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It is known that the chain homotopy category $K(\mathcal{A})$ for an abelian category $\mathcal{A}$ need not be abelian. For example, $K(\mathrm{Ab})$ is not even abelian.

Are there any known conditions on $\mathcal{A}$ which ensure that $K(\mathcal{A})$ is in fact abelian? Browsing the standard references didn't turn up anything.

Camilla Vaernes
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    Why would you want/expect it to be abelian? It is usually the case that homotopy categories (not just of chain complexes) do not have (co)equalisers. – Zhen Lin Apr 22 '14 at 08:17
  • @ZhenLin I guess I'm mostly curious if there are cases where it is in fact abelian. – Camilla Vaernes Apr 22 '14 at 23:08
  • Well, there's the trivial case where $\mathcal{A}$ is the trivial abelian category with only one object... – Zhen Lin Apr 23 '14 at 07:11

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$K(\mathcal{A})$ will be abelian if and only if all short exact sequences split in $\mathcal{A}$.

More generally, in any triangulated category $\mathcal{T}$, the only epimorphisms are the split ones (and dually, the only monomorphisms are split). So if $\mathcal{T}$ were abelian, then every map in $\mathcal{T}$ would have to be of the form $\mu\varepsilon$, where $\epsilon$ is a split epimorphism and $\mu$ a split monomorphism.

To prove the claim about epimorphisms in triangulated categories, suppose $X\stackrel{\alpha}{\rightarrow}Y$ is an epimorphism, and complete to an exact triangle $$X\stackrel{\alpha}{\rightarrow}Y\stackrel{\beta}{\rightarrow}Z$$ Since $\alpha$ is an epimorphism and $\beta\alpha=0$, $\beta$ must be the zero map, and so $\alpha$ is a split epimorphism.

Jeremy Rickard
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    Actually, pretty much the same argument was previously posted for this related question: http://math.stackexchange.com/questions/189769/when-is-the-derived-category-abelian – Jeremy Rickard Apr 23 '14 at 10:25