The proof I am trying to get at is exercise 2.2.43
in Hatcher, and I can't even get it started so I'm asking just about the first part here. It's not an actual homework assignment but I've tagged it as such since it is such an exercise. First, the complete thing I am trying to show:
If $X$ is a CW complex with finitely many cells, then $H_n(X,G)$ is a direct sum of the following:
- A copy of $G$ for each $\mathbb{Z}$ summand of $H_n(X)$.
- A copy of $G/mG$ for each $\mathbb{Z}_m$ summand of $H_n(Z)$.
- A copy of the kernel of $G\xrightarrow{m} G$ for each $\mathbb{Z}_m$ summand of $H_{n-1}(X)$
Now the first part of the proof which I need some help on.
a) Show the chain complex of free abelian groups $C_n$ splits as a direct sum of subcomplexes $0\to L_{n+1}\to K_n\to 0$ (Hint: Show the short exact sequence $0\to Ker(\partial) \xrightarrow{i} C_n \xrightarrow{j} Im(\partial) \to 0$ splits and take $K_n=Ker(\partial)$).
Proof: It's easy to show this sequence splits; for instance, $Im(\partial)$ is free. I don't quite know "which $\partial$" these spaces represent – I think
$$Im(\partial_n)=\{a\in C_{n-1}|a=\partial w \mbox{ for some } w\in C_n\}$$
$$Ker(\partial_n)=\{v\in C_n | \partial v =0 \}$$
So then the map $i$ is inclusion and $j(v)=\partial v=a$. So by using multiple rows of the chain, I should have the exact sequence
$$0\to Ker(\partial_n)\xrightarrow{i} Ker(\partial_n)\oplus Im(\partial_{n}) \xrightarrow{\partial} Ker(\partial_{n-1})\oplus Im (\partial_{n-1})\to Im(\partial_{n-1})\to 0$$
This is exact because $im(i)=(v,0)$ and $\partial(v,w)=(0, a)$ so $im(i)=ker(\partial)$. So apparently the intention is that $K_n$ is part of $Ker(\partial_n)\oplus Im(\partial_n)$ with the $Im(\partial_n)$ peeled off somehow, but I don't know how to make sense of that.
Thanks for the help so far, any other hints? I'm clearly missing something obvious.
Best Answer
Hint: You are trying to desribe a chain complex, so you have to use the boundary map $C_{n+1} \to C_n$ in your construction of $K_n$ and $L_{n+1}$, and you have to use the fact that $\partial_n\circ \partial_{n+1} = 0$. You haven't input these two ingredients yet.