The Stacks project

Proof. Recall that a finite type morphism is quasi-finite if and only if it has relative dimension $0$, see Morphisms, Lemma 29.29.5. By Lemma 37.34.9 applied with $d = 0$ we reduce to the case where $X$ and $Y$ are Noetherian. We may choose an open immersion $X \to X'$ such that $X' \to Y$ is finite, see Algebra, Lemma 10.123.14. Note that if we have $Y' \to Y$ and $U'$ as in (2), then

is open immersion between schemes finite over $Y'$ and hence is closed as well. We conclude that $U'$ corresponds to an idempotent in

whose corresponding open and closed subset is contained in the open $Y' \times _ Y X$. Let $Y'_{univ} \to Y$ and idempotent

be the pair constructed in Lemma 37.76.7 for the ring map $\Gamma (Y, \mathcal{O}_ Y) \to \Gamma (X', \mathcal{O}_{X'})$ (here we use that $Y$ is Noetherian to see that $X'$ is of finite presentation over $Y$). Let $U'_{univ} \subset Y'_{univ} \times _ Y X'$ be the corresponding open and closed subscheme. Then we see that

is a closed subset of $U'_{univ}$ and hence has closed image $T \subset Y'_{univ}$. If we set $Y_{univ} = Y'_{univ} \setminus T$ and $U_{univ}$ the restriction of $U'_{univ}$ to $Y_{univ} \times _ Y X$, then we see that the lemma is true. $\square$