During the course of trying to solve Vakil Exercise 13.1.F I decided I wanted to prove the following: Suppose $X$ is a ringed space with structure sheaf $\mathscr{O}_X$. Suppose $\mathscr{F}, \mathscr{E}$ are $\mathscr{O}_X$-modules and $\mathscr{E}$ is locally free of finite rank. Then

$$ \mathscr{F} \otimes \mathscr{E}^\vee \cong \mathcal{Hom}(\mathscr{E}, \mathscr{F}) $$

Vakil 13.7.B has us prove this in less generality, but it also mentions that the result above should hold. This should follow from the corresponding fact for modules. I couldn't find this result on the Stacks Project or elsewhere online. How do you prove it?

(I am almost certain Vakil doesn't intend 13.1.F to be solved the way I am trying to solve it given the fact that he has us prove a less general version of the result later.)


3 Answers 3


I am going to expalin how I think of giving very formal and organised proofs for these kind of results. There are two steps:

  1. Show that there exists a morphism from left to right or the other way using universal properties
  2. Show that this morphism is an isomorphism on stalks or that it is an isomorphism on a covering

This "method" can be repeated on other proofs, for example if one tries to show that $\mathscr E^{\vee \, \vee} = \mathscr E$ for a locally free sheaf $\mathscr E$.

On the left you have a tensor product. Tensor products $\mathscr F \otimes \mathscr G$ are defined as sheafifications of some presheaf, which I will denote $\mathscr F \otimes_{pSh} \mathscr G$ so they are a little bit difficult to deal with. However, remember the universal property of sheafification:

$$\text{Hom}(\mathscr F^+, \mathscr G) = \text{Hom}(\mathscr F , \mathscr G)$$

which, in more down to earth words says that to give a morphism from the sheafification of a presheaf $\mathcal F$ to another sheaf $\mathcal G$ is the same as giving one from $\mathcal F$ to $\mathcal G$. Therefore, we have the task of defning a morphism $\varphi : \mathscr{F} \otimes_{psH} \mathscr{E}^\vee \to \mathcal{Hom}(\mathscr{E}, \mathscr{F})$. Note that we know the value of both presheaves on any open set (because we are now dealing with the presheaf tensor product), so we just need to define a natural $\mathcal O(U)$-linear map

$$\mathscr{F}(U) \otimes_{\mathcal O(U)} \text{Hom}_{\mathcal O_U}(\mathscr{E}_U, \mathcal O_U) \to \text{Hom}_{\mathcal O_U}(\mathscr{E}_U, \mathscr{F}_U)$$

Which is very natural:

$$ (*) \qquad s \otimes f \longmapsto \, (t \mapsto f(t) s) \in \text{Hom}_{\mathcal O_U}(\mathscr{E}_U, \mathscr{F}_U)$$

These maps clearly constitute a morphism of sheaves (they look so natural, its is clear that they commute with restrictions) and therefore putting everything together we obtain the desired morphism $\hat\varphi:\mathscr{F} \otimes \mathscr{E}^\vee \to \mathcal{Hom}(\mathscr{E}, \mathscr{F})$, corresponing to $\varphi$

Before the edit, I was going to prove that the map is an isomorphism on stalks, but one needs to prove first that $(\mathcal{Hom}(\mathscr{E}, \mathscr{F}))_p = \text{Hom}_{\mathcal O_p}(\mathscr{E}_p, \mathscr{F}_p)$ and this is not trivial. However, it is enough to show that $\varphi$ is an isomorphism on a covering of $X$. Since $\mathscr{E}$ is locally free, for the remainder of the proof it is enough to assume that $\mathscr{E}=\mathcal O_X^{\oplus n}$. In this case, we can factor $\varphi$ as

$$\mathscr{F} \otimes_{pSh} \mathscr{E} ^\vee \stackrel{\cong}{\longrightarrow} \left( \mathscr{F} \otimes_{pSh} \mathcal O_X ^\vee\right)^{\oplus n} \longrightarrow \left( \mathcal{Hom} (\mathcal O _X ,\mathscr F)\right)^{\otimes n} \stackrel{\cong}{\longrightarrow} \mathcal{Hom}(\mathscr{E}, \mathscr{F}),$$

