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:

$$\begin{align*} & H_2(A) \rightarrow H_2(X) \rightarrow H_2(X,A) \\ \rightarrow \; & H_1(A) \rightarrow H_1(X) \rightarrow H_1(X,A) \\ \rightarrow \; & H_0(A) \rightarrow H_0(X) \rightarrow H_0(X,A) \rightarrow 0 \end{align*}$$

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](https://math.stackexchange.com/questions/999806/rank-nullity-theorem-for-free-mathbb-z-modules), 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 $n > 2$.