Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I was reading about the internal hom functor for simplicial sets, and the construction is very "localized" (nothing to do with localization, just the english word). It seems like there should be a general construction for any presheaf category that would be similar to this. That is, an actual construction, not just the existence of the functor provided by the theorem that every Grothendieck topos is a Lawvere topos and therefore cartesian closed. Does such a construction exist, and if so, can you give a reference?

share|improve this question
add comment

1 Answer

up vote 6 down vote accepted

The formula for the internal hom between presheaves $F\colon C^{op}\to Set$ and $G\colon C^{op}\to Set$ can be derived from the Yoneda lemma. Given $c\in C$, we know that we must have $G^F(c) \cong Hom(y(c), G^F) \cong Hom(y(c) \times F, G)$ so we can simply define $G^F(c) = Hom(y(c) \times F, G)$, which is evidently a presheaf on $C$. The isomorphism $Hom(H,G^F)\cong Hom(H\times F, G)$ for non-representable $H$ then follows from the fact that every presheaf $H$ is canonically a colimit of representables, and $Hom(-,G^F)$ and $Hom(-\times F,G)$ both preserve colimits (the former by definition of colimits, and the latter by that and since limits and colimits in presheaf categories are computed pointwise and products in $Set$ preserve colimits).

This is Proposition I.6.1 in "Sheaves in geometry and logic."

share|improve this answer
    
Accepted and +1, but is there any way to obtain a more concrete description? –  Harry Gindi Jan 1 '10 at 22:28
2  
You can spell out the definition of homs in presheaf categories in terms of an end in Set, if you like: $G^F(c) = Hom(y(c)\times F, G) = \int_{c'} Hom(C(c,c')\times F(c'), G(c'))$. You could then invoke the construction of limits in Set, so that $G^F(c)$ is the set of tuples $(h_{c'})_{c'\in C}$, where $h_{c'}\colon C(c,c')\times F(c')\to G(c')$, such that for any $\gamma\colon c'\to c''$ in $C$ we have $G(\gamma) \circ h_{c'} = h_{c''} \circ (C(c,\gamma) \times F(\gamma))$. –  Mike Shulman Jan 1 '10 at 22:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.