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It's well-known that, for lots of concrete categories (but by no means all), we can think of the objects as themselves being small categories, and morphisms are the functors between these categories. Examples include Grp, Ab, Top... When we apply such a construction, we turn a 1-category into a (strict, I think?) 2-category. But 2-categories carry some extra structure, namely the notion of a natural transformation. When we "decategorify" back down, where does this extra structure go?

I can work it out in some specific cases; for instance, if we categorify Ab in the obvious way, there are no nontrivial natural transformations. I don't have a characterization for when two morphisms of groups are naturally isomorphic as functors between the underlying categories, although I have a feel for how the question behaves.

Are there any sort of general results on what natural transformations between morphisms look like if we categorify thusly? Is it at least independent of how we realize the objects as small categories? (I suspect the answer to the second question is "no," but don't have the skills to construct a counterexample. I hope I'm wrong, though.)

More generally, if we can categorify an n-category into an (n+k)-category forgetting the higher morphisms, do the higher morphisms go downstairs to the n-category in any nice way?

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    $\begingroup$ What "obvious way" of categorifying Ab do you have in mind? I think there are several possibilities. $\endgroup$ Commented Dec 12, 2009 at 22:46
  • $\begingroup$ My "obvious way" was as the, um, obvious subcategory of Grp. I guess there might be others, but that's the one that leaps to mind. $\endgroup$ Commented Dec 12, 2009 at 23:23
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    $\begingroup$ What are you thinking of for Top? I don't see that as a category of small categories in any obvious way. $\endgroup$ Commented Dec 13, 2009 at 1:07
  • $\begingroup$ @Mike: Top I copied from Qiaochu's question; I don't know what he had in mind, and it could be wrong. You could replace a space with its fundamental groupoid, but of course that's not enough for the purposes of this question. $\endgroup$ Commented Dec 13, 2009 at 2:05
  • $\begingroup$ Top is kind of problematic. The construction I was thinking of uses the Grothendieck topology construction, which isn't actually enough to distinguish all topologies (in the usual definition). $\endgroup$ Commented Dec 15, 2009 at 14:53

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Here is a counterexample for your next-to-last question. Let S be a set with more than one element and consider the two full subcategories of Cat on, respectively, the single category which is the discrete category on S, and the single category which is the codiscrete category on S. In each case, when viewing Cat as a 1-category, the resulting full subcategory has a single object with endomorphisms Hom(S, S). However, if we view Cat as a 2-category, the former subcategory has no nontrivial natural transformations and thus really is BHom(S, S), while the latter has a unique natural transformation between any two functors and thus is actually • up to 2-equivalence.

Cat-the-1-category and Cat-the-2-category are very different constructs which unfortunately usually go by the same name. Even though they have "the same" objects, I suggest thinking of their objects as being different kinds of things. An object of Cat-the-1-category has more information than an object of Cat-the-2-category; we may talk about the cardinality of its set of objects, not just the cardinality of its set of isomorphism classes of objects. (This shouldn't seem too strange, since an object of Cat-the-0-category is a "specific" category, of which we may talk about the actual set of objects.) Put differently, an object of Cat-the-1-category is a "monoid with many objects", while an object of Cat-the-2-category is what we more often think of when thinking about categories (especially large ones).

In your example, you expressed Ab as a full subcategory of Cat-the-1-category. The full subcategory of Cat-the-2-category on the same objects is not Ab, since it has nontrivial natural automorphisms, as others have pointed out. It only becomes Ab after truncation—replacing each Hom-category by its set of isomorphism classes of objects. For Grp, the situation is worse, since distinct group homomorphisms may be naturally isomorphic as functors. The usual way to repair this is to work with "pointed categories", as described at this nlab page. But of course this is a kind of extra structure on a category, and if I'm allowed to introduce arbitrary extra structure then the question is too easy. Anyways, I'm not sure that one should expect various concrete categories to naturally be full subcategories of either Cat-the-1-category or Cat-the-2-category.

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  • $\begingroup$ Hmm, it took me a while but I think I follow the counterexample. Not too surprising, but a nice one nonetheless. I think your second paragraph hinges on what notions of 1- and 2-category you're using -- right? Pointed categories sound fun, although I'm more and more convinced that I need to actually learn category theory rather than solely working with the weird ad-hoc intuitions I've built up. Extra structure's fine with me, though. $\endgroup$ Commented Dec 15, 2009 at 8:19
  • $\begingroup$ Strict 2-categories or bicategories? I feel like with strict 2-categories where all of the diagrams commute on the nose, we have enough, no? $\endgroup$ Commented Dec 19, 2009 at 10:45
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Two morphisms of groups are isomorphic as functors between the related categories if and only if they differ by an inner automorphism of the target group. The choice of a particular isomorphism of such functors is equivalent to the choice of an element of the target group such that the conjugation with this element identifies the two morphisms.

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  • $\begingroup$ I thought I had a counterexample to that, but looking at it now I see I was confused. $\endgroup$ Commented Dec 13, 2009 at 1:02
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    $\begingroup$ In particular, if H is an abelian group, then the only natural transformations between functors from G to H are automorphisms, and the group of automorphisms of any such functor is H itself. $\endgroup$ Commented Dec 13, 2009 at 1:06

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