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
0 \to \mathcal{I} \to \mathcal{O}_{X \times _ R X} \to \mathcal{O}_\Delta \to 0
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
0 \to \text{pr}_{12}^*\mathcal{I} \otimes \text{pr}_{23}^*\mathcal{K} \to \text{pr}_{23}^*\mathcal{K} \to \text{pr}_{12}^*\mathcal{O}_\Delta \otimes \text{pr}_{23}^*\mathcal{K} \to 0
on X \times _ R X \times _ R Y, see Modules, Lemma 17.20.4. Thus, by Lemma 56.5.10 we obtain a surjection
\text{pr}_{13, *}(\text{pr}_{23}^*\mathcal{K}) \to \text{pr}_{13, *}( \text{pr}_{12}^*\mathcal{O}_\Delta \otimes \text{pr}_{23}^*\mathcal{K})
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
\text{pr}_{13, *}( \text{pr}_{12}^*\mathcal{O}_\Delta \otimes \text{pr}_{23}^*\mathcal{K}) = \text{pr}_{13, *} (\Delta \times \text{id}_ Y)_* \mathcal{K} = \mathcal{K}
which finishes the proof. The first equality holds for example by Cohomology, Lemma 20.54.4 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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