Lemma 69.17.1. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Assume $f$ is finite, surjective and $X$ locally Noetherian. Let $i : Z \to X$ be a closed immersion. Denote $i' : Z' \to Y$ the inverse image of $Z$ (Morphisms of Spaces, Section 67.13) and $f' : Z' \to Z$ the induced morphism. Then $\mathcal{G} = f'_*\mathcal{O}_{Z'}$ is a coherent $\mathcal{O}_ Z$-module whose support is $Z$.
Proof. Observe that $f'$ is the base change of $f$ and hence is finite and surjective by Morphisms of Spaces, Lemmas 67.5.5 and 67.45.5. Note that $Y$, $Z$, and $Z'$ are locally Noetherian by Morphisms of Spaces, Lemma 67.23.5 (and the fact that closed immersions and finite morphisms are of finite type). By Lemma 69.12.9 we see that $\mathcal{G}$ is a coherent $\mathcal{O}_ Z$-module. The support of $\mathcal{G}$ is closed in $|Z|$, see Morphisms of Spaces, Lemma 67.15.2. Hence if the support of $\mathcal{G}$ is not equal to $|Z|$, then after replacing $X$ by an open subspace we may assume $\mathcal{G} = 0$ but $Z \not= \emptyset $. This would mean that $f'_*\mathcal{O}_{Z'} = 0$. In particular the section $1 \in \Gamma (Z', \mathcal{O}_{Z'}) = \Gamma (Z, f'_*\mathcal{O}_{Z'})$ would be zero which would imply $Z' = \emptyset $ is the empty algebraic space. This is impossible as $Z' \to Z$ is surjective. $\square$
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