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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.

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  • $\begingroup$ 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 $X \to Y \to Z \to X[1].$ $\endgroup$ Commented Feb 11, 2013 at 22:27
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    $\begingroup$ For (2) see also math.stackexchange.com/q/189769 $\endgroup$
    – Martin
    Commented Feb 11, 2013 at 23:19

1 Answer 1

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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?

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    $\begingroup$ Well, since there typically are many non-split monorphisms in Kom(A) there is little hope to find any exotic triangulated structures. $\endgroup$
    – Martin
    Commented Feb 11, 2013 at 23:50
  • $\begingroup$ @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 $A$. $\endgroup$
    – Li Zhan
    Commented Feb 12, 2013 at 0:52

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