Hatcher Exercise 2.1.17 We compute H_n(X,A) in each of the following scenarios: Throughout, we will reference the long exact sequence: \dots \rightarrow H_n(A) \xrightarrow{i_*} H_n(X) \xrightarrow{j_*} H_n(X,A) \xrightarrow{\delta} H_{n-1}(A) \rightarrow \dots \rightarrow H_0(X,A) \rightarrow 0 (a): X = S^2, A is a finite set of k points. Clearly, for n > 2, we have H_n(X) = H_{n-1}(A) = 0, so it must be the case that H_n(X,A) = 0. Consider the LES in low dimensions: % Filling this in with known values, we have: 0 \rightarrow \mathbb{Z} \rightarrow H_2(X,A) \rightarrow 0 \rightarrow 0 \rightarrow H_1(X,A) \rightarrow \mathbb{Z}^k \xrightarrow{\varphi} \mathbb{Z} \rightarrow H_0(X,A) \rightarrow 0 First, we must have H_2(X,A) \cong \mathbb{Z}. \varphi is actually the inclusion H_0(A) \xrightarrow{i_*} H_0(X). Then it must be the map sending every generator of \mathbb{Z}^k to 1 since every generator of H_0(A) (one of the points in A) is homologous to the single generator of H_0(X) when considered in the (connected) space X. This map is clearly surjective. Since it’s a map between free modules over a PID, and \im \varphi is rank 1, we can conclude that \ker \varphi is free of rank k-1, i.e. \ker \varphi \cong \mathbb{Z}^{k-1}. Knowing the details of \varphi, we can find the rest of the unknown groups from repated applications of exactness and the first isomorphism theorem. H_0(X,A) \cong \mathbb{Z} / \im \varphi = \mathbb{Z} / \mathbb{Z} = 0. Likewise we find H_1(X,A) \cong \ker \varphi \cong \mathbb{Z}^{k-1}. (b): X = S^1 \times S^1 and A is a finite set of k points. This is very similar to the last problem. The first part proceeds identically so that we have H_n(X,A) = 0 for n > 2. Now, the LES is low dimensions (starting at H_2(A)) is: 0 \rightarrow \mathbb{Z} \rightarrow H_2(X,A) \rightarrow 0 \rightarrow \mathbb{Z}^2 \rightarrow H_1(X,A) \xrightarrow{\psi} \mathbb{Z}^k \xrightarrow{\varphi} \mathbb{Z} \rightarrow H_0(X,A) \rightarrow 0 Again, we must have H_2(X,A) \cong \mathbb{Z}. The map \varphi is the same one from the previous part. As such, H_0(X,A) = 0 follows by the same argument. Since H_1(X, A) is surrounded by free groups in the exact sequence, it must be free (citation needed). Thus, by the same sort of reasoning as before, we conclude that its rank must be the sum of that of \im \psi = \ker \varphi and \ker \psi, which is isomorphic to \mathbb{Z}^2 since the map to the left of H_1(X, A) is injective. That is, must be free of rank (k-1) + 2 and we get H_1(X, A) \cong \mathbb{Z}^{k+1}. Note: We can do part (a) and (b) in an easier way – though arguably against the spirit of the problem. Since X is a manifold and A is just a finite set of points, they certainly form a good pair (appropriately small open neighborhoods around each point deformation retract onto the point). Then we have H_n(X,A) = \wt{H}_n(X/A). In either case, X/A is homotopy equivalent to the wedge sum of X with k-1 copies of S^1. The best way to see this is inductively. Take X, and identify two of the points a_1, a_2 \in A. This is homotopy equivalent to attaching an external arc outside of X connected to a_1 and a_2. Then, take some path in X connecting a_1 and a_2, and contract it – we’re left with a wedge sum of X and S^1. For each remaining point a, add another arc and contract along the path from the wedge base point to a to add another copy of S^1 to the wedge sum. Continuing in this way, we end up with X \vee \bigvee_{k-1} S^1. Then, since this is a wedge sum of manifolds, each pair, consisting of each summand with the wedge base point, is a good pair and we can apply Corollary 2.25 in Hatcher to directly compute the homology based on the homology of the summands. When X = S^2, we get: H_0(X,A) = 0 \qquad H_1(X,A) = 0 \oplus \bigoplus_1^{k-1} \mathbb{Z} = \mathbb{Z}^{k-1} \qquad H_2(X,A) = \mathbb{Z} \oplus \bigoplus_1^{k-1} 0 = \mathbb{Z} And when X = T^2: H_0(X,A) = 0 \qquad H_1(X,A) = \mathbb{Z}^2 \oplus \bigoplus_1^{k-1} \mathbb{Z} = \mathbb{Z}^{k+1} \qquad H_2(X,A) = \mathbb{Z} \oplus \bigoplus_1^{k-1} 0 = \mathbb{Z} Which agrees with what we get from considering the long exact sequence. (c) X = T^2 \# T^2 and A is the copy of S^1 used in performing the connect sum (the loop dividing X into two connected components). Again A is a deformation of a small annalus around A, so (X,A) is a good pair. Then, it’s easy to see that X/A is homotopy equivalent to T^2 \vee T^2. Then, H_n(X,A) = \wt{H}_n(X/A), which is easy to compute again using Corollary 2.25. We have: H_0(X,A) = 0 \qquad H_1(X,A) = \mathbb{Z}^2 \oplus \mathbb{Z}^2 = \mathbb{Z}^4 \qquad H_2(X,A) = \mathbb{Z} \oplus \mathbb{Z} = \mathbb{Z}^2 With H_n(X,A) = 0 for n > 2. (d) X is T^2 \# T^2, and A is a copy of S^1 subdividing one of the individual tori into two connected components. The same conditions apply, so we have H_n(X,A) = \wt{H}_n(X/A). Considering T^2 / A for one of the tori, observe that this space is homotopy equivalent to the sphere with two points identified (contract A to a point, and then split that point into two), which is homotopy equivalent to S^2 \vee S^1. So then the space X is homotopy equivalent to T^2 \# (S^2 \vee S^1), but T^2 \# S^2 = T^2, so we have X \cong T^2 \vee S^1. So, we have: H_0(X,A) = 0 \qquad H_1(X,A) = \mathbb{Z}^2 \oplus \mathbb{Z} = \mathbb{Z}^3 \qquad H_2(X,A) = \mathbb{Z} \oplus 0 = \mathbb{Z} And H_n(X,A) = 0 for n > 2.