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Let $\mathcal{C}$ and $\mathcal{D}$ be categories, where $\mathcal{C}$ is an abelian category. We want to say that $\mathcal{C}^\mathcal{D}$ is also an abelian category. However, if $\mathcal{C}$ and $\mathcal{D}$ is big enough, $\hom(F, G)$ in $\mathcal{C}^\mathcal{D}$ is too big to be a set, and hence, we cannot really say it has the structure of an abelian category (at least, in the usual sense).

So, my question is: what are the ways to fix it? I am aware of the option of using universes, but is there any other way(s)?

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You could just restrict to only considering categories $\mathcal{C}$ that are locally small and $\mathcal{D}$ that are small to prevent any such issues? – Chris Heunen Mar 18 '11 at 20:24
Thanks. I'm aware of this option, but it sounds a bit restrictive. Is there any other way? – Brian Mar 18 '11 at 20:25
Are you encountering this problem in a specific context? If so, some extra information might help us help you. – S. Carnahan Mar 18 '11 at 20:59

One can use "big abelian categories". A "big abelian group" is a class that satisfies the same properties as an ordinary abelian group except that the underlying class doesn't need to be a set. Then a big abelian category is a category with the same properties as an ordinary abelian category except that the hom's aren't abelian groups but big abelian groups (see: Mitchell: "Theory of Categories", VII.1, page 164).

The usual proof that $C^D$ is abelian if $C$ is abelian and $D$ is small carries over without difficulties to show that $C^D$ is (big) abelian if $C$ is abelian and $D$ is any category.

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How much of homological algebra can you do in this setting? It seems to me that most homological arguments require the axiom of choice (to lift elements along epis). Do you run into problems when working with these enormous groups? – K.J. Moi Mar 19 '11 at 8:24
That's a good point. In order to be practical, I would consider two cases. The first case is, when only the internal addition within the hom's is used: Consider for example the statement: In a (big) abelian category a hom $f$ is mono iff $fg = 0$ implies $g = 0$ whenever $fg$ is definied. To prove this statement it would be an unnecessary strong restriction to require that hom's are sets. The second case is when the hom's are used as objects for itself. For example, if you want to form the automorphism group, factor groups, etc. of a hom. Then I would tend to require that hom's are sets. – Ralph Mar 21 '11 at 0:16
(continue) Concerning your question about AC in homological algebra: I don't think that most homological arguments require AC. Category theory is based on the notion of morphisms and universal properties. So, in general there is actually no need to choose inverse images. For example, Snake Lemma can be proved entirely with universal properties without using AC. <p>BTW: There is an axiom of choice for classes, called "Axiom of global choice". – Ralph Mar 21 '11 at 0:17
I believe one does use AC in for instance working with, or at least constructing, projective and injective resolutions. The usual proof that free modules are projective uses AC; see also… . I would guess that probably the axiom of global choice would be sufficient in most "locally-large abelian categories" arising in practice. – Mike Shulman Mar 22 '11 at 0:38
@Mike: Thank you very much. You are completely right: Of course, there is some kind of choice if one takes a projective/injective presentation/resolution etc. (independently whether the category is locally small or locally large). – Ralph Mar 22 '11 at 2:10

A way for keep the Hom as set, is request for the category $\mathcal{D}$ to have a set of objects $\mathcal{G}$ such that for any object $X$ of $\mathcal{C}$ there exist a section from $s:X\to G$ (i.e exists $r: G\to X$ with $r\circ s=1$) for some some $G\in\mathcal{G}$. Then any trasformation is uniquely determunated by its restriction to $\mathcal{G}$.

In another way $\mathcal{G}$ is simply a generator and we admit only $Epi$-preserving functors.

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