Solution: We can cut the 2-simplex in two, perform the identification and then stitch it back up. The best way to see this is with the picture:
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1 Samuel Lee Algebraic Topology Homework #6 May 11, 2016 Problem 1: ( 2.1: #1). What familiar space is the quotient -complex of a 2-simplex [v 0, v 1, v 2 ] obtained by identifying the edges [v 0, v 1 ] and [v 1, v 2 ], preserving the ordering of vertices? Solution: We can cut the 2-simplex in two, perform the identification and then stitch it back up. The best way to see this is with the picture: It s a mobius band! Problem 2: ( 2.1: #4). Compute the simplicial homology groups of the triangular parachute obtained from 2 by identifying its three vertices to a single point. Notation: For these problems, S denotes the free abelian group generated by elements of S. Solution: Denote A = [v 0, v 1, v 2 ], a = [v 0, v 1 ], b = [v 1, v 2 ], c = v 0, v 2 ] and then identify v 0 v 1 v 2 =: v. We have the chain complex C 2 (X) 2 C 1 (X) 1 C 0 (X) 0 0 where C 2 (X) = A, C 1 (X) = a, b, c and C 0 (X) = v. Then from 2 (A) = a + b c and 1 (a) = 1 (b) = 1 (c) = v v = 0
2 we can deduce that ker 2 = {0}, Im 2 = a + b c, ker 1 = a, b, c = C 1 (X), Im 1 = {0} and ker 0 = C 0 (X). The homology groups are H 2 (X) = ker 2 = {0} H 1 (X) = ker 1 Im 2 = a, b, c a + b c = Z Z H 0 (X) = ker 0 Im 1 = v = Z. Problem 3: ( 2.1: #5). Compute the simplicial homology groups of the Klein bottle using the -complex structure: Solution: Similar to the previous problem, we have the chain complex C 2 (X) 2 C 1 (X) 1 C 0 (X) 0 0 but this time with C 2 (X) = U, L, C 1 (X) = a, b, c and C 0 (X) = v. We have and so 2 (U) = a + b c 2 (L) = a b + c 1 (a) = 1 (b) = 1 (c) = v v = 0 ker 2 = {0} ker 1 = a, b, c Im 2 = 2a, a + b c Im 1 = {0} The homology groups are then given by H 2 (X) = ker 2 = {0} H 1 (X) = ker 1 Im 2 = a, b, c 2a, a + b c H 0 (X) = ker 0 Im 1 = v = Z.
3 Problem 4: ( 2.1: #7). Find a way of identifying pairs of faces of 3 to produce a -complex structure on S 3 having a single 3-simplex, and compute the simplicial homology groups of this -complex. Solution: Consider the -complex X obtained from [v 0, v 1, v 2, v 3 ] by identifying [v 1, v 2, v 3 ] [v 0, v 2, v 3 ] and [v 0, v 1, v 3 ] [v 0, v 1, v 2 ] by folding along [v 2, v 3 ] and [v 0, v 1 ], respectively. The orientations are given for the first identification via and the orientations for the second by With this gluing we have the identifications [v 0, v 3 ] [v 1, v 3 ] and [v 0, v 2 ] [v 1, v 2 ] [v 0, v 3 ] [v 0, v 2 ] and [v 1, v 3 ] [v 1, v 2 ]. [v 0, v 2, v 3 ] [v 1, v 2, v 3 ] and [v 0, v 1, v 3 ] [v 0, v 1, v 2 ] (1) [v 0, v 3 ] [v 1, v 3 ] [v 1, v 2 ] [v 0, v 2 ] [v 0 ] [v 1 ] and [v 2 ] [v 3 ] and will compute the homology groups from via the chain where 0 C 3 (X) 3 C 2 (X) 2 C 1 (X) 1 C 0 (X) 0 0 (2) C 3 (X) = [v 0, v 1, v 2, v 3 ] C 2 (X) = [v 0, v 1, v 3 ] [v 0, v 2, v 3 ] C 1 (X) = [v 0, v 3 ] [v 2, v 3 ] [v 0, v 1 ] C 0 (X) = [v 0 ] [v 2 ]. First, we should substantiate the claim that X is indeed S 3. To see this we consider the barycenter v of 3 and then decompose it into smaller 3-simplices [v, v 0, v 1, v 2 ], [v, v 0, v 2, v 3 ], [v, v 1, v 2, v 3 ] and [v, v 0, v 1, v 3 ]:
