e.g. is $\widehat{\mathbf{SET}}$ locally small?
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$\begingroup$ doesn't this depend on the precise definition of locally small and SET (e.g. with Grothendieck universes)? I think there are varying answers, depending which definitions you use. $\endgroup$– Martin BrandenburgCommented May 13, 2010 at 22:47
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No, not in general. Todd gave a useful general answer and here is a concrete counter-example.
Let $\mathbf{C}$ be a category whose objects are sets and where the only morphisms are the identities. Consider the functors $F, G : \mathbf{C} \to \mathbf{Set}$ given by $F(X) = 1 = \lbrace 0 \rbrace$ and $G(X) = 2 = \lbrace 0,1\rbrace$. Because $\mathbf{C}$ is discrete, every family of maps $(\eta_X : F(X) \to G(X))_{X \in \mathbf{C}}$ is natural. Therefore, natural transformations $F \to G$ are in bijective correspondence with classes:
a natural transformation $\eta : F \to G$ corresponds to the class $\lbrace X \in \mathbf{Set} \mid \eta_X(0) = 1\rbrace$.
a class $C$ corresponds to the natural transformation $\eta : F \to G$ given by
$\eta_X(0) = 1$ if $X \in C$ and $\eta_X(0) = 0$ if $X \not\in C$.
Thus the presheaf category $\widehat{\mathbf{C}}$ is not locally small (note that $\mathbf{C}^{op} = \mathbf{C}$).
I cannot think at the moment of concrete presheaves $F$ and $G$ on $\mathbf{Set}$ for which the natural transformations $F \to G$ form a proper class. Someone please help.
answered May 13, 2010 at 22:52
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$\begingroup$ Andrej, did you take a look at the Freyd-Street paper? Because their construction is quite explicit and concrete: see their definition of a functor "T". $\endgroup$ Commented May 14, 2010 at 2:39
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$\begingroup$ I looked after you posted your answer (I was writing mine when yours arrived). I just wanted to give as easy a counter-example as possible. $\endgroup$ Commented May 14, 2010 at 6:55
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$\begingroup$ I meant not with regard to your counter-example, but with regard to your plea at the end (but maybe you understood that's what I meant! :-) $\endgroup$ Commented May 14, 2010 at 10:46
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There's a theorem due to Freyd and Street that says that if $C$ is locally small and the presheaf category $Set^{C^{op}}$ is locally small, then $C$ is equivalent to a small category. See this paper.
answered May 13, 2010 at 22:37
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$\begingroup$ @Todd Trimble: Is this simple argument wrong? Consider a non-empty hom-set $\widehat{C}(F,G)$, and let $\tau$ be in this hom-set. If the hom-set is small, then by transitivity of the universe, $\tau$ is small too. But $\tau$ is just a function with domain $\operatorname{obj}(C)$ (so $\tau$ is a triple with $\operatorname{obj}(C)$ as its first component). But then from transitivity again we get $\operatorname{obj}(C)\in U$, a contradiction. (So, in general,the answer to the original question is ``no'') $\endgroup$– user2734Commented May 13, 2010 at 22:52
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$\begingroup$ Even if $C$ is large, the collection of transformations $F \to G$ between presheaves can be essentially small (in definable isomorphism with a set). Consider for example $F$ representable and apply Yoneda. $\endgroup$ Commented May 13, 2010 at 23:39
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$\begingroup$ Thank you very much for your answer! I hope it is OK that I ask another silly question: Is my comment above correct, but just useless, because (for some reason that I still don't understand) "locally small" can refer to non-small hom-sets that are in bijection with small sets? [I'm assuming a single universe, as in Mac Lane.] $\endgroup$– user2734Commented May 14, 2010 at 12:49
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$\begingroup$ Yes, under the sorts of set-theoretic encodings you appear to have in mind, the argument looks to me impeccable but still "morally wrong"! (It would mean $\hom(F, G)$ is not small even if $G$ is terminal.) It's an excellent illustration of the kind of sophistry that's possible by encoding everything as sets and sets as membership trees. More satisfactory would be a foundations where this type of argument cannot even be formulated (cf. discussion of structuralist vs. materialist forms of set theory in the nLab). Suffice it to say that the problem is not insurmountable. :-) $\endgroup$ Commented May 14, 2010 at 12:54
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$\begingroup$ Thank you very much! Having such tips from an expert is extremely helpful. I have a million more question to ask, but I guess these comments aren't the right place for such a mini course... $\endgroup$– user2734Commented May 14, 2010 at 13:09