I'm trying to compute the homology of the pair $(X, A)$ where $X=S^2$ and $A=\{\pm x\}$ is a pair of antipode points. Can anyone check if I'm doing it right?
What I tried:
1. $A$ has two path components $\{x\}$ and $\{-x\}$ hence $$H_n(A)\simeq H_n(\{x\})\oplus H_n(\{-x\}),$$ for all $n\geq 0$. Hence, $$H_0(A)\simeq H_0(\{x\})\oplus H_0(\{-x\})\simeq \mathbb Z^2.$$ and $H_1(A)=H_2(A)=\{0\}$.
2. I already know $H_0(X)\simeq \mathbb Z$, $H_1(X)=0$ and $H_2(X)\simeq \mathbb Z$.
(a) Consider the part $H_0(A)\longrightarrow H_0(X)\longrightarrow H_0(X, A)\longrightarrow 0$ of the long exact sequence of the pair $(X, A)$: $$\mathbb Z^2\longrightarrow \mathbb Z\longrightarrow H_0(X, A)\longrightarrow 0.$$ The map $\mathbb Z\longrightarrow H_0(X, A)$ is surjective for the above sequence is exact. Hence $$H_0(X, A)\simeq \frac{\mathbb Z}{\textrm{ker}(\mathbb Z^2\longrightarrow H_0(X, A))}.$$ I suppose $\textrm{ker}(\mathbb Z^2\longrightarrow H_0(X, A))\simeq\mathbb Z$ (is that right?) so we would get $H_0(X, A)\simeq 0$.
(b) Consider the part $H_1(X)\longrightarrow H_1(X, A)\longrightarrow H_0(A)\longrightarrow H_0(X)\longrightarrow H_0(X, A)$ of the long exact sequence of the pair $(X, A)$. If what I said in $(a)$ is right then we get the exact sequence: $$0\longrightarrow H_1(X, A)\longrightarrow \mathbb Z^2\longrightarrow \mathbb Z\longrightarrow 0.$$ Since $\mathbb Z$ is a free abelian group this sequence splits hence $$\mathbb Z^2\simeq H_1(X, A)\oplus \mathbb Z,$$ therefore, $H_1(X, A)\simeq \mathbb Z$, right?
(c) Finally, considering the part $H_2(A)\longrightarrow H_2(X)\longrightarrow H_2(X, A)\longrightarrow H_1(A)$ of the long exact sequence of the pair $(X, A)$. We find $$0\longrightarrow \mathbb Z\longrightarrow H_2(X, A)\longrightarrow 0.$$ Therefore, $H_2(X, A)\simeq \mathbb Z$.
To summarize: $$H_n(X, A)\simeq \left\{\begin{array}{ccc}\mathbb Z&\textrm{if}&n=1, 2\\ 0&\textrm{if}&n\neq 1, 2\end{array}\right..$$ Questions:
(i) is $\textrm{ker}(\mathbb Z^2\longrightarrow H_0(X, A))\simeq \mathbb Z$?
(ii) Why the requeriment of the points of $A$ to be antipode? For what I did would work for any pair of points (I guess..)
The map $H_0(A)\to H_0(X)$ is a map between free Abelian groups. Recall that $H_0(X)$ is free Abelian on the path components of $X$, and its generators are $[x_i]$ where $i$ ranges over the path components $X_i$ and $x_i\in X_i$. If $A_j$ are the path components of $A$, then the generator $[a_j]$ is mapped to $[a_j]=[x_i]$, where $X_i$ contains $A_j$. In particular, the map is surjective iff each path component of $X$ is met by $A$.
So, yes, $\text{Im}(\Bbb Z^2→\Bbb Z)=\text{ker}(\Bbb Z→H_0(X,A))=\Bbb Z$ and $H_0(X,A)=0$
Your computations are all correct. The short exact sequence $$0⟶H_1(X,A)⟶\Bbb Z^2⟶\Bbb Z⟶0$$ also gives you a generator. Note that the kernel of the last map is generated by $x-(-x)$. The embedding $H_1(X,A)\hookrightarrow H_0(A)$ as the kernel of the last map sends the generator of $H_0(X,A)$ to $x-(-x)$, so any $1$-simplex $\sigma$ joining $x$ with $-x$ generates $H_1(X,A)$
Of course this would work if the points are not antipodal, as long as they are distinct