4 Performing the identification (1) glues the back simplex [v, v 1, v 2, v 3 ] to the left simplex [v, v 0, v 2, v 3 ] via [v 0, v 2, v 3 ] [v 1, v 2, v 3 ] and similarly the right simplex [v, v 0, v 1, v 3 ] to the bottom simplex [v, v 0, v 1, v 2 ] with the identification [v 0, v 1, v 3 ] [v 0, v 1, v 2 ]: Looking at the left of these objects, there are two faces [v, v 2, v 3 ] which in the original diagram were the one 2-simplex between [v, v 1, v 2, v 3 ] and [v, v 0, v 2, v 3 ]. Topologically the union of these forms a disc, which when contracted to a point, say v, leaves us with at solid ball D 3. By the same process, the right object can also be seen to be a D 3, and it is not hard to see that with all of the identifications made, these two D 3 s are glued along their boundaries. Hence the resulting space is X S 3. Now we compute the homology for X. Recalling the identifications (1) the values for the
5 boundary maps are 3 ([v 0, v 1, v 2, v 3 ]) = [v 1, v 2, v 3 ] [v 0, v 2, v 3 ] + [v 0, v 1, v 3 ] [v 0, v 1, v 2 ] = 0 2 ([v 0, v 1, v 3 ]) = [v 1, v 3 ] [v 0, v 3 ] + [v 0, v 1 ] = [v 0, v 1 ] 2 ([v 0, v 1, v 2 ]) = [v 1, v 2 ] [v 0, v 2 ] + [v 0, v 1 ] = [v 0, v 1 ] 1 ([v 0, v 3 ]) = [v 3 ] [v 0 ] = [v 2 ] [v 0 ] 1 ([v 2, v 3 ]) = [v 3 ] [v 2 ] = 0 1 ([v 0, v 1 ]) = [v 1 ] [v 0 ] = 0 0 ([v k ]) = 0 Now we can easily compute the homology groups for X: H 3 (X) = ker 3 = [v 0, v 1, v 2, v 3 ] = Z H 2 (X) = ker 2 Im 3 = {0} {0} = {0} H 1 (X) = ker 1 Im 2 = [v 0, v 1 ] [v 2, v 3 ] [v0, v 1 ] [v 2, v 3 ] = {0} H 0 (X) = ker 0 Im 1 = [v 0] [v 2 ] [v2 ] [v 0 ] = Z This agrees with the previously derived fact that H i (S n ) = Z exactly when i {0, n}. Problem 5: ( 2.1: #11). Show that if A is a retract of X then the map H n (A) H n (X) induced by the inclusion A X is injective. Proof. Let r : X A be a retract and i : A X be inclusion. Then r i = 1 and so the induced map of the composition is r i = 1 which implies that i : H n (A) H n (X) must be injective. Problem 6: ( 2.1: #15). For an exact sequence A e B f C g D h E (3) show that C = 0 iff the map A B is surjective and D E is injective. Hence for a pair of spaces (X, A), the inclusion A X induces isomorhpism on all homology groups iff G n (X, A) = 0 for all n. Proof. Denote the maps by e, f, g and h, respectively. If C = 0, then Im e = ker f = B so e is surjective. Also, for g to be a homomorphism g(0) = 0 and so the exactness of the sequence implies that ker h = Im g = {0} which in turn implies that h is injective.
