I am trying to calculate the singular homology $H_q(D^n,S^{n-1}) \quad \forall q$.
With respect to the cases $q=0,1$ and $n=1$ I have the following exact sequence $$\dots \rightarrow H_1(S^0) \rightarrow H_1(D^1) \rightarrow H_1(D^1,S^0) \rightarrow H_0(S^0) \overset{i_{*0}}{\longrightarrow} H_0(D^1) \rightarrow H_0(D^1,S^0)\rightarrow 0 $$ So, substituting the known homologies,
$$0 \rightarrow H_1(D^1,S^0) \rightarrow \mathbb{Z} \oplus \mathbb{Z} \overset{i_{*0}}{\longrightarrow} \mathbb{Z} \rightarrow H_0(D^1,S^0)\rightarrow 0 $$
If I were able to say that $i_{*0}$ is surjective I would be able to finish, because then I would conclude that, by sequence factorization, $H_0(D^1,S^0)=0$ and $H_1(D^1,S^0)=\mathbb{Z}$.
But I don't undestrand why is $i_{*0}$ surjective. Have you any suggestions?
If $X$ is a space, $H_0(X)$ is a free Abelian group consisting of a direct sum of one copy of $\Bbb Z$ for each path component $C$ of $X$. This $\Bbb Z$ has a canonical generator which I write as $[C]$. In the map $H_0(X)\to H_0(Y)$ induced by $f:X\to Y$, $[C]$ maps to $[D]$ where $D$ is the path component of $Y$ containing $f(C)$.
Here $S^0$ has two path components and $D^1$ has one. Both the canonical generators of $H_0(S^0)$ map to the canonical generator of $H_0(D^1)$. Your map $i_0^*$ is therefore surjective.