where the left and right arrows are isomorphisms, so one reduces to the case $n=1$. In this case, the isomorphism can be checked on any open subset $V$, and using that

$$\mathcal{Hom}(\mathcal O_X , \mathscr F) (V) = \text{Hom}_{\mathcal O_X(V)}(\mathcal O_X(V) , \mathscr F (V))$$ $$\mathcal{Hom}(\mathcal O_X , \mathcal O_X) (V) = \text{Hom}_{\mathcal O_X(V)}(\mathcal O_X(V) , \mathcal O_X(V))$$

and the notation $M=\mathscr F(V)$, $R= \mathcal O_X(V)$, it is equivalent to checking that the $R$-module homomorphism

$$M \otimes_R \text{Hom}_R(R, R) \to \text{Hom}_R(R, M)$$

Given in a similar fashion as in $(*)$ is an isomorphism

  • 1
    $\begingroup$ There is a problem with this proof: You need to define an $\mathscr{O}_X$-linear map $\mathscr{F}(U) \otimes \hom(\mathscr{E}|_U, \mathscr{O}_X|_U) \to \hom(\mathscr{E}|_U, \mathscr{F}|_U)$, not a map $\mathscr{F}(U) \otimes \hom(\mathscr{E}(U), \mathscr{O}_X(U)) \to \hom(\mathscr{E}(U), \mathscr{F}(U))$. $\endgroup$
    – Josiah
    Jan 27 at 3:21
  • 1
    $\begingroup$ Also, while I don't give a full proof either, you do need to prove that $\mathcal{Hom}(\mathscr{E}, \mathscr{F})_p \cong \hom(\mathscr{E}_p, \mathscr{F}_p)$ for locally free finite rank $\mathscr{E}$. $\endgroup$
    – Josiah
    Jan 27 at 3:27
  • $\begingroup$ You are right. The two issues have been solved after the edit. $\endgroup$ Jan 27 at 15:05
  • $\begingroup$ $(*)$ is not well defined as stated. An element of $\hom(\mathscr{E}|_U, \mathscr{F}|_U)$ is not a morphism of sets but rather a morphism of sheaves, i.e. a natural transformation. You need to give a definition of the morphism on each open $V \subseteq U$ and verify naturality. $\endgroup$
    – Josiah
    Jan 27 at 23:51
  • $\begingroup$ @Josiah I am making an abuse of notation but it makes sense. If $f : \mathscr E_U \to \mathcal O_U$ (ie $f$ is a collection of maps $f_V$) and $s \in \mathscr F(U)$, the image of $s \otimes f$ is the morphism such that, on each open $V\subset U$ is given by $t \mapsto f_V(t) \cdot s_V$, where $s_V$ is the restriction of $s$ to $V$. You can check that it constitutes a morphism of sheaves but it is a trivial computation $\endgroup$ Jan 28 at 0:38

I'm also a new learner in algebraic geometry. I'm so sorry if my understanding on this is incorrect.

Actually the dual sheaf $\mathcal{E}^{\vee}$ is the dual object in the symmetric monoidal category of $\mathcal{O}_X$-modules when $\mathcal{E}$ is locally a direct summand of a finite free $\mathcal{O}_X$-module. (See Stacks Project Tag 0FNU and 0FNV at https://stacks.math.columbia.edu/tag/0FNU) Note that this condition on $\mathcal{E}$ is more general than Vakil's one.

