For level-wise free chain-complexes a quasi-isomorphism is always induced by homotopy equivalence Let $C_{\cdot}, D_{\cdot}$ be level-wise free chain-complexes, i.e. such that each $C_n$ and $D_n$ is a free abelian group.
Let $f:C_{\cdot} \to D_{\cdot}$ be a chain-map and a quasi-isomorphism. Thus we find a map $\tilde{g}:H_{\cdot}(D) \to H_{\cdot}(C)$ such that $H_n(f) \circ \tilde{g} = \operatorname{id}_{H_{\cdot}(D)}$ and $\tilde{g} \circ H_{\cdot}=\operatorname{id}_{H_{\cdot}(D)}$.
Prove that there exists a map $g:D_{\cdot} \to C_{\cdot}$ such that $g \circ f$ and $f \circ g$ are chain homotopic to the respective identites.
My ideas: I tried to write all the boundary maps down as explicit linear maps, as well as $f$, and somehow calculate $g$. This did not work out.
Any hints or references would be welcome.
Edit: I would also be very interested in hints/a solution to the problem under the additional assumption that both complexes are bounded from below.
 A: One approach is to use the mapping cone of a chain complex. For some details, see these notes, for example, and in particular Corollary 2.5 is getting to the result you want. Addendum: the cited notes are really just replicating section 1.5 of Weibel's book An Introduction to Homological Algebra. Corollary 1.5.4 is the relevant one.
Given a chain map $f: C_{\bullet} \to D_{\bullet}$, define its mapping cone $C(f)_{\bullet}$ to be the chain complex with $n$th term $C_{n-1} \oplus D_{n}$, boundary map $\begin{pmatrix} -\partial^{C} & 0 \\ f & \partial^{D} \end{pmatrix}$. Then there is a natural map $D_{\bullet} \to C(f)_{\bullet}$ and the maps $C_{\bullet} \to D_{\bullet} \to C(f)_{\bullet}$ induce a long exact sequence in homology. If the map $f$ is a quasi-isomorphism, then $C(f)_{\bullet}$ is free (that's the corollary from the notes), and furthermore it is degree-wise free. If it is bounded below, then it must be contractible (chain-homotopy equivalent to zero). This is not hard to prove by induction. (See also https://mathoverflow.net/questions/73687/when-mapping-cone-is-contractible and https://mathoverflow.net/questions/73687/when-mapping-cone-is-contractible, which both at least peripherally address the case of chain complexes, in addition to topological spaces.)
Now I claim that the chain homotopy between $1$ and $0$ on $C(f)_\bullet$ gives a chain homotopy equivalence between $C_\bullet$ and $D_\bullet$. This requires some checking.
