Proof that $H_1(D^2, S^1) = 0$ I was trying to prove that $H_1(D^2, S^1) = 0$ and I came up with the following proof.
Proof: Since $S^1 \subseteq D^2$ we have the following short exact sequence
$$0 \to H_1(S^1) \xrightarrow{i_*} H_1(D^2) \to H_1(D^2, S^1) \to 0$$
where $i_*$ is the induced monomorphism from $H_1(S^1)$ to $H_1(D^2)$ induced by the inclusion map $i ; C_1(S^1) \to C_1(D^2)$.
Now we know that $H_1(D^2) = 0$ thus we obtain the following exact sequence 
$$0 \to H_1(D^2, S^1) \to 0$$
 and so we have $H_1(D^2, S^1) = 0$ as desired. $\square$

But now I believe that my above proof is errornous since we would also obtain an exact sequence $0 \to H_1(S^1) \to 0$ implying that $H_1(S^1) = 0$ a contradiction since we know by the Hurewicz map that $H_1(S^1) \cong \pi_1(S^1) \cong \mathbb{Z}$.
What have I done wrong in my proof?
 A: We have the long exact sequence
$$
\dots \to H_n(S^1) \to  H_n(D^2) \to H_n(D^2,S^1) \to \dots
$$
Since $H_n(S^1) = 0$ for $n > 1$, we have that $H_n(D^2,S^1) = 0$ for $n > 2$ and thus we are left with the following exact sequence
$$
0 \to H_2(D^2,S^1) \to H_1(S^1) \to H_1(D^2) \to H_1(D^2,S^1) \to H_0(S^1) \to H_0(D^2) \to H_0(D^2,S^1) \to 0
$$
Note that only $H_1(D^2)$ is immediately zero, but we don't know about the rest. So you don't have the short exact sequence you claim. A possible simplification is to consider the reduced long exact sequence 
$$
\dots \to \tilde{H}_n(S^1) \to  \tilde{H}_n(D^2) \to H_n(D^2,S^1) \to \dots
$$
from which we have
$$
0 \to H_2(D^2,S^1) \to \tilde{H}_1(S^1) \to \tilde{H}_1(D^2) \to{H}_1(D^2,S^1) \to 0 
$$
and we also get that $H_0(S^1,D^2) = 0$. Now we can use your idea: note that as you say, we have 
$$
H_1(D^2,S^1) =  0
$$
but there is no contradiction, since the other part of the sequence is
$$
0  \to H_2(D^2,S^1) \to \tilde{H}_1(S^1) \to 0 
$$
Moreover, we get for free that $H_2(D^2,S^1) \simeq \tilde{H}_1(S^1) \simeq \mathbb{Z}$.
A: You are assuming that $\iota : C(S^1) \rightarrow C(D^2)$ induces an injection on homology but this in general fails, e.g. the vertical maps in
$\require{AMScd}$
\begin{CD} 
0 @>>> \mathbb{Z} @>\cdot n>> \mathbb{Z} @>>> 0 @>>> 0 \\
@VVV @V\text{id}VV @V\text{id}VV @VVV @VVV \\
0 @>>> \mathbb{Z} @>\cdot n>> \mathbb{Z} @>>> \mathbb{Z}/n @>>> 0
\end{CD}
make up an injective chain map (between the horzintal complexes) which induces $\mathbb{Z}/n \rightarrow n\mathbb{Z}/n\mathbb{Z} = 0$ on the homology of the middle part.
However, you can still use the long exact sequence for the pair $(D^2,S^1)$:
$$0 = H_1(D^2) \rightarrow H_1(D^2,S^1) \xrightarrow{\delta} H_0(S^1) \rightarrow H_0(D^2). $$
Since the right map is an isomorphism we have $\delta = 0$ by exactness, hence ker $\delta = H_1(D^2,S^1)$. Again using exactness we have that the left map is surjective, hence $H_1(D^2,S^1)=0$.
