[Math] Internal hom of sheaves

ag.algebraic-geometryct.category-theorysheaf-theory

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" $U\mapsto \hom_{PreShv}(X\times \hom_{Sets}(-,U),Y)$.

Best Answer

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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