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Dec 2 at 20:05 comment added Martin Brandenburg The statement by Marc is not correct. See math.stackexchange.com/questions/3654695/…
Nov 3, 2016 at 21:07 vote accept Mikhail Bondarko
Oct 29, 2016 at 16:20 answer added ACL timeline score: 4
Oct 29, 2016 at 10:25 answer added user337830 timeline score: 3
Oct 29, 2016 at 9:41 comment added Mikhail Bondarko Yes, my "main" results concern locally small categories only. However, for a (locally small) subcategory $A$ of the category of functors from $C$ into $Ab$ it is rather convenient to say that $A$ is an exact subcategory of the "big abelian" category $AddFun(C,Ab)$.
Oct 29, 2016 at 9:33 history edited Mikhail Bondarko CC BY-SA 3.0
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Oct 29, 2016 at 9:25 comment added HeinrichD In my experience the category of all functors $\mathsf{Ab} \to \mathsf{Ab}$ is not really useful and not well-behaved. This is because many results in category theory, when applied to specific examples, require locally small categories. Instead, we should look at subcategories of functors $\mathsf{Ab} \to \mathsf{Ab}$ which are locally small, for example the accessible functors. Actually, one can do a lot of category theory only with locally small categories, which makes me wonder why we should need anything else. And many authors define categories as locally small per definition.
Oct 28, 2016 at 16:10 comment added Philippe Gaucher @MikhailBondarko A category does not have to be locally small. The only precaution you have to take is not asserting that a proper class is a set if it is not a set.
Oct 28, 2016 at 15:41 comment added Mikhail Bondarko Cannot $Fun(C.D)$ have "too many" objects?
Oct 28, 2016 at 13:10 comment added Marc Hoyois Being locally small is invariant under equivalences of categories. In particular if C is essentially small and D is locally small then Fun(C,D) is locally small.
Oct 28, 2016 at 10:07 history asked Mikhail Bondarko CC BY-SA 3.0