# How does $C$ small imply $Set^{C^{op}}$ locally small?

I read in some notebook that $C$ small implies $Set^{C^{op}}$ locally small, but I don't see what is the reasoning used, because the Yoneda lemma is not mentioned so that it is probably not needed...

Could somebody propose me a track?

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When $C$ is a small category and $D$ is a locally small category, then $D^C$ is locally small. The reason is that for functors $F,G : C \to D$ the set of natural transformations $F \to G$ embeds into $\prod_{x \in \mathrm{ob}(C)} \hom(F(x),G(x))$, which is a product of sets, therefore also a set.
Now $\hom(F(x),G(x))$ is in $D$. So, this argument also uses that $D$ is locally small. E.g. let $D$ be a big monoid, and $C:=1$ the terminal category... –  Berci Apr 9 '13 at 19:11
Thank you very much Martin and Berci! Yes, I agree with Berci, this last part uses the fact that $D = Set$ so that its $hom$ are sets. –  almaus Apr 9 '13 at 19:19