6 Suppose that e is surjective and h is injective. So Im e = B and ker h = 0. Then by exactness ker f = Im e = B and Im g = ker h = 0. The first of these says that the image of f is 0 latter of these says that the ker g = C. By exactness, 0 = Im f = ker g = C. Problem 7: ( 2.1: #17). (a) Compute the homology groups H n (X, A) when X is S 2 or S 1 S 1 and A is a finite set of points in X. Proof. To compute the relative homology groups H n (X, A), we examine the long exact sequence... H 2 (A) i H 2 (X) j H 2 (X, A) H 1 (A) i H 1 (X) j H1 (X, A) H 0 (A) i H 0 (X) j H 0 (X, A) 0. (4) To figure out the H n (X, A), we want to fill in as much of the sequence as possible and then exploit its exactness. First we will do the case where X = S 1. As established in class, H i (S n ) = Z when i = n and is the trivial group otherwise, in this case H 2 (X) = Z. We also have that H 0 (Y ) for a space Y is just r k=1 Z where r is the number of path components of Y. Hence H 0(A) = A k=1 Z and H n (A) = 0 otherwise due to proposition 2.8. Using this information we can form a short exact sequence in the first half of (4): H 2 (A) i H 2 (X) j H 2 (X, A) H 1 (A) i 0 i j Z H2 (X, A) 0 i. (5) The fact that Im (Z) = ker = H 2 (X, A) implies that H 2 (X, A) = Z. To find H 1 (X, A) we look at the reduced homology groups of the latter half of (4). The reduced homology for a space Y is isomorphic to the regular homology for n 0 in which case H 0 (Y ) = H 0 (Y ) Z. So H 0 (A) = A 1 k=1 Z. This also means that H 1 (X) = H 0 (X) = 0. The end of the exact sequence is i H 1 (X) j H 1 (X, A) H 0 (A) i H 0 (X) j H 0 (X, A) 0 i 0 j H1 (X, A) Z A 1 i j 0 H0 (X, A) 0 Again, the short exact sequence between H 1 (X) and H 0 (X) implies that the two middle groups are isomorphic, namely that H 1 (X, A) = H 1 (X, A) = Z A 1. Lastly, each other H i (X, A) will be between two trivial groups in the long exact sequence. Thus the 0 = Im j = ker = H n (X, A) for n 1, 2. In summary: H n (S 2, A) = Z A 1 n = 1 Z n = 2 0 Otherwise. Now let X = T = S 1 S 1 and again we will use the sequence (4). To compute H 2 (T, A), it happens that (5) still holds and by the exact same reasoning, so H 2 (T, A) = Z. We are left with
7 the remainder of the sequence H 1 (A) i H 1 (T ) j H 1 (T, A) T 0 (A) i H 0 (T ) j H 0 (T, A) 0. By previous remarks, H 1 (A) = 0 and H 0 (A) = Z A and we also know (from Hatcher) that H 1 (T ) = Z Z. Since T is path connected, H 0 (T ) = Z. So 0 i Z Z j H 1 (T, A) Z A i j Z H0 (T, A) 0. Moving to reduced homology, on the right hand side of this sequence we have i j 0 H0 (T, A) 0 which tells us that H 0 (T, A) = 0 and therefore that H 0 (T, A) = 0. Lastly, to identify H 1 (T, A) notice that (T, A) is certainly a good pair since A is a deformation retract of open discs around the points that comprise it giving the isomorphism H 1 (T, A) = H 1 (T/A). Furthermore, the first homology group of a space is isomorphic to the abelianization of it s fundamental group, and I claim that π 1 (T/A) = A 1 Z. To see this we look at the topology of T/A; it is homotopic to (in fact a deformation retract of) S 1 S 1 CA with CA being the cone A I C, C identifies A 1 and C 0 having the obvious gluing to the set A of points on the torus. We can deformation retract one of the legs of the cone, a I to a point and the resulting space is the wedge of the torus with A 1 circles: ( S 1 S 1) A 1 S 1. By Van Kampens theorem, the fundamental group of this space is π 1 ( S 1 S A 1 1) S 1 = (Z Z) A 1 1=1 Z. H 1 (T/A) is the abelianization of this, and so is the group A +1 Z. To summarize: 0 n = 0 H n (T, A) = Z A +1 n = 1 Z n = 2 (6) Problem 8: ( 2.1: #18). Show that for the subspace Q R, the relative homology group H 1 (R, Q) is free abelian and find a basis. Proof. We can use the tail end of the reduced relative sequence H 1 (R) H 1 (R, Q) H 0 (Q) H 0 (R). We know that H 1 (R) = H 1 (R) = 0 since R deformation retracts to a point and H 0 (R) Z = H 0 (R) = Z implies that H 0 (R) = 0. Also H 0 counts path components so H 0 (Q) Z = H 0 (Q) = Z Q gives H 0 (Q) = Z Q 1 = Z Q. Hence the sequence above becomes 0 H 1 (R, Q) Z Q 0
