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I have heard the claim that the derived category of an abelian category is in general additive but not abelian. If this is true there should be some toy example of a (co)kernel that should be there but isn't, or something to that effect (for that matter, I could ask the same question just about the homotopy category).

Unless I'm mistaken, the derived category of a semisimple category is just a ℤ-graded version of the original category, which should still be abelian. So even though I have no reason to doubt that this is a really special case, it would still be nice to have an illustrative counterexample for, say, abelian groups.

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The following nicely does the trick I think...

Lemma Every monomorphism in a triangulated category splits.
Proof: Let $T$ be a triangulated category and suppose that $f\colon x\to y$ is a monomorphism. Complete this to a triangle
$x \stackrel{f}{\to} y \stackrel{g}{\to} z \stackrel{h}{\to} \Sigma x$
then $f\circ \Sigma^{-1}h = 0$ as we can rotate backward and maps in triangles compose to zero. Since $f$ is a monomorphism we deduce that $\Sigma^{-1}h$ and hence $h$ are zero. But this implies that $y\cong x\oplus z$ (a proof of this can be found in the first part of my answer here so that $f$ is a split monomorphism. █

Since every kernel is a monomorphism we get the following counterexample. The map
$\mathbb{Z}/p^2\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$ does not have a kernel in $D(Ab)$ by virtue of the fact that $\mathbb{Z}/p^2\mathbb{Z}$ is indecomposable. Of course the same thing works in the homotopy category.

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    $\begingroup$ In fact the lemma shows that every abelian triangulated category is semisimple (cf. Gelfand and Manin, Exercises to IV 1). $\endgroup$
    – JS Milne
    Commented Feb 18, 2010 at 3:07
  • $\begingroup$ @Greg: I think there might be simpler proof for this lemma. Consider the abelianization of triangulated category. Then the objects of triangulated category is injectives in the correspondence frobenius abelian category. Then a monomorphism from injective object splits. $\endgroup$
    – Shizhuo Zhang
    Commented Feb 18, 2010 at 3:16
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    $\begingroup$ @Shizhuo: I fail to see how your proof is easier than Greg's :-) $\endgroup$
    – Andrea Ferretti
    Commented Feb 18, 2010 at 11:11
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    $\begingroup$ @Andrea: because it is direct. $TC=(C,\theta ,Tr)\xrightarrow[]{abelianization}(C_{a},\theta _{a})$. the right hand side is a frobenius abelian category. Then this functor maps any pair of objects with monomorphism: $M\rightarrow N$ in $TC$ to $C_{a}$. There is an obsevation that this functor is fully embedding as full subcategory of $C_{a}$, and, each object is injective in $C_{a}$. Then in $C_{a}$, monomorphism from injective object splits. $\endgroup$
    – Shizhuo Zhang
    Commented Feb 18, 2010 at 13:43
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    $\begingroup$ Ok, thanks, but I assume that for the homotopy category you have to elaborate more the argument, because having nonzero cohomology only at degree zero doesn't mean that the complex is concentrated on degree zero... $\endgroup$
    – Miguel
    Commented Mar 4, 2013 at 15:25
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Even the homotopy category K(A) is not abelian. given any f in C(A) the 1: cone(f) ---> cone(f) is homotopic to 0 morphism. but homotopy for their kernels doesn't make sense. Well, this is kind of wague, but can be made precise.

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