Let $R$ be a ring, and suppose that we have a short exact sequence of chain complexes over $R$: $$0 \rightarrow A \xrightarrow{i} B \xrightarrow{p} C \rightarrow 0$$ and a chain map $r\colon B \rightarrow A$.
If $ri = \mathrm{id}_A$ (i.e. if $r$ is a retraction of our short exact sequence) then the sequence splits and the map $i \oplus p\colon B \simeq A \oplus C$ is an isomorphism. A way to prove this is to use the decomposition: $$B = \mathrm{Im}(i) \oplus \mathrm{Ker}(r)$$ and using the exactness in $B$ to show that the restriction of $p$ on the second term of the sum is an isomorphism.
Is it true that if we relax our hypothesis, by only asking for a chain homotopy $ri \simeq \mathrm{id}_A$, then $i \oplus p$ is an chain homotopy equivalence $B \simeq A \oplus C$?
The problem is that in this case, $\mathrm{Im}(i)$ and $\mathrm{Ker}(r)$ are no longer supplementary in $B$ in general, and so I have a hard time constructing a section $C \rightarrow B$. I have the intuition that I need to replace $B$ with the cone of some morphism and to construct a real retraction using $r$ somehow but all my attempts have failed (mainly because I do not see how to preserve the homotopy type of $B$ by doing so). I think it would help if our short exact sequence was termwise split, so that it gives rise to an exact triangle in $\mathbf{K}(R)$, and use the fact that $i$ must be an monomorphism in this category (thanks to $r$), but even then I don't see how to conclude as I am inexperienced with triangulated categories.
Any insight would be greatly appreciated ! Maybe the result is false, but that would make me very sad.
Thank you for taking the time of reading me.