# Is homotopy category of chain complexes abelian

I would like to show that homotopy category of chain complexes of any abelian category is not necessarily an abelian category. Here's my attempt: Assume homotopy category of chain complexes is abelian than any chain map $f:X \rightarrow Y$ splits into $p_2 \cdot p_1$ where $p_1:X \rightarrow Z$ $p_2: Z \rightarrow Y$ and $Z$ is both coim(f) and im(f) and thus $p_1$ is an epimorphism and $p_2$ is a monomorphism. Let $C(f)$ denote a mapping cone of $f$.Then chain map $X \rightarrow Y \rightarrow C(f)$ is homotopic to zero but $p_1$ is an epimorphism, so $Z \rightarrow Y \rightarrow C(f)$ is also homotopic to zero. We can combine this map with $a: C(f) \rightarrow Y$ ( it's just a projection from $X^{n+1} + Y^{n}$ to $Y^{n}$) and now it means that $Z \rightarrow Y$ is homotopic to zero, which gives a contradiction. Is this right approach to this task? It looks suspiciously simple.

The problem with your approach is that the homotopy category $K (\mathcal{A})$ is almost never abelian, but it seems like you are trying to show that it's not abelian for any $\mathcal{A}$, which is not true (for instance, if $\mathcal{A}$ is the category of vector spaces over a field, then $K (\mathcal{A})$ is abelian; or in general $K (\mathcal{A})$ is abelian iff all short exact sequences in $\mathcal{A}$ split).
A counterexample should use something specific about $\mathcal{A}$ (essentially a non-split short exact sequence in $\mathcal{A}$). For $\mathcal{A} = \mathbb{Z}\textit{-Mod}$ it's easy to deduce a contradiction. See for instance here: