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 (56.5.1.1).

Proof. Let $\mathcal{K}$ be an object of $\mathit{QCoh}(\mathcal{O}_{X \times _ R Y})$ and $F_\mathcal {K}$ the functor (56.5.1.1). 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 (56.5.1.1). 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. $\square$

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