Does the hom-tensor adjunction hold for $O_X$ modules also? With sheaf hom and sheaf tensor product, the statement would consist of a natural transformation $Hom_O (M \otimes_O N, K)\cong_{nat} Hom_O(M, Hom_O(N, K))$, $O = O_X$ is the structure sheaf and $M,N,K$ are sheaves of $O_X$ modules.

If true I think I can check this by hand, by describing sheaves in terms of compatible germs, defining this adjunction on the germs and checking that they glue together. It is easier in the quasi-coherent sheaves on a scheme case because then one can just work in an affine cover and the distinguished base.

I am little bit concerned that one might need a finite presentation hypothesis somewhere, in order for $Hom$ to localize well. ($Hom_{R[S^{-1}]}(M[S^{-1}], N[S^{-1}]) \cong Hom_R(M,N)[S^{-1}]$ needs $M$ finitely presented.)


Everything actually works for arbitrary ringed spaces and arbitrary sheaves of $\mathcal O_X$-modules, cf. e.g. Tag 01CM of the Stacks Project (where the proof of this lemma is omitted).

Firstly, you should be aware of three subtleties:

  • $M \otimes_{\mathcal O_X} N$ is the sheafification of the presheaf $$U \mapsto M(U) \otimes_{\mathcal O(U)} N(U).$$ To avoid confusion, I will write $M \odot_{\mathcal O_X} N$ for this presheaf tensor product.
  • The sheaf $\mathscr Hom_{\mathcal O_X}(M,N)$ is given on $U$ by $\operatorname{Hom}_{\mathcal O_U}(M|_U, N|_U)$, not $\operatorname{Hom}_{\mathcal O(U)}(M(U),N(U))$. This makes it slightly harder to define the 'obvious map', as we will see below.
  • For $M, N$ both quasicoherent, it is not in general true that $\mathscr Hom_{\mathcal O_X}(M, N)$ is quasicoherent (this is the remark you make on the bottom; it is enlightening to try to think of a counter-example). However, it does work when $M$ is coherent (and $X$ is Noetherian, or use the correct definition of coherent sheaves on an arbitrary scheme or even ringed space).

I will define an obvious (but not so obvious) isomorphism $$f \colon \mathscr Hom_{\mathcal O_X}(M, \mathscr Hom_{\mathcal O_X} (N, K)) \to \mathscr Hom_{\mathcal O_X}(M \otimes_{\mathcal O_X} N, K).$$ We have to construct compatible isomorphisms $$f_U \colon \operatorname{Hom}_{\mathcal O_U} (M|_U, \mathscr Hom_{\mathcal O_U} (N|_U, K|_U)) \to \operatorname{Hom}_{\mathcal O_U}((M \otimes_{\mathcal O_X} N)|_U, K|_U)$$ for all $U \subseteq X$. Let $U$ be fixed from now on. I will break down what both sides are.

Right hand side. Since sheafification commutes with restriction to opens, the right hand side is $$\operatorname{Hom}_{\mathcal O_U}(M|_U \otimes_{\mathcal O_U} N|_U, K_U) = \operatorname{Hom}_{\mathcal O_U}^{\operatorname{pre}} (M_U \odot_{\mathcal O_U} N|_U, K|_U).$$ An element of this is a compatible family of maps (for all $V \subseteq U$) $$\psi_V \colon M(V) \otimes_{\mathcal O(V)} N(V) \to K(V).$$ However, as pointed out by the OP in the comments, you cannot now use the classical tensor-hom adjunction to say that this is the same as giving compatible maps $$M(V) \to \operatorname{Hom}_{\mathcal O(V)}(N(V), K(V)).$$ See also the remark below.

Left hand side. On the other hand, the left hand side consists of giving a compatible system of maps $$\phi_V \colon M(V) \to \operatorname{Hom}_{\mathcal O_V}(N|_V, K|_V).$$ The map $f$. Thus, there is an obvious map from left to right given by taking global sections: given $\phi_V$ and $m \in M(V)$, the map $\phi_V(m)\colon N|_V \to K|_V$ induces a map $\chi_V(m)\colon N(V) \to K(V)$ by taking global sections. This gives $\chi_V \colon M(V) \to \operatorname{Hom}_{\mathcal O(V)}(N(V), K(V))$. Using the classical tensor-hom adjunction, this in turn gives a map $$\psi_V \colon M(V) \otimes_{\mathcal O(V)} N(V) \to K(V).$$ However, having a map $\chi_V(m) \colon N(V) \to K(V)$ does not give you a map $\phi_V(m) \colon N|_V \to K|_V$ in general. It works when $V$ is an affine scheme and $N$ is quasicoherent, but that is not satisfactory. Thus, defining an inverse is not quite so easy.

The inverse of $f$. This is where the compatibility of the various $\psi_V$ comes in. That is, given a system $\psi_V$ from the right hand side, for any $m \in M(V)$, we have to construct a morphism of sheaves $$\phi_V(m) \colon N|_V \to K|_V.$$ For any $W \subseteq V$, we define \begin{align*} \phi_V(m) \colon N(W) &\to K(W)\\ n &\mapsto \psi_W(m|_W \otimes n). \end{align*} One checks that these are compatible for varying $W$, so we get $\phi_V(m)$ as claimed. Then check that the $\phi_V$ are compatible for varying $V$, etc. $\square$

I also omitted the verifications that anything was $\mathcal O$-linear, or even that the maps defined above are actually inverses. All of this is relatively straightforward.

Remark. Given a commutative diagram $$\begin{array}{ccc} M(V) \otimes_{\mathcal O(V)} N(V) & \to & K(V) \\ \downarrow & & \downarrow \\ M(W) \otimes_{\mathcal O(W)} N(W) & \to &\ K(W), \end{array}$$ we do not get a commutative diagram $$\begin{array}{ccc} M(V) & \to & \operatorname{Hom}_{\mathcal O(V)}(N(V), K(V)) \\ \downarrow & & \downarrow \\ M(W) & \to &\ \operatorname{Hom}_{\mathcal O(W)}(N(W), K(W)).\end{array}$$ In fact, we don't even get the right vertical map, because of the contravariance of $\operatorname{Hom}$ is the first variable. Thus, we cannot simplify the description of the right hand side.

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    $\begingroup$ I think there must be something I am misunderstand about the classical adjunction, because I don't see why it sends a restriction compatible system of maps to another compatible system (the last line in Right Hand Side). There is something confusing about the covariance/contravariance of these functors - well anyway the compatible system diagram isn't induced by functoriality, so I'm not sure why a natural isomorphism will preserve it. Does my confusion make sense? Thank you for the otherwise very clear answer. $\endgroup$ – Lorenzo Najt Dec 29 '15 at 16:50
  • $\begingroup$ By 'compatibility', I do mean functoriality. Morphisms of sheaves are just morphisms of presheaves, i.e. of functors $\mathfrak{Top}_X^{\operatorname{op}} \to \underline{\operatorname{Ab}}$. The classical adjunction $\operatorname{Hom}_R(M \otimes_R N, K) = \operatorname{Hom}_R(M, \operatorname{Hom}_R(N, K))$ can be seen as an isomorphism of trifunctors (covariant in $K$ and contravariant in $M$ and $N$). $\endgroup$ – Remy Dec 29 '15 at 17:29
  • $\begingroup$ Ah, so it's the contravariance that's the problem. I have fixed the proof and added this as a remark. $\endgroup$ – Remy Dec 29 '15 at 18:50

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