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I have been studying category theory and have been exploring hom functors. I've come across an interesting question and after spending several hours thinking about it, haven't gotten anywhere.

Let $X$ be a category. Then let $F:X\to Set^{X^{op}}$ denote the covariant functor taking each $A\in X$ to the contravariant functor $\hom(-, A)$. I want to prove the $F$ is a continuous functor. I suspect that this should follow easily from the continuity of $\hom(A,-)$, but haven't been able to move very far in that direction, possibly due to my own inexperience with categories. Any help would be greatly appreciated.

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    $\begingroup$ My eyes switched the letters in my first read. I read Mom Functor. $\endgroup$ – Billy Rubina Dec 18 '14 at 23:38
  • $\begingroup$ The key statement is that limits in functor categories are computed pointwise. $\endgroup$ – Qiaochu Yuan Dec 19 '14 at 9:26
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Let $A_i$ be the objects of a diagram and $L$ the a limit of the diagram. Pushing through your functor gives the diagram $\hom(-, A_i)$ and you want to show that $\hom(-, L)$ is it's limit. So suppose you have a contravariant functor $F$ and natural transformations $F \to \hom(-, A_i)$. I'm guessing you need to know how to define $F \to \hom(-, L)$.

Well, take an object $C$, then $F(C)$ is a set and you need a map $F(C) \to \hom(C, L)$. Pick $c \in F(C)$. From $F(C) \to \hom(C, A_i)$ you get morphisms $C \to A_i$. You'll have to show that these maps commute with the maps in the diagram, but then you get $C \to L$ because $L$ was a limit. So let that be the image of $c$ in $F(C) \to (C, L)$. Now you need to show that this is natural (so that $F$ is a well defined functor) and that the choice is unique (so that you have universality).

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