Lemma 108.7.4. Let $B$ be an algebraic space. Let $\pi : X \to Y$ be an open immersion of algebraic spaces which are separated and of finite presentation over $B$. Then $\pi $ induces an open immersion $\mathrm{Hilb}_{X/B} \to \mathrm{Hilb}_{Y/B}$.
Proof. Omitted. Hint: If $Z \subset X_ T$ is a closed subscheme which is proper over $T$, then $Z$ is also closed in $Y_ T$. Thus we obtain the transformation $\mathrm{Hilb}_{X/B} \to \mathrm{Hilb}_{Y/B}$. If $Z \subset Y_ T$ is an element of $\mathrm{Hilb}_{Y/B}(T)$ and for $t \in T$ we have $|Z_ t| \subset |X_ t|$, then the same is true for $t' \in T$ in a neighbourhood of $t$. $\square$
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