I've been doing some more exercises in Hatcher, in particular the following:
Show that $H_0(X,A) = 0$ iff $A$ meets each path-component of $X$.
"$\Leftarrow$": Let $x_i \in A \cap X_i \neq \emptyset \forall $ path-components $X_i$. Then for any $x \in X_i$ there is a path $\gamma$ from $x_i$ to $x$. They therefore differ by a boundary: $\partial \gamma = x - x_i$ and therefore \{ x \} = \{ x_i \} in $H_0(X,A) $. Therefore $H_0(X,A) = 0$.
"$\Rightarrow$" Now for this direction I'm not so sure and I'd be glad if you could give me a hint. Let $H_0(X,A) = 0$. By definition, the relative homology group is calculated from the following sequence of chain groups:
$ 0 \rightarrow C_0(A) \rightarrow C_0(X) \xrightarrow{\partial_1} C_0(X) / C_0(A) \xrightarrow{\partial_0} 0$
Then $H_0(X, A) = H_0(C_0(X) / C_0(A)) = ker \partial_0 / im \partial_0 = (C_0(X) / C_0(A) ) / im \partial_1 = 0$.
There are two cases where this equality holds:
(i) $C_0(X) / C_0(A) = 0$
(ii) $im \partial_1 = C_0(X) / C_0(A)$
In case (i), $A \cap X_i \neq \emptyset$ for all $X_i$.
In case (ii) I'm not sure how to proceed. Can you give me a hint? Many thanks for your help!
Edit Or maybe I could do the second direction like this:
$0 = H_0(X,A) = \oplus_i H_0(X_i, A) \implies H_0(X_i, A) = 0 \forall i$
$ \implies A \cap X_i \neq \emptyset \forall i$?