homotopically equivalent. The space X is not homeomorphic to S because it is not compact in R2 since it is not bounded, however, the infinite wedge is compact in R2 because it is closed and bdd. Since homeomorphisms preserve compactness these two spaces cannot be homeomorphic. 7

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