Let A be an Abelian category.

From this category, we can form the chain complex category Ch(A). The objects of Ch(A) are chain complexes of objects of A. The morphisms of Ch(A) are chain maps. Ch(A) is an Abelian category for every Abelian category A.

Now from Ch(A), we can form the chain homotopy category K(A). The objects of K(A) are just objects of Ch(A), but the morphisms of K(A) are chain homotopy classes of chain maps. Thus, K(A) is a quotient of Ch(A).

It turns out that K(A) is an additive category for any Abelian category A. From asking people, I seem to get the impression that K(A) is not always abelian, but I've had a hard time showing this. If all objects of A are projective (e.g. if A is the category of vector spaces over some field k), then K(A) will be Abelian.

I've been trying to show that K(Ab) is not Abelian (where Ab is the category of Abelian groups). More specifically, I've been trying to show this by showing the following (which may or may not be true):

Let X be a chain complex with the group of integers in dimension 0, and zero in every other dimension. Let f be the chain map from X to X, that sends each integer x to 2x. I've been trying to show that in K(Ab), the homotopy class of this chain map does not have a cokernel.

Any answers, suggestions, hints, or comments would be greatly appreciated!


1 Answer 1


You might want to have a look at my answer to this question.

  • 1
    $\begingroup$ But your answer is about $D(Ab)$, not $K(Ab)$? $\endgroup$
    – VA.
    Apr 29, 2010 at 1:33
  • 2
    $\begingroup$ The same thing works in the homotopy category by using the lemma, which is valid for any triangulated category, to produce a counterexample (for instance the one I give there works). I guess one does need to prove that $K(Ab)$ is triangulated first though... $\endgroup$ Apr 29, 2010 at 1:44

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