Lemma 108.7.5. 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 $\pi $ induces a closed immersion $\mathrm{Hilb}_{X/B} \to \mathrm{Hilb}_{Y/B}$.
Proof. Since $\pi $ is a closed immersion, it is immediate that given a closed subscheme $Z \subset X_ T$, we can view $Z$ as a closed subscheme of $X_ T$. Thus we obtain the transformation $\mathrm{Hilb}_{X/B} \to \mathrm{Hilb}_{Y/B}$. This transformation is immediately seen to be a monomorphism. To prove that it is a closed immersion, you can use Lemma 108.5.8 for the map $\mathcal{O}_ Y \to \mathcal{O}_ X$ and the identifications $\mathrm{Hilb}_{X/B} = \mathrm{Quot}_{\mathcal{O}_ X/X/B}$, $\mathrm{Hilb}_{Y/B} = \mathrm{Quot}_{\mathcal{O}_ Y/Y/B}$ of Quot, Lemma 99.9.2. $\square$
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