2
$\begingroup$

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

$\endgroup$
3
  • 3
    $\begingroup$ "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. $\endgroup$
    – Alon Amit
    Jun 3 '11 at 2:45
  • $\begingroup$ @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)? $\endgroup$
    – Juan S
    Jun 3 '11 at 2:52
  • $\begingroup$ 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. $\endgroup$
    – Alon Amit
    Jun 3 '11 at 7:26
5
$\begingroup$

(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!

$\endgroup$
2
  • $\begingroup$ thank you for the correction, that was a bad mistake. I will have a think about your question. $\endgroup$
    – Juan S
    Jun 3 '11 at 2:40
  • $\begingroup$ 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... $\endgroup$
    – Juan S
    Jun 3 '11 at 3:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.