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Let $F: Set^{op} \rightarrow Set$ be a functor such that for corresponding functor $\overline{F}: Set \rightarrow Set^{op}$ we have $\overline{F} \dashv F$. With corresponding functor I mean that $F$ and $\overline{F}$ are just basically the same functor (just written differently).

An exercise says that from this information, it follows that $F$ is naturally isomorphisc to $Set(-,A)$. Clearly we need to use Yoneda's Lemma, but I cant really see how. I have no clue how to show this $A$ exists at all or where it comes from. Anybody have an idea?

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Yoneda's lemma doesn't help to characterize when a given functor is representable or not.

Here just write the fact that you have an adjunction : $$Hom_{Set^{op}}(\overline{F}(A),B) = Hom_{Set}(A,F(B))$$ which means in $Set$: $$Hom(B,F(A)) = Hom(A,F(B)).$$

Now take $B = \ast$ (a point) : $$F(A)\simeq Hom(\ast,F(A)) = Hom(A,F(\ast))$$ so you get a natural isomorphism $$F\simeq Hom(\bullet,F(\ast)).$$

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  • $\begingroup$ You use Yoneda in the fact that $F(A) \cong Hom(*,F(A))$ right? Thnx for the answer! $\endgroup$ – Rico May 25 '16 at 18:27
  • $\begingroup$ Oh nvm, ofc you use property of sets, map from * to F(A) can be seen as element of F(A). $\endgroup$ – Rico May 25 '16 at 20:03

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