A complex of free abelian groups and its homology

Question

Let $L=\{d_i: L_i \rightarrow L_{i-1}\}$ be a complex of free abelian groups. $H(L)=\{H_p(L) \}$ is the homology group of $L$. Then $H(L)$ can be regarded as a complex with differentials zero. I see in a book that as $L$ free, there is a chain transfromation $f:L \rightarrow H(L)$ which is the identity on homology. I am confused on how to get $f$? For example, $H_i(L)=Kerd_i/ Im d_{i+1}$ is a subquotient of $L_i$ and how to define $f_i: L_i \rightarrow H_i(L)$? $Kerd_i$, as a subgroup of $L_i$, is free abelian. Then I don't know what to do next.

Answer 1

The key observation is not that $\ker(d_i)$ is free but rather that $L_i/\ker(d_i)\cong \operatorname{im}(d_i)$ is free, since it is a subgroup of $L_{i-1}$. Consequently, the short exact sequence $0\to \ker(d_i)\to L_i\to\operatorname{im}(d_i)\to 0$ splits, so you can write $L_i=\ker(d_i)\oplus K_i$ for some $K_i\subseteq L_i$. Now you can define $f_i:L_i\to H_i(L)$ to be the quotient map on $\ker(d_i)$ and $0$ on $K_i$. (You then have to check that this actually is a chain map, but that is easy: the composition $f_{i-1}d_i$ is $0$ since $f_{i-1}|_{\ker(d_{i-1})}$ is the quotient map that kills $\operatorname{im}(d_i)$.)