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Briefly: do they agree?

In more detail: denote by $\mathbf{Finset}$ the category of finite sets, and by $\mathbf{Set}$ the category of sets. I want to know what the functor categories $[\mathbf{Finset},\mathbf{Set}]$ and $[\mathbf{Finset}^{\text{op}},\mathbf{Set}]$ look like. Are they both equivalent to $\mathbf{Set}$?

How should I think about these so that I at least conjecture the answer (and then try to construct a functor that is an equivalence)? I am really struggling with thinking about (explicit) functor categories intuitively.

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    $\begingroup$ $\mathbf{Set}$, $[\mathbf{FinSet}, \mathbf{Set}]$ and $[\mathbf{FinSet}^\mathrm{op}, \mathbf{Set}]$ are all distinct. $\endgroup$ – Zhen Lin Apr 22 '16 at 13:42
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A set $X$ determines very special functors from finite sets to sets, contravariantly by sending $S$ to the set of functions $f:S\to X$ and, dually, covariantly. These are restrictions of representable functors on the full category of sets, so for instance the contravariantly one send colimits to limits. There's no reason at all why a general functor should do that-how about the constant functor $S\mapsto \{a,b\}$?

In fact, a functor determined by a set in this way is determined by its value at a singleton, since every finite set is a coproduct of singletons. This shows that there are intuitively many more functors on finite sets than there are sets. Indeed, the category of sets is equivalent to the category of colimit-preserving functors on finite sets in this way.

Now to prove the categories aren't actually equivalent, we need more than that. The important result is that the Yoneda embedding of any category into its contravariant set-valued functors is a free cocompletion of the category, so that the Yoneda embedding is the initial functor into a cocomplete category. But the category of sets is very far from being the free cocompletion of finite sets-this is because the embedding of finite sets into sets preserves many colimits, in fact all colimits, whereas the Yoneda embedding preserves essentially none. Being a free cocompletion is a universal property, so there can't possibly be an equivalence between these categories under the standard embeddings of finite sets. In fact there's no equivalence at all, for instance because in the functor categories not every object is a coproduct of final objects.

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