The Stacks project

Lemma 56.5.2. Let $R$ be a ring. Let $X$ and $Y$ be schemes over $R$ with $X$ affine. There is an equivalence of categories between

  1. the category of $R$-linear functors $F : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ Y)$ which are right exact and commute with arbitrary direct sums, and

  2. the category $\mathit{QCoh}(\mathcal{O}_{X \times _ R Y})$

given by sending $\mathcal{K}$ to the functor $F$ in (

Proof. Let $\mathcal{K}$ be an object of $\mathit{QCoh}(\mathcal{O}_{X \times _ R Y})$ and $F_\mathcal {K}$ the functor ( By the discussion in Example 56.5.1 we already know that $F$ is $R$-linear and commutes with arbitrary direct sums. Since $\text{pr}_2 : X \times _ R Y \to Y$ is affine (Morphisms, Lemma 29.11.8) the functor $\text{pr}_{2, *}$ is exact, see Cohomology of Schemes, Lemma 30.2.3. Hence $F$ is right exact as well, in other words $F$ is as in (1).

Let $F$ be as in (1). Say $X = \mathop{\mathrm{Spec}}(A)$. Consider the quasi-coherent $\mathcal{O}_ Y$-module $\mathcal{G} = F(\mathcal{O}_ X)$. The functor $F$ induces an $R$-linear map $A \to \text{End}_{\mathcal{O}_ Y}(\mathcal{G})$, $a \mapsto F(a \cdot \text{id})$. Thus $\mathcal{G}$ is a sheaf of modules over

\[ A \otimes _ R \mathcal{O}_ Y = \text{pr}_{2, *}\mathcal{O}_{X \times _ R Y} \]

By Morphisms, Lemma 29.11.6 we find that there is a unique quasi-coherent module $\mathcal{K}$ on $X \times _ R Y$ such that $F(\mathcal{O}_ X) = \mathcal{G} = \text{pr}_{2, *}\mathcal{K}$ compatible with action of $A$ and $\mathcal{O}_ Y$. Denote $F_\mathcal {K}$ the functor given by ( There is an equivalence $\text{Mod}_ A \to \mathit{QCoh}(\mathcal{O}_ X)$ sending $A$ to $\mathcal{O}_ X$, see Schemes, Lemma 26.7.5. Hence we find an isomorphism $F \cong F_\mathcal {K}$ by Lemma 56.2.6 because we have an isomorphism $F(\mathcal{O}_ X) \cong F_\mathcal {K}(\mathcal{O}_ X)$ compatible with $A$-action by construction.

This shows that the functor sending $\mathcal{K}$ to $F_\mathcal {K}$ is essentially surjective. We omit the verification of fully faithfulness. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FZD. Beware of the difference between the letter 'O' and the digit '0'.