Hatcher Problem 2.1.16 (b) I am trying to do the stated problem in Hatcher:


Show $H_1(X,A) = 0$ iff $H_1(A) \to H_1(X)$ is surjective and each path component of $X$ contains at most one path component of $A$.


Now I have reduced the problem to showing that $i_\ast : H_0(A) \to H_0(X)$ injective iff each path component of $X$ contains at most one path component of $A$. This comes from looking at the end of the LES of the pair $(X,A)$:
$$\ldots \to H_1(X) \to H_1(X,A) \to H_0(A) \to H_0(X) \to H_0(X,A) \to 0$$
Now one direction I have shown, the other that is giving me trouble is the converse. That is if $i_\ast$ is not injective then there is a path component of $X$ that contains at least two path components of $A$. I have the following:
Suppose $i_\ast$ is not injective. Then there is a $\tau \in C_0(A)$ such that $[\tau \circ i] = 0$ but $[\tau] \neq 0$. That is to say, $\tau \circ i = \partial(\sigma)$ for some $\sigma \in C_1(X)$ but $\tau$ is not the boundary of any $\sigma'\in C_1(A)$. However I'm confused because to me the only way for $\tau \circ i$ to be the boundary of a singular $1$ - simplex $\sigma$ in $X$ is if $\sigma$ is a loop. What's wrong here?
Thanks.
 A: So first, as you have remarked, if $H_1(X,A) = 0$ then it is immediate from the long exact sequence that $H_1(A) \to H_1(X)$ is surjective and that $H_0(A) \to H_0(X)$ is injective. Conversely, suppose that $H_1(X,A) \neq 0$, so let $\sigma \neq 0 \in H_1(X,A)$. Then you have two possibilities:


*

*Either $\partial \sigma = 0 \in H_0(A)$. Then by exactness, there exists $\gamma \in H_1(X)$ such that $j_*\gamma = \sigma$. But then $i_* : H_0(A) \to H_0(X)$ cannot be surjective: if you have $\gamma = i_*\alpha$, then $j_*\gamma = j_*i_*\alpha = 0$ by exactness.

*Or $\partial \sigma$ is nonzero. But then $i_*(\partial\sigma) = 0$ by exactness, so $i_* : H_0(A) \to H_0(X)$ is not injective.


So all that's left to prove is that $i_* H_0(A) \to H_0(X)$ is not injective iff there is a path component of $X$ containing at least two path components of $A$.
Again, one direction is clear. Suppose a path component of $X$ contains two path components of $A$. Pick points $a,b \in A$ in the two distinct path components. Then $[a]-[b] \in H_0(A)$ is nonzero, but $i_*[a] = i_*[b]$ thus $i_*([a]-[b]) = 0$.
Conversely suppose that $i_* : H_0(A) \to H_0(X)$ is not injective. Let $0 \neq \alpha \in H_0(A)$ be such that $i_*\alpha = 0$. Write $\alpha = \sum_{k \in I} n_k [a_k]$ as a sum of vertices. Moreover assume that all the $a_i$ are in different path components.
Of course, $i_*(\alpha) = \sum_{k \in I} n_k [a_i]$ viewed as $0$-cycles in $H_0(X)$. If the path components corresponding to the $a_i$ were all different in $X$ then this wouldn't be possible unless $n_i = 0$ for all $i$ (here you use the fact that $H_0(X)$ is the free abelian group on the path components of $X$). But $\alpha \neq 0$, so this isn't possible. It follows that at least two of the $a_i$ are in the same path component in $X$.
