Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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

1 Answer 1

up vote 2 down vote accepted

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 $i$?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.