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Looking at the nCatLab page on chain complexes, it is implicitly assumed at the start of the page that one is working in an additive category. However, the only structure required to define chain complexes is that one be working in a pointed category, so that the notion of a zero morphism makes sense. However, I have not been able to find any papers which consider chain complexes even in categories such as preadditive or semiadditive categories, much less arbitrary pointed categories.

I understand that many of the results of homological algebra rely on an Abelian structure to work. Is it simply the case that non-additive categories don't have enough structure to yield any interesting results about these complexes? Or is there a concrete issue that prevents chain complexes from making sense at all in more general pointed categories? Do these issues still arise even when working in preadditive or semiadditive categories?

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    $\begingroup$ You need abelianness to define kernels, cokernels, quotients, and so on (after all, you are very interested in the homology of such complex!) $\endgroup$ – Pedro Tamaroff Jul 30 '16 at 5:06
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    $\begingroup$ At least in your first brush with homological algebra, it might be worthwhile to ask "what is it good for" (not absolutely nothing!) rather than "how can we generalize it". Lang suggests pretending all abelian categories are categories of modules on your first reading of his two chapters on homological algebra. I wholeheartedly agree. $\endgroup$ – Noah Olander Jul 30 '16 at 5:31
  • $\begingroup$ Don't ask what you can generalize; ask what the generalization can do for you. $\endgroup$ – Ittay Weiss Jul 30 '16 at 5:38
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    $\begingroup$ @Noah: well, but on the other hand, one of the most charming features of abelian categories is that they're closed under taking opposites, so every theorem about abelian categories has a corresponding dual theorem. That's very far from being true of categories of modules. $\endgroup$ – Qiaochu Yuan Jul 30 '16 at 5:45
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    $\begingroup$ The most general setting, in some sense, in which homological algebra works well is that of a homological category, where groups are the standard nonabelian example: groups have all the standard diagram lemmas etc. But already chain complexes of groups are a bit hairier than those we're used to, insofar as not every subgroup is normal. $\endgroup$ – Kevin Carlson Jul 30 '16 at 6:06
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You can write down definitions all day long, but at the end of the day those definitions should do something for you or else there was no point in writing them down. The point of chain complexes is to do homological algebra, and to do homological algebra you need short exact sequences, kernels, cokernels, etc. Our honored ancestors worked out long ago that abelian categories were a very convenient setting with essentially everything we needed to do homological algebra, so that's typically where we do it.

(And yes, it's true that you can define kernels and cokernels in a category with zero morphisms, without also requiring that it be additive. But kernels and cokernels don't have the properties you'd expect them to have unless you're in an additive category or thereabouts; for example, it's not true in general that if a morphism has zero kernel then it's a monomorphism.)

Homological algebra in turn is just the "linear" version of homotopical algebra. Here, instead of chain complexes we use simplicial objects, which can be thought of as "nonlinear" generalizations of chain complexes. The theorem that makes this idea precise is the Dold-Kan theorem, which also requires more than just zero morphisms to work. Even if you can write down, say, a definition of chain complex of pointed sets, that's not what you need to actually do "homological algebra" with pointed sets.

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