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If $\mathcal{A}$ and $\mathcal{B}$ are small categories (i.e. objects are sets, not proper classes) and $F,G:\mathcal{A} \to \mathcal{B}$ are both (contra/co)variant functors, then the the 'collection' of natural transformations $Nat(F,G)$ is also a set.

Say we now have arbitary categories $\mathcal{C}$ and $\mathcal{D}$, with functors $S,T$ such that $S,T:\mathcal{C} \to \mathcal{D}$. Then it seems logical that $Nat(S,T)$ may be a proper class, and not a set.

Can anyone provide any examples?

Moreover (this is a bit more philosophical) - are such examples pathological, or do non-small categories arise naturally, outside of say, just studying higher category theory?

I am guessing in something like higher homotopy theory, one could possibly get in trouble?

This thread seems somewhat related

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"objects are sets, not proper classes" is not what "small category" means. It means the class of all objects is a set, not that each object is one. –  Alon Amit Jun 3 '11 at 2:45
@Alon - thank you for that. Is that why the category of sets is not a small category - there is no collection of all sets (in ZF set-theory)? –  Juan S Jun 3 '11 at 2:52
Yes. The same goes for the other common categories Mariano mentions - there's no set of all groups, no set of all topological spaces, etc. –  Alon Amit Jun 3 '11 at 7:26
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1 Answer 1

(1) You mean $\operatorname{Nat}(F,G)$ and $\operatorname{Nat}(S,T)$...

Let $\mathcal C$ be the category whose objects are the sets and where for each pair of objects $x$, $y$ we have $\hom(x,y)=\emptyset$ if $x\neq y$ and $\hom(x,x)=\mathbb Z_2$, with composition given by addition. Let $I:\mathcal C\to\mathcal C$ be the identity functor.

Can you see what $\operatorname{Nat}(I,I)$ is?

(2) Non-small categories are all over the place: the category of sets, of vector spaces, of abelian groups, of topological spaces... There is absolutely no need to go look for them in exotic places like higher category theory!

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thank you for the correction, that was a bad mistake. I will have a think about your question. –  Juan S Jun 3 '11 at 2:40
Hi Mariano, I am really not sure how to approach your question. If $\operatorname{hom}(x,y)=\emptyset$ for $x \ne y$, then we can't have the usual diagram for a natural transformation unless $x = y$, as the morphism $I(x) \to I(Y) = x \to y$ does not exist... –  Juan S Jun 3 '11 at 3:13
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