Topology Hmwk 6 All problems are from Allen Hatcher Algebraic Topology (online) ch 2

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1 Topology Hmwk 6 All problems are from Allen Hatcher Algebraic Topology (online) ch 2 Andrew Ma August 25, Proof. Please refer to the attached picture. We have the following chain complex δ 3 C 2 =< T > δ 2 C 1 =< a, b, c > δ 1 C =< v > δ where each C i is an abelian group of chains generated by the stated elements. Then compute for the following functions. δ 2 : T a + b c Kerδ 2 =, Imδ 2 =< a + b c > δ 1 : a v v =, b v v =, c v v = Kerδ 1 =< a, b, c >, Imδ 1 = Knowing this we can compute the homology groups. H n (X) = for n 3. H 2 (X) = Ker δ 2 = H 1 (X) = Ker δ 1 Im δ 2 = <a,b,c> <a+b c> =< a, b >= Z2 H (X) = Ker δ Im δ = <v> = Z Proof. Following the picture of the Klein bottle given on page 12. We have the following chain complex: δ 3 δ C 2 δ 2 1 δ C1 C <U,L> <v> <a,b,c> Next compute the following for the boundary maps. δ 2 : U a + b c, L a b + c Since these two are linearly independent Ker δ 2 = 1

2 and Im δ 2 =< a + b c, a b + c > δ 1 : a v v =, b, c Thus Ker δ 1 =< a, b, c > and Imδ 1 =. Thus we have the following for the homology groups. H n (K) = for n 3 H 2 (K) = Ker δ 2 =< a + b c, a b + c >= Z 2 H 1 (K) = Ker δ 1 <a,b,c>. Since this is an abelian group we can say that < a + Im δ 2 = <a+b c,a b+c> b c, a b + c >=< a + b c > + < a b + c > and compute the quotient iteratively <a,b,c> by the third isomorphism theorem. First <a+b c> =< a, b, a + b >=< a, b >. Next we quotient by the group <a b+c> <a,b> <a+b c> =< a b + a + b >=< 2a >. Then <2a> = Z Z 2. We can check this computation by comparing it to the abelianization of π 1 (K) (which this equals). H (K) = Ker δ Im δ = <v> = Z. *note to self - What helped me was computing this problem first for n = 3* Also I will attach a picture for this problem. First label the simplexes. Fix a tetrahedron i.e. a 4-simplex, call this T 1 and let T i be the ith tetrahedron iterated counterclockwise in the provided picture in Hatcher. For T i in the picture in hatcher there are two faces which are exposed on the outside of the figure and two faces which are inside the figure. Label the left - inner face Fi 1 and the top face Fi 2. Next label the edges. In the figure in Hatcher the perimeter of the figure will all be one edge in the quotient so label this x and label one verticle edge inside the figure y. For each T i label the edge shared by Fi 1 and Fi 2 as E i. Finally notice that there are exactly two vertices in the quotient of the figure provided in Hatcher. Label the vertex on the outside V 1 and the vertex at the center of the figure V 2. Now consider the chain complex and notice that δ 3 δ C 2 δ 2 1 δ C1 C C 3 =< T 1,..., T n > C 2 =< F 1 1, F2 1,..., F1 n, F 2 n > C 1 =< x, y, E 1,..., E n : x n = y n = 1 > C =< V 1, V 2 > Where these are all abelian groups. Claim: H 3 (X) = Z Proof. Consider the map δ 4 = since C 4 = so that Im δ 4 =. Then consider the map δ 3 : T i F 1 i + F 2 i F 1 i+1 F2 i+1 2

3 where the index i is taken modulo n i.e. if i = n then I mean i + 1 = 1. Now I want to show that ker δ 3 =< T T n >. To see this, suppose that the element i=1 n c it i ker δ 3. This element i=1 n c it i (c 1 c n )(F1 1 + F2 1 ) + (c 2 c 1 )(B 1 + B 2 ) + + (c n c n 1 )(Fn 1 + Fn) 2 =. This implies that c 1 = c 2 = = c n s.t. this element is really in the span of T T n. Clearly T T n ker δ 3, so ker δ 3 =< T T n >. Thus the homology group H 3 (X) = <T 1 + +T n > = Z. Claim: H 2 (X) = Proof. I got the idea to use rank - nullity from an online solution at uiuc. Note that the map δ 2 acts as follows: { F 1 δ 2 = i F 2 i y + E i E i 1 x + E i 1 E i By direct computation one can check that that Fi 1 + Fi 2 Fj 1 Fj 2 ker δ 2. Take the subgroup of the kernel < F1 1 + F2 1 F1 2 F2 2,..., F1 n 1 + F2 n 1 F1 n Fn 2 >. It is straightforward to check that these elements are linearly independent (just write down a linear combination of these elements, set equal to zero, and solve for coefficients). Next see that the images of the faces F 1 1, F1 2,..., F1 n 1 are linearly independent in C 1 by doing a similar process to the one described previously (the computations are easy to do for n = 3). By Rank - Nullity thm we have that dim ker δ 2 + dim Im δ 2 = dim C 2 We have shown that dim Im δ 2 n 1 and we have that dim C 2 = 2n hence dim ker δ 2 n 1. Since we have n 1 linearly independent elements of the kernel these must generate the kernel. Finally H 2 (X) = ker δ 2 Im δ 3 but we have that ker δ 2 =< F F2 1 F1 2 F2 2,..., F1 n 1 + F2 n 1 F1 n F 2 n > which also happens to be the image of δ 3. Hence the homology group is. Claim: H 1 (X) = Z n Proof. Consider the map δ 1 : x V 1 V 1 = δ 1 = y V 2 V 2 = E 1,..., E n V 1 V 2 Thus it s clear that ker δ 1 =< x, y, E 1 E n, E 2 E 1,..., E n E n 1 >. Now consider the map δ 2 which was defined in the previous claim. I wts that Im δ 2 =< y (E n 3

4 E 1 ), (E n E 1 ) x,..., y (E n 1 E n ), (E n 1 E n ) x >. This is true by looking at the generators of C 2 and where they map to. F 1 1 y (E n E 1 ) F 2 1 (E n E 1 ) x. F 1 n y (E n 1 E n ) F 1 n (E n 1 E n ) x Since H 1 (X) = ker δ 1 Im δ 2 =< x > but recall that in C 1, x n = 1. Hence this group is isomorphic to Z n. Claim: H (X) = Proof. This idea is borrowed from Alicia from class. While this homology group may be computed directly, it s easier to see that the space is a quotient of a path connected space hence it is path connected. By Prop 2.7 the homology group is Z Proof. Consider the chain complex of X which is δ n+1 Cn < n > δ n δ 3... C2 < 2 > δ 2 C1 < 1 > δ 1 C < > δ where each C i =< i > for i n because of the definition of X. Also note that in this case every map δ i = because when i n each i is a cycle (since it s boundary is the i 1 simplex). Thus i n, ker δ i =< i >, Im δ i =. This gets that each homology group H i (X) = < i > = Z for i n and otherwise Proof. Given the induced map i : H n (A) H n (X) s.t. i ([α]) = [i α] and refer the notation used in the picture of chain complexes at the top of page 111 in 2.1. Begin by noting that if i : A X is injective then by the definition on page 111 the induced homomorphism on chain complexes, i, will be injective. Then do a diagram chase to show that the Kernel i is trivial. Assume the homolgy class [α] ker i then i ([α]) Im δ = [i α] Im δ. Therefore β C n+1 (X) s.t. i α = δβ. If δβ = then by injectivity we get α = and would be done, so assume that δβ =. By applying i 1 β we can produce an element α C n+1 (A). By the commutivity of the chain homomorphisms we have that ı δα = δ i α but the RHS of this equation is δβ = i α ı δα = i α. By applying inverse maps of i we conclude that 4

5 δα = α = α Im δ. This shows that [α] = [] H n (A) so ker i = and injectivity is shown (a) Proof. By induction. Base case: n = : In this case X = a pt. Then H (X) = Z which is clearly free and H i (X) = i > by prop 2.8. Inductive step: Assume the result holds for n 1 and show it holds for n. First I ll show that H n (X) is free. Note that if X is dimension n then X = X n we ll always have the following long exact sequence by thm 2.13 H n (X n 1 ) i H n (X n ) j H n (X n /X n 1 ) (usually these are reduced homologies but since n > this is equivalent to ordinary homology) By our induction hypothesis we have that H n (X n 1 ) = and so ker j = Im i = j is injective. Additionally, since X n /X n 1 is a wedge of S n spheres the homology group H n (X n /X n 1 ) splits as a direct sum of H n (S n ) by corollary Furthermore, H n (S n ) = Z by corollary Thus H n (X n /X n 1 ) is a free group and since H n (X n ) injects into this group it may be viewed as a subgroup of a free group and by a previous exercise or example in the graph section we know that subgroups of free groups are free. Next I want to show that H i (X n ) = for i > n. This result comes from again examining an exact sequence H i (X n 1 ) H i (X n ) H i (X n /X n 1 ) H i 1 (X n 1 ) If i > n then H i (X n 1 ) =, H i 1 (X n 1 ) =. So H i (X n ) = H i (X n /X n 1 ) by exactness but H i (X n /X n 1 ) = since this is a direct sum of H i (S n ) terms and each will be by corollary So H i (X n ) = when i > n. (b) Proof. I ll first show part (c) of Lemma 2.34 of the next section because I think this result is easy enough to see by tools of this section. The claims is for k < n, H k (X n ) = H k (X) where X is a finite dimensional CW - complex. To get this result examine the LES H k+1 (X n+1 /X n ) H k (X n ) H k (X n+1 ) H k (X n+1 /X n ) By applying part (a) of this problem we get that H k+1 (X n+1 /X n ) = H k (X n+1 /X n ) =, = H k (X n ) = H k (X n+1 ). By inductively repeating this argument and assuming 5

