1

If $\phi:C_0(K)\to \mathbb{Z}$ is an epimorphism such that $\phi\circ \partial_1=0$ then show that

$$H_0(K)\cong \frac{ker\phi}{im\partial_1}\oplus\mathbb{Z}.$$

My working is since $C_0(K)$ is abelian group and $\phi$ is an epimorphism then there exists an infinite cyclic subgroup $H$ such that

$$C_0(K)=ker\phi\oplus H.$$ Hence $$H_0(K)=\frac{C_0}{im\partial_1}=\frac{ker\phi\oplus H}{im\partial_1}=\frac{ker\phi}{im\partial_1}\oplus H=\frac{ker\phi}{im\partial_1}\oplus \mathbb{Z}.$$

This proof leads to reduced homology that has been asked here. So I just wanna ask where do I use the fact that $\phi\circ \partial_1=0$? I did not use it but everything seemed just fine. I there anything wrong from my proof?

Community
  • 1
Chen M Ling
  • 702
  • in the last line, after the third equal sign: you are using that the image of $\partial_1$ is inside the kernel of phi - otherwise that "splitting" wouldn't be possible! – nelv May 16 '16 at 11:05
  • @nelv Thanks for the comment, it has similar reasoning with the below answer :) – Chen M Ling May 16 '16 at 11:26

1 Answers1

4

You used it when you wrote this equation: $$\frac{\ker \phi \oplus H}{\operatorname{im} \partial_1} = \frac{\ker \phi}{\operatorname{im} \partial_1} \oplus H$$ You're implicitly using the fact that $\operatorname{im} \partial_1 \subset \ker \phi \iff \phi \circ \partial_1 = 0$. Otherwise taking the quotient $\ker \phi / \operatorname{im} \partial_1$ doesn't even make any sense!

Also your proof isn't really satisfactory. You say that since $\phi$ is an epimorphism onto $\mathbb{Z}$ and $C_0(K)$ is abelian, then there exists some $H \cong \mathbb{Z}$ such that $C_0(K) \cong \ker \phi \oplus H$. This isn't always true. Consider the epimorphism $f : \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ (reduction modulo $2$). Then there is no subgroup $H \subset \mathbb{Z}/4\mathbb{Z}$ isomorphic to $\mathbb{Z}/2\mathbb{Z}$ such that $\mathbb{Z}/4\mathbb{Z} \cong \ker f \oplus H$ (a nice exercise).

Here it actually is true because the image of $\phi$ is $\mathbb{Z}$, which is a projective abelian group. This implies that the short exact sequence: $$0 \to \ker \phi \to C_0(K) \xrightarrow{\phi} \mathbb{Z} \to 0$$ splits on the right (i.e. there is some splitting $s : \mathbb{Z} \to C_0(K)$ such that $\phi \circ s = \operatorname{id}_\mathbb{Z}$), and since $C_0(K)$ is abelian it implies your claim. But you really need $\mathbb{Z}$ to be projective (free abelian implies projective) here... So it needs to appear somewhere in your proof.

Najib Idrissi
  • 54,185
  • Why is $\phi \circ \partial_1 = 0 \Rightarrow \operatorname{im} \partial_1 \subset \ker \phi$? – Chen M Ling May 16 '16 at 11:16
  • @ChenMLing Just think about it... Write down explicitly what it means to have $\operatorname{im} \partial_1 \subset \ker \phi$. – Najib Idrissi May 16 '16 at 11:18
  • It is obvious that $im\partial_1\subset ker\phi$ because in reduced homology of zero-th dimensional we use the quotient of this. – Chen M Ling May 16 '16 at 11:25
  • Oh I see now... Thank you for asking me to write down. It's clear now. Many thanks, Najib. Also I think I will change my proof a littl bit since the reasing for the abelian group and epimorphism is not quite true, thanks for letting me know that :) – Chen M Ling May 16 '16 at 12:05