Let $(C, \partial), (C',\partial')$ be two chain complexes, and $f: C \rightarrow C'$ be a chain map. In general we know that if $f$ is zero-homotopic (i.e. $\exists$ a collection of maps $s_k: C_k \rightarrow C_{k+1}'$ with $f_k = \partial' s_{k} + s_{k-1}\partial$), then $f$ induces the zero map on homology. I think the converse is in general false. But I was able to prove the following:
If $C$ is a free complex, $C'$ is exact and both complexes are zero in degree $< 0$, then every chain map which induces the zero map on homology is in fact zero-homotopic.
Proof We construct the $s_k$ inductively. First let $k=0$. Because $C'$ is exact and $C'_{-1} = 0$ we know that $\partial_1: C'_1 \rightarrow C'_0$ is surjective. If $\{b_i\}$ is a basis of $C_0$, we may just choose preimages $c_i \in C_1'$ for each $f(b_i)$, and define $s_0(b_i) := c_i$.
Now let $k > 0$ and suppose for $j < k$ we already have defined $s_j$. To define $s_k$, consider the map $g := f_k - \partial_k' s_{k-1}$. To choose preimages of $g(b_i)$ we need to show that $g(C_k) \subset \text{im}(\partial'_{k+1}) = \text{ker}(\partial'_k)$. But
\begin{align} \partial'_k g &= \partial_k' f - \partial_k' s_{k-1} \partial_k = \partial'_kf - \partial'_k s_{k-1} \partial_{k} \\ &= \partial'_kf - (f_{k-1} - s_{k-2}\partial_{k-1})\partial_k \\&= \partial'_kf_k - f_{k-1} \partial_{k} = 0 \end{align}
My questions are:
- Is my reasoning correct?
- Can some of the hypothesis be weakened?