If $M$ is equivalent to zero, must $C$ and $C/M$ be equivalent? Let $C$ be a chain complex over a commutative ring $R$. Let $M$ be a subchain of $C$ such that $M$ is chain equivalent to the zero chain complex. Must $C/M$ and $C$ be chain equivalent ?
Thank you

Motivation: I m just comparing the situation  to the category of modules over a ring or the category of groups to the category of chain complexes (with morphisms mod out by chain homotopies). In category of modules over a ring or the category of groups this statement obviously holds.

Responding to comments asking for clarification:
Two chain complexes $A$ and $B$ are called chain equivalent iff there exists chain maps $f\colon A\to B$ and $g\colon B\to A$ such that there exists a chain homotopy between $gf$ and $id_A$ and a chain homotopy between $fg$ and $id_B$
As for effort: I tried to prove the statement but couldn't as there seems to be no natural way of having a chain map from $C/M$ to $C$. Since I expected that if there is a counterexample to this "natural" question  it would be pathological counterexample so I came to ask here to get help from people who are more knowledgeable than me about chain complexes.

Take complexes of abelian groups:
$$C:= \dots\longrightarrow0\longrightarrow\mathbb{Z}\stackrel{\times2}{\longrightarrow}\mathbb{Z}\longrightarrow0\longrightarrow\dots,$$
$$M:= \dots\longrightarrow0\longrightarrow\mathbb{Z}\stackrel{\times2}{\longrightarrow}2\mathbb{Z}\longrightarrow0\longrightarrow\dots,$$
$$C/M:= \dots\longrightarrow0\longrightarrow0\longrightarrow\mathbb{Z}/2\mathbb{Z}\longrightarrow0\longrightarrow\dots.$$
Then M is isomorphic to the complex
$$\dots\longrightarrow0\longrightarrow\mathbb{Z}\stackrel{\text{id}}{\longrightarrow}\mathbb{Z}\longrightarrow0\longrightarrow\dots,$$
which is contractible (homotopy equivalent to the zero complex), but there are no nonzero maps $C/M\to C$, so $C$ and $C/M$ are not homotopy equivalent.