8 and so which is free abelian. H 1 (R, Q) = H 1 (R, Q) = Z Q I claim that B = {[p, 0] p Q} {[0, q] q Q} is a basis. We can consider these as images of 2-simplices, and are all certainly contained in the kernal of since the endpoints are rational numbers hence elements of C 0 (Q). Suppose that m n 0 = a j [p j, 0] + b i [0, q i ] j=1 and we can also assume that p m <..., < p 1 < 0 < q 1 < < q n. We can rewrite the sum of the positive intervals [0, q j ] as n b i [0, q i ] = b 1 [0, q 1 ] + + b n 1 [0, q n 1 ] + b n [0, q n ] = b 1 [0, q 1 ] + + (b n + b n 1 )[0, q n 1 ] + b n [q n 1, q n ] (7). = (b n + + b 1 )[0, q 1 ] + (b n b 1 )[q 1, q 2 ]+ + (b n + b n 1 )[q n 2, q n 1 ] + b n [q n 1, q n ] and do the same for the sum of the negative intervals [p j, 0]. Since the q i s are ordered and any two of the [q l 1, q l ] intersect in at most a point, we know that 0 = b k + + b 1 = 0 for every k because they sum to zero. Starting from the last coefficient in the last line of (7), we have b n = 0, which implies that b n + b n 1 = b n 1 = 0 and repeating this process recursively gives b 1 = = b n = 0. The same process can be applied to the negative intervals and similarly a 1 = = a m = 0. Hence B is a set of linearly independent elements whose span is H 1 (R, Q). Problem 9: ( 2.1: #23). Show that the second barycentric subdivision of a -complex is a simplicial complex. Namely, show that the first barycentric subdivision produces a -complex with the property that each simplex has all its vertices distinct, then show that for a -complex with this property, barycentric subdivision produces a simplicial complex. Proof. Let X be a -complex and first consider the first barycentric subdivision X 1. Let σ : n X be a simplex [v 0,..., v n ] in X. After the subdivision occurs, we give barycenter b α each k-dimensional simplex [v i0,..., v ik ] and form the resulting n-simplices. Each of these simplices has only one of it s original vertices. Each of the barycenters b α is distinct because they lie in a unique face of σ, but also is clearly distinct from all of the v i by it s coordinate description thus it follows that the vertices of each n-simplex in the subdivision is distinct. Assuming that X 1 is a -complex with the above property that all it s vertices are distinct, let X 2 be it s barycentric subdivision and suppose that σ 1 and σ 2 are two n-simplices having the same set of vertices in X 2. Since they come from the subdivision, their vertex set consists of one
9 vertex of a simplex σ in the original -complex and barycenters of faces of σ. Since each point in X is in the image of exactly one restriction of a simplex, each point on the interior of Im σ must be in exactly one n-simplex of rhe subdivision. Hence σ 1 = σ 2 which implies that X 2 is a simplicial complex. Problem 10: ( 2.2: #4). Construct a surjective map S n S n of degree zero, for each n 1. Solution: Consider the sphere S n = {x R n+1 : x = 1} in R n+1 and let p : S n D n 1 be projection that forgets the last coordinate sending (x 1,..., x n, x n+1 ) (x 1,..., x n, 0) and let q : D n 1 Dn 1 S n 1 = S n be the quotient map of D n 1 by it s boundary. The composition p q : S n S n then has degree zero since the induced map can be written H n (S n ) p H n (D n 1 ) q H n (S n ) where H n (D n 1 ) = {0} since D n 1 is homotopic to a point. Then we have H n (S n ) p 0 q H n (S n ) and so the image of q is {0}, hence 1 0 so the degree is zero.
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