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

Question

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

score 2 · Accepted Answer · edited Apr 13 '17 at 12:58

2

For (2) see MO/15658.

(1) isn't well-defined, because a triangulated category is not just a category with extra properties. It also comes equipped with extra structure, namely the class of distinguished triangles. A more precise question would include a definition of the distinguished triangles and then ask why it doesn't satisfy the axioms of a triangulated category. Or are you interested in pathological / arbitrary triangulated structures?

edited Apr 13 '17 at 12:58

Community

1

answered Feb 11 '13 at 23:08

Martin Brandenburg

163,620

1

Well, since there typically are many non-split monorphisms in Kom(A) there is little hope to find any exotic triangulated structures. – Martin Feb 11 '13 at 23:50
@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 00:52

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

1 Answers1