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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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Tyler gave an example for the homotopy category here: mathoverflow.net/questions/10364/categorical-homotopy-colimits/… –  Reid Barton Feb 18 '10 at 2:48
    
I realized I probably misinterpreted "homotopy category", but I'll leave the comment up as the question in the link is closely related. –  Reid Barton Feb 24 '10 at 23:33

2 Answers 2

up vote 28 down vote accepted

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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In fact the lemma shows that every abelian triangulated category is semisimple (cf. Gelfand and Manin, Exercises to IV 1). –  JS Milne Feb 18 '10 at 3:07
    
@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. –  Shizhuo Zhang Feb 18 '10 at 3:16
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@Shizhuo: I fail to see how your proof is easier than Greg's :-) –  Andrea Ferretti Feb 18 '10 at 11:11
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@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. –  Shizhuo Zhang Feb 18 '10 at 13:43
    
@Andrea: why the objects in $TC$ is injectives in $C_{a}$? This is because Yoneda embedding to $Funct(TC^{op},Sets)$ factors through C_{a}. Therefore the image of objects of $TC$ in $C_{a}$ is representable(so by Yoneda lemma, projective), but $C_{a}$ is frobenius, therefore projectives automatically becomes injectives. –  Shizhuo Zhang Feb 18 '10 at 13:50

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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