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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.

flag	asked Feb 18 2010 at 2:27 Harold Williams 1,788●6●17

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Tyler gave an example for the homotopy category here: mathoverflow.net/questions/10364/… – Reid Barton Feb 18 2010 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 2010 at 23:33

Greg Stevenson · Accepted Answer · 2010-02-18 02:53:50Z UTC

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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.

edited Feb 18 2010 at 2:53

answered Feb 18 2010 at 2:43

Greg Stevenson
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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 2010 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 2010 at 3:16

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@Shizhuo: I fail to see how your proof is easier than Greg's :-) – Andrea Ferretti Feb 18 2010 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 2010 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 2010 at 13:50

show 2 more comments

Sunil Rana · Answer 2 · 2010-02-24 21:33:45Z UTC

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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.

answered Feb 24 2010 at 21:33

Sunil Rana
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Miguel · Answer 3 · 2013-03-01 16:22:22Z UTC

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I follow Greg's argument, but I don't understand why you reach a contradiction, since $\mathbb{Z}/p^2\mathbb{Z}$ is idecomposable in the category of abelian groups, but in the homotopy category it may not be...

answered Mar 1 at 16:22

Miguel
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	if it's indecomposable in the abelian category it will also be in the derived category. think about it: if it were a direct sum of complexes not concentrated in degree zero then it would have to have cohomology outside of zero, which it doesn't. – Jacob Bell Mar 1 at 16:50
	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... – Miguel Mar 4 at 15:25

How do I know the derived category is NOT abelian?

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