3

3

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?

flag

1 Answer

6

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."

link|flag
Accepted and +1, but is there any way to obtain a more concrete description? – Harry Gindi Jan 1 2010 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 2010 at 22:38

Your Answer

Get an OpenID
or

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