$H_1(X,A) = 0 \iff H_1(A) \rightarrow H_1(X)$ surjective and $X_i$ contains no more than one path-component of $A$

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 $\operatorname{im}{ g} = 0 = \operatorname{ker}{ h}$.

$H_1(X,A) / \operatorname{ker}{ h} = \operatorname{im}{ h} = H_1(X,A) = \operatorname{ker}{i} = 0$.

-
Matt: a small thing on formatting. Could you please use \operatorname{ker}{g} and \operatorname{im}{g} so that these look like $\operatorname{ker}{g}$ and $\operatorname{im}{g}$ instead of $kerg$ and $img$? – t.b. Sep 6 '11 at 10:16
Yes, of course! I didn't know I could do that, thanks! – Rudy the Reindeer Sep 6 '11 at 10:39
@Theo: Why not \ker? Something wrong about it? :) – J. M. Sep 6 '11 at 15:37
@J.M.: No, nothing wrong about it. It's just that I use both $\operatorname{Ker}$ and $\operatorname{ker}$ (one for the object, one for the morphism) and I never know which one is implemented in standard LaTeX. Moreover, some ... had the idea to define \Im to be what it is... – t.b. Sep 6 '11 at 15:39
@Theo, yeah, I do use \Re and \Im as a pair most times, so I understand that choice at least... – J. M. Sep 6 '11 at 15:47

There are a few awkward phrasings in your proof (for example, $A_i$ must be defined, and there is probably a typo in $H_1(X, A) = 0 = \ldots$ in line 10).
$$H_1(A) \stackrel{f}{\to} H_1(X) \to H_1(X, A) \to H_0(A) \stackrel{i}{\to} H_0(X).$$
If $H_1(X,A) = 0$, what is implied about $f$ and $i$?