Lemma 108.4.6. Let $B$ be an algebraic space. Let $\pi : X \to Y$ be a closed immersion of algebraic spaces which are separated and of finite presentation over $B$. Then the morphism $\mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B}$ of Lemma 108.4.4 is a closed immersion.
Proof. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be the sheaf of ideals cutting out $X$ as a closed subspace of $Y$. Recall that $\pi _*$ induces an equivalence between the category of quasi-coherent $\mathcal{O}_ X$-modules and the category of quasi-coherent $\mathcal{O}_ Y$-modules annihilated by $\mathcal{I}$, see Morphisms of Spaces, Lemma 67.14.1. The same, mutatis mutandis, is true after base by $T \to B$ with $\mathcal{I}$ replaced by the ideal sheaf $\mathcal{I}_ T = \mathop{\mathrm{Im}}((Y_ T \to Y)^*\mathcal{I} \to \mathcal{O}_{Y_ T})$. Analyzing the proof of Lemma 108.4.4 we find that the essential image of $\mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B}$ is exactly the objects $\xi = (T \to B, \mathcal{F})$ where $\mathcal{F}$ is annihilated by $\mathcal{I}_ T$. In other words, $\xi $ is in the essential image if and only if the multiplication map
is zero and similarly after any further base change $T' \to T$. Note that
Hence the vanishing of the multiplication map on $T'$ is representable by a closed subspace of $T$ by Flatness on Spaces, Lemma 77.8.6. $\square$
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