The Stacks project

Proof. Let $X \to Y' \to Y$ be as in the conclusion of Lemma 76.34.2. By Limits of Spaces, Lemma 70.9.7 we can write $\nu _*\mathcal{O}_{Y'} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{A}_ i$ as a directed colimit of finite quasi-coherent $\mathcal{O}_ X$-algebras $\mathcal{A}_ i \subset \nu _*\mathcal{O}_{Y'}$. Then $\pi _ i : T_ i = \underline{\mathop{\mathrm{Spec}}}_ Y(\mathcal{A}_ i) \to Y$ is a finite morphism for each $i$. Note that the transition morphisms $T_{i'} \to T_ i$ are affine and that $Y' = \mathop{\mathrm{lim}}\nolimits T_ i$.

By Limits of Spaces, Lemma 70.5.7 there exists an $i$ and a quasi-compact open $U_ i \subset T_ i$ whose inverse image in $Y'$ equals $f'(X)$. For $i' \geq i$ let $U_{i'}$ be the inverse image of $U_ i$ in $T_{i'}$. Then $X \cong f'(X) = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_{i'}$, see Limits of Spaces, Lemma 70.4.1. By Limits of Spaces, Lemma 70.5.12 we see that $X \to U_{i'}$ is a closed immersion for some $i' \geq i$. (In fact $X \cong U_{i'}$ for sufficiently large $i'$ but we don't need this.) Hence $X \to T_{i'}$ is an immersion. By Morphisms of Spaces, Lemma 67.12.6 we can factor this as $X \to T \to T_{i'}$ where the first arrow is an open immersion and the second a closed immersion. Thus we win. $\square$