Lemma 81.10.5. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $Z \subset X$ be a closed subspace. Assume $f^{-1}Z \to Z$ is an isomorphism and that $f$ is flat in every point of $f^{-1}Z$. For any $Q$ in $D_\mathit{QCoh}(\mathcal{O}_ Y)$ supported on $|f^{-1}Z|$ we have $Lf^*Rf_*Q = Q$.
Proof. We show the canonical map $Lf^*Rf_*Q \to Q$ is an isomorphism by checking on stalks at $\overline{y}$. If $\overline{y}$ is not in $f^{-1}Z$, then both sides are zero and the result is true. Assume the image $\overline{x}$ of $\overline{y}$ is in $Z$. By Lemma 81.10.1 we have $Rf_*Q_{\overline{x}} = Q_{\overline{y}}$ and since $f$ is flat at $\overline{y}$ we see that
\[ (Lf^*Rf_*Q)_{\overline{y}} = (Rf_*Q)_{\overline{x}} \otimes _{\mathcal{O}_{X, \overline{x}}} \mathcal{O}_{Y, \overline{y}} = Q_{\overline{y}} \otimes _{\mathcal{O}_{X, \overline{x}}} \mathcal{O}_{Y, \overline{y}} \]
Thus we have to check that the canonical map
\[ Q_{\overline{y}} \otimes _{\mathcal{O}_{X, \overline{x}}} \mathcal{O}_{Y, \overline{y}} \longrightarrow Q_{\overline{y}} \]
is an isomorphism in the derived category. Let $I_{\overline{x}} \subset \mathcal{O}_{X, \overline{x}}$ be the stalk of the ideal sheaf defining $Z$. Since $Z \to X$ is locally of finite presentation this ideal is finitely generated and the cohomology groups of $Q_{\overline{y}}$ are $I_{\overline{y}} = I_{\overline{x}}\mathcal{O}_{Y, \overline{y}}$-power torsion by Lemma 81.10.4 applied to $Q$ on $Y$. It follows that they are also $I_{\overline{x}}$-power torsion. The ring map $\mathcal{O}_{X, \overline{x}} \to \mathcal{O}_{Y, \overline{y}}$ is flat and induces an isomorphism after dividing by $I_{\overline{x}}$ and $I_{\overline{y}}$ because we assumed that $f^{-1}Z \to Z$ is an isomorphism. Hence we see that the cohomology modules of $Q_{\overline{y}} \otimes _{\mathcal{O}_{X, \overline{x}}} \mathcal{O}_{Y, \overline{y}}$ are equal to the cohomology modules of $Q_{\overline{y}}$ by More on Algebra, Lemma 15.89.2 which finishes the proof. $\square$
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