Exactness in homology I've been thinking about the exactness "axiom" which says that 
For any topological pair $(X,A)$ the following sequence is exact:
$$ \dots \rightarrow H_n(X) \rightarrow H_n(X,A) \rightarrow H_{n-1}(A) \rightarrow \dots$$
I thought I could use this to compute the relative homology so I tried the following example:
$H_n(D^2, S^1) = ?$
By exactness the following sequence is exact:
$$ \dots \rightarrow H_n(D^2) \xrightarrow{j_\ast} H_n(D^2,S^1) \xrightarrow{\partial_\ast} H_{n-1}(S^1) \rightarrow \dots$$
Then I though I could use the following 2 facts to find $H_n(D^2, S^1) $:
$H_n(D^2) = H_n(\{\ast\}) = 0 (n>0) $ and $\mathbb{Z}(n=0)$ 
$H_n(S^1) = 0 (n \geq 2) $ and $\mathbb{Z}(n=0,1)$
But somehow knowing $im j_\ast = ker\partial_\ast$ etc. doesn't give me any useful information to find $H_n(D^2, S^1) $. What am I doing wrong? Is this not a good example of when to apply the exactness axiom? If no, can someone please show me a better one? Many thanks for your help!
Edit
OK, after Theo's comment I produced the following exact sequence:
$$ \dots 0 \xrightarrow{a} H_2(D^2, S^1) \xrightarrow{b} \mathbb{Z} \rightarrow 0 \xrightarrow{c} H_1(D^2, S^1) \xrightarrow{d} \mathbb{Z} \rightarrow \mathbb{Z} \xrightarrow{e} H_0(D^2, S^1) \xrightarrow{f} 0$$
I assume that $H_n(D^2, S^1) = 0$ for $n > 2$ because the space doesn't have any $n$ cells in it but I'm not entirely sure that's rigorous. Anyway, with the exactness of the sequence I get the following:
(i) $im a = 0 = ker b $ $\implies b$ is injective
(ii) $ker f = H_0(D^2,S^1) = im e$
(iii) $im c = 0 = ker d$ $\implies d$ injective

Since $H_n(D^2) = H_n(S^1) = 0$ for $n \geq 2$, we also have $H_n(D^2, S^1) = 0$ for $n> 2$. So we must focus on $H_n(D^2,S^1)$ for $n=0,1,2$.
For $n=0$, we have
$$
\dots \longrightarrow H_0(S^1) \longrightarrow H_0(D^2) \longrightarrow H_0(D^2,S^1) \longrightarrow 0
$$
which is
$$
\dots \longrightarrow \mathbb{Z} = \mathbb{Z} \longrightarrow H_0(D^2,S^1) \longrightarrow 0 \ .
$$
Hence 
$$
H_0(D^2,S^1) = 0 \ .
$$
For $n=1$, the situation is
$$
\dots  \longrightarrow H_1(D^2) \longrightarrow H_1(D^2,S^1) \longrightarrow H_0(S^1) \longrightarrow H_0(D^2) \longrightarrow \dots
$$
That is,
$$
0 \longrightarrow H_1(D^2,S^1) \longrightarrow \mathbb{Z} = \mathbb{Z}
$$
Which again means 
$$
H_1(D^2, S^1) = 0 \ .
$$
Finally, for $n=2$,
$$
\dots \longrightarrow H_2(D^2) \longrightarrow H_2(D^2,S^1) \longrightarrow H_1(S^1) \longrightarrow H_1(D^2) \longrightarrow \dots
$$
Which is
$$
0 \longrightarrow H_2(D^2, S^1) \longrightarrow \mathbb{Z} \longrightarrow 0
$$
So, 
$$
H_2(D^2, S^1) = \mathbb{Z}  \ .
$$