6 that X is finite dimensional we can conclude that H k (X n ) = H k (X). Now I will apply the result to prove the original claim. By the result H n (X n+1 ) = H n (X). By assumption X n+1 = X n s.t. H n (X n ) = H n (X) so I will simply compute H n (X n ). Look at the exact sequence H n (X n 1 ) H n (X n ) H n (X n /X n 1 ) H n 1 (X n 1 ) By part (a) of this problem H n (X n 1 ) = and by assumption X n 2 = X n 1 s.t. H n 1 (X n 1 ) = H n 1 (X n 2 ) = by part (a) again. By exactness of this sequence H n (X n ) = H n (X n /X n 1 ). The group H n (X n /X n 1 ) is freely generated by generators in bijection with the number of n-cells because this is the homology group of a wedge of S n. Thus the claim is shown. (c) Proof. I ll use the result of part (c) in Lemma 2.34 which I proved in the previous part. Consider the exact sequence H n (X n 1 ) H n (X n ) j H n (X n /X n 1 ) Since H n (X n 1 ) = by part (a) the map j is injective and H n (X n /X n 1 ) is generated by k-elements (discussed in the previous parts of this problem) then H n (X n ) is generated by at most k elements. Next I want apply Lemma 2.34 (c) to state that H n (X n+1 ) = H n (X). Now I ll relate H n (X n ) and H n (X n+1 ) by the following exact sequence: H n (X n ) i H n (X n+1 ) j H n (X n+1 /X n ) Since H n (X n+1 /X n ) = (seen by breaking it up as a wedge of copies of S n+1 ) we have that H n (X n+1 ) = ker j = Im i hence H n (X n ) is surjective. This shows that H n (X n+1 ) is generated by at most k elements and so then is H n (X) There s a great link to explain how to compute boundary maps in the lecture notes of Maxim: maxim/topnotes1.pdf maxim/topnotes2.pdf (a) Proof. The quotient of S 2 with two anitpodal points identified is homotopically equivalent to the wedge S 2 S 1 (I believe this was explained in ch. ) and call this space X. At this point you could break up the homology of the wedge as the direct sum of 6

7 the homology of the parts for a quick answer, but out of anger at my midterm I will compute this with cellular homology. This CW - complex can be seen as the combination of 1 e, 1 e 1, and 1 e 2. We can see immediately that H (X) = Z because X is path connected (it s a quotient of a path connected space). We can compute the other groups by applying cellular homology and the properties of Lemma Here is the cellular complex: Z d 2 Z d 1 Z d Now the map d 1 is the trivial map because by the cellular homology isomorphism H (X) = ker d Im d 1 = Z by the earlier discussion. We can next use the cellular boundary formula to see that the disc D 2 is not attached along the 1-cell so that the map d 2 must be the trivial map. This gets homology groups H 1 = H2 = Z. (b) Proof. Let X be the space S 1 (S 1 S 1 ) and note that space is similar to two torii attached along the outside by a copy of S 1 (for example nested doughnuts or two stacked bagels). We can build this CW - complex using 1 e, 3 e 1, and 2 e 2 cells. Again H (X) = Z because there is only one path connected component (X is a quotient of a single torus, which is path connected). I ll use cellular homology to compute the rest. d 3 Z 2 d 2 Z 3 d 1 Z d By reasoning similar to part (a) we can conclude that d 1 is the trivial map and by directly computing the boundary map d 2 following Maxim s notes one can see that for example one of the e 2 cells is attached along the 1-cells by aca 1 c 1 (if the loops are labeled a, b, c). Then whether you collapse a, b, or c either way the resulting boundary map becomes has degree. The same is true for the other 2-cell we are attaching and so we may conclude that d 2 is the trivial map. Putting this all together we can compute the homology groups (using kernels and images of this chain complex) to get that H 1 = Z 3, H 2 = Z 2. (c) Proof. Since this hmwk assignment was not collected I didn t complete these problems but... All we need to do is to put a CW - complex on this. I motivated this by computing if there was only one hole removed (which is a torus). So I see this as label the boundaries of all 3 circles as, a, a 1-cell. Connect these copies of a by 1-cells b and c. Then attach 7

