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Consider a topos, i.e. the category $Shv$ of sheaves on a Grothendieck site $T$ with values in abelian groups. The category $Shv$ is symmetric monoidal with $\otimes$, the tensor product in every degree. It has also an internal hom (meaning that it is closed symmetric monoidal). There are two "definitions" of this internal hom in the literature which I can't bring together (it can also be that they are different and I have missed the point). Fix objects $X$ and $Y$ of $Shv$. The first definition of the internal hom is the presheaf $$ U\mapsto \hom_{Shv}(X|_U, Y|_U) $$

which turns out to be sheaf. The second one is the sheafification of the presheaf

$$ U\mapsto \hom_{PreShv}(X\times \hom(-,U),Y) $$

Do they coincide? Perhaps I am totally blind so I apologize in either way.

An edit: Perhaps I was too hasty and thoughtless when I formulated the question. So let's look at set-valued sheaves. A third definition is the sheafification of the presheaf $$ U\mapsto \hom_{Sets}(X(U),Y(U)). $$ I do not even see (after playing around with Yoneda's lemma) why this coincides with the "second definition" $U\mapsto \hom_{PreShv}(X\times \hom_{Sets}(-,U),Y)$.

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I don't understand the notation for the second definition. – Andrea Ferretti Mar 16 '10 at 11:48
I made an edit. Perhaps it is better now – roger123 Mar 16 '10 at 13:10
The third definition you suggest (in the edit) doesn't work: that formula does not define a presheaf; there is no way to define the restriction maps. – Omar Antolín-Camarena Mar 16 '10 at 15:54
You are right, Omar. But this is strange: Look at Shafarevic, Basic Algebraic Geometry 2, VI.3.2. Example 2 (page 88). "the dual sheaf ... is the sheafication of the presheaf $G(U)=Hom(F(U),O_X(U))$". – roger123 Mar 17 '10 at 11:48
The "second definition" is from "Sheaves in geometry and logic" "Mac Lane, Moerdijk" I. Proposition 1. and page 136. I don't remember where the first definition is from. – roger123 Mar 18 '10 at 11:41

Well, here is a partial answer. The category of abelian group-valued sheaves is not a topos, the category of set-valued sheaves is. And I think you should look at set-valued presheaves, at least $hom(-,U)$ is one:

When the site comes from a topological space, you can see as follows that your two definitions coincide: When you insert into the $hom$ in your second expression an open set $V$, you either get $hom(V,U)=$the one-element-set containing only the inclusion, if $V \subseteq U$, or $hom(V,U)=$empty set, if not, so taking a product with the set-valued presheaf $hom(-,U)$ either leaves X as it is or "deletes" it (i.e. transforms it into the (presheaf with value the) empty set).

So a natural transformation in $hom_{PreShv}(X \times hom(-,U), Y)$ is given just by its components for all $V \subseteq U$ and for other $V$ it extends to the unique map from the initial object into Y. This is the same as a natural transformation of restricted presheaves.

Edit: It now occurred to me that you were probably actually speaking of the sheaves of group homomorphisms, not just any natural transformations. You get that, when you apply the "free group"-functor to the set-valued sheaf "hom(-,U)" and take tensor product instead of product. The free group over the empty set ist the trivial group, thus the initial object in groups and tensoring with it trivializes $X$. Tensoring with the free group over one element doesn't change anything, so the same things happen as in the set case...

For more general sites, the expression $X| _U$ probably means that you look at $X$ as a functor defined on the slice category given by maps on your site into its object $U$. But you can also see $X$ as a sheaf not just on the site, but on all of its ambient sheaf category (as a hom-functor, and where you give to topos an appropriate topology - you sometimes do this in topos theory). Then looking at restricted sheaves is the same as passing to the slice category $Shv / hom(-,U)$, and $X \times hom(-,U)$ is the image of $X$ under the canonical functor $Shv \rightarrow Shv/hom(-,U)$ (which is given by taking product with $hom(-,U)$), so it actually $is$ $X| _U$. This still doesn't make it clear to me why the two functors in question coincide, but I guess it may be a way to look at it to find it out...

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I believe I have the answer in the setting of sheaves of sets.

Let us first do this for presheaves, $Set^{C^{op}}$. This category is Cartesian-closed. This can be seen by setting $Y^X(U):=Hom(X \times U, Y)$, where I have identified $U$ with the representable presheaf $Hom(\cdot,U)$. This is easy to verify. It suffices to show that $Y^X$ agrees with the presheaf $U \mapsto Hom(X|_U,Y|_U)$.

As Peter pointed out, we have the functor $l_U:Set^{C^{op}} \to Set^{C^{op}}/U$ which sends a presheaf $X \mapsto \left(X \times U \to U\right)$, which is right adjoint to the corresponding forgetful functor $Set^{C^{op}}/U \to Set^{C^{op}}$. Here, $X|_U:=l_U(X)$. To explain the notation, note that we have an equivalence of categories $Set^{C^{op}}/U \cong Set^{\left(C/U\right)^{op}}$, so we can think of $l_U$ as restricting $X$ to a presheaf over the slice category $C/U$. Now, given $X$ and $Y$ presheaves on $C$, $Hom(l_U(X),l_U(Y))\cong Hom(X \times U, Y)$ since $l_U$ is a right adjoint to the forgetful functor and the forgetful functor applied to $l_U(X)$ is simply $X \times U$. Hence, we see that $U \mapsto Hom(X|_U,Y|_U)$ agrees with the functor $U \mapsto Hom(X \times U, Y)$.

I claim the same works for a Grothendieck topos:

For this, it suffices to prove that the functor $U \mapsto Hom(X \times U, Y)$ is a sheaf whenever $Y$ is. Let $\left(s_i:U_i \to U\right)_i$ be a cover of the object $U$. Note that $\left(s_i \times id:U_i\times X \to U\times X\right)_i$ is a cover of $U\times X$.

So $\varprojlim \left( \prod \limits_i Y^X(U_i) \rightrightarrows \prod \limits_{i,j} Y^X(U_i\times_U U_j)\right)\cong \varprojlim \left( \prod \limits_i Hom(U_i\times X,Y) \rightrightarrows \prod \limits_{i,j} Hom(U_i\times_U U_j \times X,Y)\right)$

and this is in turn:

$\varprojlim \left( \prod \limits_i Hom(U_i\times X,Y) \rightrightarrows \prod \limits_{i,j} Hom(\left(U_i\times X\right)\times_{U\times X} \left(U_j \times X\right),Y)\right) \cong Hom(U \times X, Y)$

since $Y$ is a sheaf.

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The Latex shows up fine when I preview it but it's not showing up when I post... strange. – David Carchedi Mar 18 '10 at 3:42
is this a latex problem that is magically solved by surround everything by back ticks ` ? – David Steinberg May 14 '10 at 20:09

The answer is also discussed at nLab: closed monoidal structure on presheaves

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