Lemma 56.5.11. In Lemma 56.5.7 let $F$ and $\mathcal{K}$ correspond. If $X$ is separated and flat over $R$, then there is a surjection $\mathcal{O}_ X \boxtimes F(\mathcal{O}_ X) \to \mathcal{K}$.

**Proof.**
Let $\Delta : X \to X \times _ R X$ be the diagonal morphism and set $\mathcal{O}_\Delta = \Delta _*\mathcal{O}_ X$. Since $\Delta $ is a closed immersion have a short exact sequence

Since $\mathcal{K}$ is flat over $X$, the pullback $\text{pr}_{23}^*\mathcal{K}$ to $X \times _ R X \times _ R Y$ is flat over $X \times _ R X$. We obtain a short exact sequence

on $X \times _ R X \times _ R Y$, see Modules, Lemma 17.20.4. Thus, by Lemma 56.5.10 we obtain a surjection

By flat base change (Cohomology of Schemes, Lemma 30.5.2) the source of this arrow is equal to $\text{pr}_2^*\text{pr}_{2, *}\mathcal{K} = \mathcal{O}_ X \boxtimes F(\mathcal{O}_ X)$. On the other hand the target is equal to

which finishes the proof. The first equality holds for example by Cohomology, Lemma 20.51.4 and the fact that $\text{pr}_{12}^*\mathcal{O}_\Delta = (\Delta \times \text{id}_ Y)_*\mathcal{O}_{X \times _ R Y}$. $\square$

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