8 2-cells along the boundaries aba 1 b 1, aca 1 c 1, and a(bc)a 1 (bc) 1 - for a total of 3 2-cells. Since there is only 1 -cell we have the following chain complex: d 3 Z 3 d 2 Z 3 d 1 Z d Again d 1 is trivial (because there is only one -cell) and by direct computation of d 2 we can see that d 2 is a trivial map too. Now all the CW homologies are computatble since we know the maps. (d) Proof. Again all we need is a CW - complex. The motivation the CW structure is to basically do what you would do for a torus i.e. S 1 S 1. In this case take two 1-cells to make loops and call them a and b. Then attach one 2-cell along a n b m a n b m. We get the complex: d 3 Z d 2 Z 2 d 1 Z d Again we realize that d 1 is trivial and directly compute d 2 which we will see is also trivial. So everything will be computable Claim: The quotient map S 1 S 1 S 2 collapsing the subspace S 1 S 1 to a point is not nullhomotopic Proof. Let q be the quotient map. Since q is surjective the induced map q : H 2 (S 1 S 1 ) H 2 (S 2 ) is surjective (to see this you can consider the simplex chain complexes that are given by CW-complexes). All that is left is to show that q is injective. Consider the exact sequence given by the thm 2.13: H 2 (S 1 S 1 ) i H 2 (S 1 S 1 ) q H 2 (S 1 S 1 /S 1 S 1 ) Here S 1 S 1 /S 1 S 1 = S 2 and H 2 (S 1 S 1 ) = because it has no 2-cells. Then Im i = = ker q = by exactness and q is injective. Thus q is an isomorphism between H 2 (S 1 S 1 ) and H 2 (S 2 ). Claim: Any map S 2 S 1 S 1 is nullhomotopic Proof. Let f be such a map and note that because π 1 (S 2 ) = then there exists a lift of f, f to a map S 2 S 1 S 1 the universal cover of S 1 S 1. Since the universal cover of the torus is R 2 and this is contractible we can construct a homotopy from f to a constant map, φ : S 2 R 2. Let f t be the homotopy of f φ. From this we can construct a homotopy of f to a constant map, namely f t = p f t. Thus the map f is nullhomotopic. 8

9 (a) Proof. Call the space in question X and note that for n 3 the homology group H n (X) = because there are many zeros in the Mayer-Vietoris sequence. Additionally H (X) = Z because X is path connected anyway. So first I ll compute the second homology group using the following exact, Mayer-Vietoris sequence: H 2 (M T) = i 1 H2 (M) H 2 (T) =Z j H 2 (X) δ H 1 (M T) i 2 H 1 (M) H 1 (T) H 2 (M T) = because M T is the wedge of two circles which has no 2-cells and by Cell complexes must have trivial second homology group. Similarly we can compute that H 2 (M) = and I believe that it was shown in the book that H 2 (T) = Z. So now we know the second term in this exact sequence. Then using knowledge of maps (in a similar manner to the computation example in the book with the Klein bottle) we know that i 2 : 1 (1, (, 2)) because circle in which Mobius strip and torus are attached is one generator loop in M but 2 generator loops in T. It is clear that i 2 is injective so that = ker i 2 = Im δ. Thus δ is the trivial map and ker δ = H 2 (X) = Im j. Since j is injective already (by examining the map i 1 ), we have that j is an isomorphism so that H 2 (X) = Z. To determine H 1 (X) use the following exact reduced Mayer-Vietoris sequence: H 1 (M T) i 2 H 1 (M) H 1 (T) j 2 H 1 (X) δ 2 H (M T) = The last term is the trivial group because I chose to look at the reduced sequence. By exactness j 2 is a surjective map. So to get an isomorphsim we simply need to quotient H 1 (M) H 1 (T) by ker j 2 = Im i 2 = (1, (, 2)) as discussed previously. Thus we have the resulting quotient being = Z Z Z 2 = H1 (X). (b) This will all be identical computation to part (a) so I won t do it Before doing this problem it is helpful to study example 2.47, the Klein bottle, to learn how to use Mayer-Vietoris. I have some comments on the proof in the book. 1. The maps are induced by the inclusion maps on chains. In particular ι : x (x, x) but these are what the element x represents in the new space. This is why 9

