How to compute $H_1(S^1, S^0) $ As in the title, my problem is to compute the relative homology group $H_1(S^1, S^0) $  of these two spheres. 
The problem is that te long exact sequence of the couple gives
$0 \rightarrow \mathbb{Z}\rightarrow H_1(S^1, S^0)\rightarrow\mathbb{Z}^2 
\rightarrow \mathbb{Z} \rightarrow 0$ 
And I think this is not so helpfull.
I know there are tricks using the cellular structure of these two spaces but there is a more basic way to see the result?
Thank you!
 A: Another way to do this, without CW complexes but with one tiny topological thought, goes like this. The key idea is to remember where those homomorphisms come from:
$$0 \rightarrow \underbrace{H_1(S^1)}_{\mathbb{Z}} \rightarrow H_1(S^1,S^0) \rightarrow \underbrace{H_0(S^0)}_{\mathbb{Z}^2} \rightarrow \underbrace{H_0(S^1)}_{\mathbb{Z}} \rightarrow \underbrace{H_0(S_1,S_0)}_{0}
$$
The map $\mathbb{Z}^2 \approx H_0(S^0) \to H_1(S^1) \approx \mathbb{Z}$ is induced by the inclusion $S^0 \hookrightarrow S^1$. Each of the two points of $S^0$ maps to a point of the path connected space $S^1$, so each of the two generators of $\mathbb{Z}^2 \approx H_0(S^0)$ maps to the same generator of $H_0(S^1) \approx \mathbb{Z}$. This tells me that the image of the map $H_1(S^1,S^0) \mapsto \mathbb{Z}^2$, which equals the kernel of the map $\mathbb{Z}^2 \mapsto \mathbb{Z}$, is isomorphic to $\mathbb{Z}$. Therefore, we get a shorter exact sequence
$$0 \mapsto \mathbb{Z} \mapsto H_1(S^1,S^0) \mapsto \mathbb{Z} \mapsto 0
$$
and it follows that the abelian group $H_1(S^1,S^0)$ is isomorphic to $\mathbb{Z}^2$.
A: It is possible to compute $H_1(S^1,S^0)$ as a group just from the long exact sequence given. Because $\mathbb{Z}$ and $\mathbb{Z}^2$ are both torsion-free, $H_1(S^1,S^0)$ must also be torsion-free. So we need only compute the rank of $H_1(S^1,S^0)$. Rank is additive over exact sequences, and so the rank must be 2. 
However, this approach doesn't give much insight into the geometry of the generators of $H_1(S^1,S^0)$. If you'd like to write down cycles representing the generators, a cell structure is likely the easiest way to go.

Edit: here is a general sketch for the torsion-free argument.

Suppose that $0 \to A \to_f B \to_g C$ is an exact sequence. If $A$ and $C$ are torsion-free, then so is $B$.

Proof: suppose that for $b \in B$ and $n>0$, $nb = 0$. Then

 $ng(b) = 0$, so $g(b) = 0$,

so $b = f(a)$ for some $a \in A$. Then

 since $f$ is injective, $nb = nf(a) = 0$ implies $na=0$, 

so $a=0$ and thus $b=0$.
A: Hello fellow Andres.
Here is a different (geometric) way to do approach the problem.

If $X$ is a connected $CW$ complex, and $A$ is a $CW$ subspace, then $H_k(X,A) \cong \tilde{H_k}(X/A)$

this can be proven by considering the map induced by projection: $q_*:H_n(X,A) \to H_n(X/A,A/A) =\tilde{H_n}(X/A)$ which is an isomorphism (and the content of Hatcher $2.22$.)
From this, one can realize the quotient $S^1/S^0$ by just pinching two points together. You can prove that the first homology here is $\mathbb Z \oplus \mathbb Z$ either by Mayer-Vietoris or realizing it as the abelianization of the fundamental group.
A: The exact sequence
$0 \rightarrow \mathbb{Z}\rightarrow H_1(S^1, S^0)\rightarrow\mathbb{Z}^2 
\rightarrow \mathbb{Z} \rightarrow 0$ 
ends in $\mathbb Z$. This is free, so the last projection splits, and you have in fact
$0 \rightarrow \mathbb{Z}\rightarrow H_1(S^1, S^0)\rightarrow\mathbb{Z} \rightarrow 0$ 
This splits again because $\mathbb Z$ is free, so your group is $\mathbb Z^2$. 
