The Stacks project

Proof. We have (1) $\Rightarrow$ (2) by Morphisms of Spaces, Lemmas 67.45.9, 67.45.8. We have (2) $\Rightarrow$ (3) by Morphisms of Spaces, Lemma 67.27.5. By definition (3) implies (4).

Assume (4). Since $f$ is universally closed it is quasi-compact (Morphisms of Spaces, Lemma 67.9.7). Pick a point $y$ of $|Y|$. We represent $y$ by a morphism $\mathop{\mathrm{Spec}}(k) \to Y$. Note that $|X_ k|$ is finite discrete as a quasi-compact discrete space. The map $|X_ k| \to |X|$ surjects onto the fibre of $|X| \to |Y|$ over $y$ (Properties of Spaces, Lemma 66.4.3). By Morphisms of Spaces, Lemma 67.34.8 we see that $X \to Y$ is quasi-finite at all the points of the fibre of $|X| \to |Y|$ over $y$. Choose an elementary étale neighbourhood $(U, u) \to (Y, y)$ and decomposition $X_ U = V \amalg W$ as in Lemma 76.33.1 adapted to all the points of $|X|$ lying over $y$. Note that $W_ u = \emptyset$ because we used all the points in the fibre of $|X| \to |Y|$ over $y$. Since $f$ is universally closed we see that the image of $|W|$ in $|U|$ is a closed set not containing $u$. After shrinking $U$ we may assume that $W = \emptyset$. In other words we see that $X_ U = V$ is finite over $U$. Since $y \in |Y|$ was arbitrary this means there exists a family $\{ U_ i \to Y\}$ of étale morphisms whose images cover $Y$ such that the base changes $X_{U_ i} \to U_ i$ are finite. We conclude that $f$ is finite by Morphisms of Spaces, Lemma 67.45.3. $\square$