10 1 gets sent to 2 because in the inclusion the generator of the intersection is actually two loops when seen in the klein bottle. 2. By exactness we know the next map, which I ll call φ is a homomorphism. It s kernel is the image of ι so by the first isomorphism theorem, if we quotient the group by the kernel, we get an isomorphism. So we just need to produce generators which quotient nicely. Now to tackle the question. Proof. I ll first compute the homology groups of X. For n 4 H 4 (X) = because X as a CW-complex doesn t have 4-cells or higher. This can be more rigorously shown by looking at the Mayer-Vietoris sequence for n 4. Since X is path connected (at least I think it is), then H (X) = Z. Next realize that H ( R) = H k (D 2 ( g 1 S1 )) = H k ( g 1 S1 ) = Z g for k = 1 and o.w. Using this now examine the Mayer-Vietoris sequence H 3 (X) H 2 (M g ) H 2 (X) ψ φ H 1 (M g ) H 1 (R) H 1 (R) H 1 (X) Z Z 2g Z g Z g Right away we see that H 3 (X) = Z by exactness and the s. Next we need to understand the maps here. By exactness ψ is injective and Im ψ = ker φ by exactness. So we just need to understand the map φ. This map is induced by the inclusion by chains and since we are examining the first homology groups which are the abelianizations we can make claims about the kernels of the map φ. In H 1 (M g ) say the elements are (x 11, x 12,..., x g1, x g2 ) where x i1 corresponds to a loop through the ith hole and x i2 correspond to a loop around the ith hole. Using the inclusion map φ, (x 11, x 12,..., x g1, x g2 ) ( (, x12,...,, x i2,...,, x g2 ), (, x 12,...,, x i2,...,, x g2 ) ) or better notated ( (x12,..., x g2 ), (x 12,..., x g2 ) ) since all loops through holes in R are homotopic to trivial loops. Hence the kernel of this map is isomorphic to Z g. By applying exactness of the sequence while examining kernels and images we can conclude that H 2 (X) = Z g. By applying exactness and the first isomorphism like in the Klein bottle example, we can compute H 1 (X) by quotienting H 1 (R) H 1 (R) by Im φ. By choosing a basis carefully, such as ((x 12,,..., ), (x 12,..., )) and ((x 12,,..., ), (,..., )) etc. we can get the quotient H 1 (X) = Z g. Thus we have computed every homology group Proof. Since SX can be seen as a double cone let A be the top cone and B be the bottom cone, taking a little extra of the other cone as necessary to ensure that the interiors of each set union to be SX. Now consider the Mayer - Vietoris sequence, specifically consider the reduced version: H n (A) H n (B) H n (SX) H n 1 (A B) H n 1 (A) H n 1 (B) 1

11 Recognize that A B can deformation retract along lines (used in the creation of the cones) to the space X so we may consider H n 1 (A B) = H n 1 (X). Furthermore since the sets we took are contractible to the point of a cone (by moving along lines to the point of cones) then H k (A) = H k (B) = s.t. H k (A) H k (B) = for any n. Hence by exactness of the sequence we get that H n (SX) = H n 1 (X) for any n. Proof. Now I ll compute the relative homology groups. This uses ideas borrowed from a place on the internet. To start, realize that for n 4 the relative homology H n (R, M g ) = by looking at the LES. For n = 3 we have the sequence H 3 (R) H 3 (R, M g ) M 2 (M g ) H 2 (R) By exactness we have an isomorphism H 3 (R, M g ) = M 2 (M g ) = Z. For n = 2 we have the sequence H 2 (R) H 2 (R, M g ) δ H 1 (M g ) Z 2g i H1 (R) Z g j By exactness we have that δ is injective. Similar to the discussion in the previous proof we know i is induced by the inclusion map, and since we are dealing with the first homology groups we understand the homology classes and how i maps. Specifically, ker i = Z g and since Im δ = ker i we have that H 2 (R, M g ) = Z g. For n = 1 look at the exact sequence H 1 (M g ) i H 1 (R) j H 1 (R, M g ) H (M g ) By exactness j is surjective and by using the first isomorphism theorem we can say that H 1 (R, M g ) = H 1 (R)/ ker j. By noting ker j = Im i. Since we know how the map i works (it induced from the inclusion and this is only the first homology group) we know that i is surjective. Therefore Im i = H 1 (R) s.t. H 1 (R, M g ) = H 1 (R)/Im i = Finally for n =, I didn t bother computing this because I think it will be easy. Probably do a similar argument with assessing understood maps with an exact sequence involving H (R, M g ). 11