4

In algebraic topology we have the result $$H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}.$$ In Massey's book, this is a result that follows from the fact that the sequence $$0 \rightarrow \tilde{H}_0(X) \xrightarrow{\xi} H_0(X) \xrightarrow{\xi_*} \mathbb{Z} \rightarrow 0$$ is exact. So let me try and make sense of this:

The sequence is exact because $\text{im}(\xi) = \text{ker}(\xi_*)$. Where does this equality come from? We have that $\varepsilon \circ \partial_1 =0$ (where we have the sequence $\cdots \rightarrow C_1(X) \xrightarrow{\partial_1} C_0(X) \xrightarrow{\varepsilon} \mathbb{Z} \rightarrow 0$), hence $\varepsilon$ vanishes on $\partial_1$. We know that $\tilde{H}_0(X)=\text{ker}(\varepsilon)/\text{im}(\partial_1)$. Why is this the kernel of $\xi_* : H_0(X) \rightarrow \mathbb{Z}$ though? Can we show both inclusions?

Once that is understood, I need to understand where the isomorphism comes from. Do we have that for any map $\phi:A \to B$, then $A \cong \text{ker}(\phi) \oplus \text{im}(\phi)$? Or what's the deal? :)

I hope that you can help me clarify these issues.

Numbersandsoon
  • 2,910

1 Answers1

4

$\require{AMScd}$

You have a commutative diagram $$\begin{CD} 0 @>>> ker(\varepsilon) @>i>> C_0(X) @>\varepsilon>> \Bbb Z @>>> 0\\ @. @VVV @VVV @| \\ 0 @>>> \tilde H_0(X) @>\bar i>> H_0(X) @>\bar\varepsilon>> \Bbb Z @>>> 0 \end{CD}$$ where the vertical maps send each element $\gamma\in C_0(X)$ to its homology class $[\gamma]$. Since $im(\partial_1)\subseteq ker(\varepsilon)$, there is an induced map $\bar\varepsilon$.
Using the exactness of the upper row, it is easy to show that the lower row is exact, as well. Moreover, since $\Bbb Z$ is free, we can construct a section $s$ for $\bar\varepsilon$, which is determined by some $[x_0]=s(1)$ An isomorphisms $H_0(X)\to\tilde H_0(X)\oplus\Bbb Z$ is given by $$[\gamma]\mapsto\left([\gamma-\varepsilon(\gamma)x_0],\varepsilon(\gamma)\right)$$

Stefan Hamcke
  • 27,733
  • 2
    So, to make it clear to the OP, in general, given $\phi:A\rightarrow B$, $A$ is not isomorphic to $\ker(\phi)\oplus \operatorname{im}(\phi)$. A counter example is provided by the map $\phi:\mathbb{Z}/4\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$ sending $\overline{1}$ to $\overline{2}$: the image and kernel are both isomorphic to $\mathbb{Z}/2\mathbb{Z}$, but $\mathbb{Z}/4\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/2\mathbb{Z} \oplus\mathbb{Z}/2\mathbb{Z}$. So, Stefan is using a very special property of $\mathbb{Z}$ to make everything work. – Jason DeVito - on hiatus Sep 17 '14 at 13:52
  • @JasonDeVito Oh right, I saw the OP was asking about that, and originally I intended to provide a counterexample, but in the end I have forgot to do that. Thanks for reminding me :) – Stefan Hamcke Sep 17 '14 at 13:56
  • @JasonDeVito. Wouldn't you rather send $\bar 1$ to $\bar 1$ via $\phi$ ? – Stefan Hamcke Sep 17 '14 at 13:58
  • Yes, what a dumb typo! – Jason DeVito - on hiatus Sep 17 '14 at 14:43