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Let $\mathcal{A}$ be a small category, ie. $\text{Ob}(\mathcal{A})$ is a set. $\text{Fun}(\mathcal{A}, \text{Set})$ is the category of functors from $\mathcal{A}$ to the category of sets, with morphisms the natural transformations between functors. I have already proven that given a functor $F: \mathcal{A} \to \text{Set}$ for an arbitrary category $N(A) = \text{Nat}(\mathcal{A}(A, -), F)$ is a functor, $\theta_{A,F}: \text{Nat}(\mathcal{A}(A,-),F) \xrightarrow{\simeq} FA$ is a bijection for all $A \in \mathcal{A}$, and $\theta_{F,A}$ is also a natural map in the variable $A$. I don't know if any of that is relevant to the question: why does it make sense to consider now $\text{Fun}(\mathcal{A}, \text{Set})$, when $\mathcal{A}$ is a small category and not otherwise?

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    $\begingroup$ Because if you're not having fun with sets, then nothing makes sense anymore! :-) $\endgroup$ – Asaf Karagila Jan 25 '16 at 23:49
  • $\begingroup$ @AsafKaragila category theory is $\text{Fun}$. $\endgroup$ – StudySmarterNotHarder Jan 25 '16 at 23:59
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    $\begingroup$ Well, that's just plain false! ;-) $\endgroup$ – Asaf Karagila Jan 26 '16 at 0:00
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In a category, the set of morphisms $Hom(X,Y)$ is a set. If $A$ is small for every functors $F,G$, the natural transformations between $F$ and $G$ is a set: it is a subset of $\prod_{X\in A}Hom_{Set}(F(X),G(X))$. This not guarantee if $A$ is not small.

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  • $\begingroup$ I see now. Thank you for your explanation! $\endgroup$ – StudySmarterNotHarder Jan 25 '16 at 23:57

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