I've done another exercise in Hatcher and was wondering if you could tell me if I did it right.
The exercise: $H_1(X,A) = 0 \iff H_1(A) \rightarrow H_1(X)$ surjective and $X_i$ contains no more than one path-component $A_i$ of $A$ where $X_i$ are the path components of $X$
"$\Rightarrow$"
Let $H_1(X,A) = 0$. Consider the exact sequence $ H_1(A) \xrightarrow{f} H_1(X) \xrightarrow{g} H_1(X,A) = 0 \xrightarrow{} H_0(A)$
Then $\operatorname{ker}{ g} = H_1(X) = \operatorname{im}{ f} \implies f $ is surjective.
Now for the other half of what needs to be shown consider: 0 \xrightarrow{f} H_0(A_i) \xrightarrow{g} H_0(X_i) \xrightarrow{h} H_0(X_i , A_i) \rightarrow 0
where the first term is zero because $H_1(X,A) = 0 = \oplus_i H_1(X_i, A_i) \implies H_1(X_i, A_i)= 0$.
Then $f = 0 = $ const. $\implies g$ is injective.
$X_i$ path connected $\implies H_0(X_i) \cong \mathbb{Z}$. If $X_i$ contained more than one path-component $A_i$, say $n$, then $\operatorname{im}{ g} \cong \mathbb{Z}^n$, which is a contradiction to $\operatorname{im}{ g} \subset H_0(X_i) = \mathbb{Z}$.
"$\Leftarrow$"
Let $H_1(A) \rightarrow H_1(X)$ be surjective and let $X_i$ be such that it contains no more than one path-component of $A$.
claim: $H_1(X,A) = 0$.
Consider the following exact sequence:
$ H_1(A) \xrightarrow{f} H_1(X) \xrightarrow{g} H_1(X,A) \xrightarrow{h} H_0(A) \xrightarrow{i} H_0(X)$
where $H_0(A) = \oplus H_0(A_i)$ and $H_0(X) = \oplus H_0(X_i)$.
Because $X_i$ cannot contain more than one $A_i$, $i$ is injective, i.e. $\operatorname{ker} i = 0$.
$f$ surjective $\implies \operatorname{ker}{ g} = H_1(X) \implies g = 0$ and $\operatorname{im}{ g} = 0 = \operatorname{ker}{ h}$.
$H_1(X,A) / \operatorname{ker}{ h} = \operatorname{im}{ h} = H_1(X,A) = \operatorname{ker}{i} = 0$.
Many thanks for your help!