# Why $Kom(\mathcal{A})$ may not be triangulated, while $D(\mathcal{A})$ may not be abelian?

Let $\mathcal{A}$ be an abelian category, let $Kom(\mathcal{A})$ be the category of complex with a shift functor $T$, and Let $D(\mathcal{A})$ be the derived category of $\mathcal{A}$.

Why:

(1) $Kom(\mathcal{A})$ may not be a triangulated category?

(2) $D(\mathcal{A})$ may not be an abelian category?

I am quite new to those derived category stuff, so any intuitions behind the counterexamples are also very welcome, and will be very important for me to understand these concepts.

• For a) it might be helpful to notice that any morphism $X \to Y$ in a tiangulated category can be completed to a distinguished triangle $X \to Y \to Z \to X[1].$ – Ehsan M. Kermani Feb 11 '13 at 22:27
• For (2) see also math.stackexchange.com/q/189769 – Martin Feb 11 '13 at 23:19

• @Martin Brandenburg Thank you for your link to the first question, I think combining with the comment given by Martin above, it answers my second question as well, because "Every monomorphism in a triangulated category splits", and in order to make $Kom(A)$ triangulated, we should put some conditions on $A$. – Li Zhan Feb 12 '13 at 0:52