Let $X = S^1 \times S^1$ and let $p_1,p_2,p_3 \in S^1$ and let $A =....$. Compute $H_i(X,A)$ $\forall i$. Let $X = S^1 \times S^1$ and let $p_1,p_2,p_3$ be distinct points in $S^1$ and let $A = (S^1 \times \{p_1\}) \cup (S^1 \times \{p_2\}) \cup (S^1 \times \{p_3\})$
Compute $H_i(X,A)$ $\forall i$.
We have a L.E.S.:
$$0 \rightarrow H_2(A) \stackrel{h}{\rightarrow} H_2(X) \stackrel{g}{\rightarrow} H_2(X,A) \stackrel{f}{\rightarrow} H_1(A) \stackrel{e}{\rightarrow} H_1(X) \stackrel{d}{\rightarrow}$$ $$ H_1(X,A) \stackrel{c}{\rightarrow} H_0(A) \stackrel{b}{\rightarrow} H_0(X) \stackrel{a}{\rightarrow} H_0(X,A) \rightarrow 0$$
We can easily compute the following: 
$H_2(A) = 0$
$H_2(X) = \mathbb{Z}$
$H_1(A) = \mathbb{Z}^3$
$H_1(X) = \mathbb{Z}^2$
$H_0(A) = \mathbb{Z}^3$
$H_0(X) = \mathbb{Z}$
Furthermore, I have concluded that $H_0(X,A) = 0$, since $b$ is a surjective map.
Also, $g$ is injective so $ker(f)=im(g)=H_2(X) = \mathbb{Z}$ 
Also, I was thinking that $im(f)=ker(e)=\mathbb{Z}$, but am not sure and would like to know how to think about this more rigorously.
And so I thought that maybe $H_2(X,A)=ker(f) \oplus im(f) = \mathbb{Z}^2$, but I'm not sure. 
And I'm quite lost for $H_1(X,A)$.
Insight appreciated!!
 A: Lord Shark is correct in the comments: $A\cong S^1 \sqcup S^1 \sqcup S^1$ (as long as the $p_i$'s are all distinct), so $H_1(A) \cong \mathbb{Z}^3\cong H_0(A)$. Moreover the images of each of these circles in $H_1(X)$ are homologous, and can be described by the Kunneth theorem as $u\otimes 1$ where $u$ is the generator of $H_1(S^1)$.  The kernel of $e$ is then the set of all $(x, y, z) \in \mathbb{Z}^3$ such that $x+y+z = 0$, which is isomorphic to $\mathbb{Z}^2$, and the kernel of $b$ is similarly described. Now we have 
$$ 0 \stackrel{h}{\to} \mathbb{Z} \stackrel{g}{\to} H_2(X, A) \stackrel{f}{\to} \mathbb{Z}^3 \stackrel{e}{\to} \mathbb{Z}^2 \stackrel{d}{\to} H_1(X, A) \stackrel{c}{\to} \mathbb{Z}^3 \stackrel{b}{\to} \mathbb{Z} \to 0 $$
To help us compute the unknown groups, here's a helpful lemma you should verify as an exercise:

Lemma: Suppose we have a long exact sequence of abelian groups
  $$\dots \to A \stackrel{a}{\to} B\stackrel{b}{\to} C \stackrel{c}{\to} D \stackrel{d}{\to} E \stackrel{}{\to} \dots $$
  then there is a short exact sequence
  $$0 \to coker(a) \to C \to im(c) \to 0 $$

Using this lemma we can isolate $H_1(X,A)$ with the short exact sequence
$$0 \to coker(e) \to H_1(X, A) \to im(c) \to 0 $$
By the above discussion, $im(c) = ker(b)$ is isomorphic to $\mathbb{Z}^2$, and also $im(e)$ is one of the $\mathbb{Z}$ summands of $H_1(X)$ so $coker(e)\cong \mathbb{Z}$, so $H_1(X,A)$ is an abelian extension of $\mathbb{Z}^2$ by $\mathbb{Z}$, and hence must be $\mathbb{Z}^3$ by the Splitting Lemma since any surjective homomorphism to $\mathbb{Z}^2$ has a section (because it is a free abelian group and hence projective). Similarly we can isolate $H_2(X, A)$ with 
$$ 0 \to coker(h) \to H_2(X, A) \to im(f) \to 0$$
and again $coker(h) \cong \mathbb{Z}$ and $im(f) = ker(e)\cong \mathbb{Z}^2$ so again $H_2(X, A)\cong \mathbb{Z}^3$.
