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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!

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    @Theo, yeah, I do use `\Re` and `\Im` as a pair most times, so I understand that choice at least...2011-09-06

1 Answers 1

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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).

But notice that both directions of the proof involve just the analysis of the exact sequence,

$ 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 $i$?