Then by following results on the dual object in symmetric monoidal category:

  1. In symmetric monoidal categories, the left dual object and the right dual object are isomorphic. (Though I haven't found this on Stacks Project unfortunately.)
  2. Tag 0FFQ of Stacks Project: https://stacks.math.columbia.edu/tag/0FFQ ,

we see that $$ \mathrm{Hom}_{\mathcal{O}_X}(\mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{E}, \mathcal{G}) = \mathrm{Hom}_{\mathcal{O}_X} (\mathcal{F}, \mathcal{G} \otimes_{\mathcal{O}_X} \mathcal{E}^{\vee}). $$ Then recall the adjointness of $-\otimes_{\mathcal{O}_X} \mathcal{E}$ and $\mathcal{H}om_{\mathcal{O}_X}(\mathcal{E}, -)$, we see that $$ \mathrm{Hom}_{\mathcal{O}_X}(\mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{E}, \mathcal{G})= \mathrm{Hom}_{\mathcal{O}_X}(\mathcal{F}, \mathcal{H}om_{\mathcal{O}_X}(\mathcal{E}, \mathcal{G})). $$ Comparing this two, we see that $\mathcal{G} \otimes_{\mathcal{O}_X} \mathcal{E}^{\vee} = \mathcal{H}om_{\mathcal{O}_X}(\mathcal{E}, \mathcal{G})$ by (the dual version of) Yoneda's lemma.

P.S. In the above discussions, all $=$ conneting "$\mathrm{Hom}$" should be natural isomorphisms.

Sorry for possible mistakes and misunderstandings.

  • $\begingroup$ Unless I misread, this approach basically proves 13.1.F in the process! Thanks! $\endgroup$
    – Josiah
    Jan 27 at 2:36
  • $\begingroup$ I haven't read Vakil's note before. Actually his hint there (considering the case when $\mathcal{E}$ is free) is essentially the method proving that $\mathcal{E}^{\vee}$ is indeed the dual object in $(\mathcal{O}_X-\mathsf{Mod}, \otimes)$ skectched in Stacks Project Tag 0FNV. Actually I first read this result on Hartshorne Exercise II.5.1. Glad that the post is helpful. :) $\endgroup$
    – Hetong Xu
    Jan 27 at 2:42

$$ \DeclareMathOperator{\Esh}{\mathscr{E}} \DeclareMathOperator{\Fsh}{\mathscr{F}} \DeclareMathOperator{\Hom}{\mathcal{Hom}} \DeclareMathOperator{\Osh}{\mathscr{O}} $$ I believe the following should suffice. We will need the fact that $\Hom(\Esh, \Fsh)_p \cong \hom(\Esh_p, \Fsh_p)$ (for locally free finite rank $\Esh$). This follows from the fact that $\Hom$ commutes with finite direct sums and direct sums commute with taking stalks.

We begin by defining the morphism of presheaves $(\Fsh \otimes \Esh^\vee)^\text{pre} \to \Hom(\Esh, \Fsh)$. Let $U$ be an arbitrary open. Then we define $\phi_U: \Fsh(U) \otimes \Esh^\vee(U) \to \hom(\Esh|_U, \Fsh|_U)$ by sending $f \otimes e$ to the morphism of sheaves $g$ given by $g_U(x)=e_U(x)f|_U$. It is easily checked that the following diagram commutes:

Compatibility condition

Hence we have defined the desired morphism. By universal property of sheafification, we get a morphism $\Fsh \otimes \Esh^\vee \to \Hom(\Esh, \Fsh)$. Now, since $\Esh$ is finite locally free, we may localize at $p$ to check isomorphism. This gives us a morphism of modules $\Fsh_p \otimes \Esh^\vee_p \to \hom(\Osh_{X,p}^n, \Fsh_p)$ such that $f \otimes e \mapsto e(x)f$. By general module theory, this is an isomorphism, hence we have the desired global isomorphism.

  • $\begingroup$ Your first line is wrong in general and can be better approached as detailed in this question. Further, one may embed commutative diagrams without needing any rep at all, see meta. $\endgroup$
    – KReiser
    Jan 27 at 1:36
  • $\begingroup$ We are assuming $\mathscr{E}$ to be locally free of finite rank, so it holds. $\endgroup$
    – Josiah
    Jan 27 at 1:48

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