Lemma 52.16.9. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Let $A \to A'$ be a flat ring map. Set $X' = \mathop{\mathrm{Spec}}(A')$, let $U' \subset X'$ be the inverse image of $U$, and denote $g : U' \to U$ the induced morphism. Set $(\mathcal{F}'_ n) = (g^*\mathcal{F}_ n)$, see Cohomology of Schemes, Lemma 30.23.9. If $(\mathcal{F}_ n)$ canonically extends to $X$, then $(\mathcal{F}'_ n)$ canonically extends to $X'$. Moreover, the extension found in Lemma 52.16.8 for $(\mathcal{F}_ n)$ pulls back to the extension for $(\mathcal{F}'_ n)$.

Proof. Let $f : X' \to X$ be the induced morphism. We have $H^0(U', \mathcal{F}'_ n) = H^0(U, \mathcal{F}_ n) \otimes _ A A'$ by flat base change, see Cohomology of Schemes, Lemma 30.5.2. Thus if $(\mathcal{G}_ n)$ in $\textit{Coh}(X, I\mathcal{O}_ X)$ is pro-isomorphic to $(\widetilde{H^0(U, \mathcal{F}_ n)})$, then $(f^*\mathcal{G}_ n)$ is pro-isomorphic to

$(f^*\widetilde{H^0(U, \mathcal{F}_ n)}) = (\widetilde{H^0(U, \mathcal{F}_ n) \otimes _ A A'}) = (\widetilde{H^0(U', \mathcal{F}'_ n)})$

This finishes the proof. $\square$

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