The Stacks project

Proof. Let $U \to X$ be an étale morphism and $u \in U$. The geometric statement (A) reduces directly to the case where $U$ and $Y$ are affine schemes. Denote $x \in X$ and $y \in Y$ the images of $u$. Since $X \to Y$ is locally quasi-finite, and $U \to X$ is locally quasi-finite (see Morphisms, Lemma 29.36.6) we see that $U \to Y$ is locally quasi-finite (see Morphisms, Lemma 29.20.12). Moreover both $X \to Y$ and $U \to Y$ are separated. Thus More on Morphisms, Lemma 37.41.5 applies to both morphisms. This means we may pick an étale neighbourhood $(V, v) \to (Y, y)$ such that

and points $w \in W$, $w' \in W'$ such that

1. $W$, $R$ are open and closed in $X \times _ Y V$,

2. $W'$, $R'$ are open and closed in $U \times _ Y V$,

3. $W \to V$ and $W' \to V$ are finite,

4. $w$, $w'$ map to $v$,

5. $\kappa (v) \subset \kappa (w)$ and $\kappa (v) \subset \kappa (w')$ are purely inseparable, and

6. no other point of $W$ or $W'$ maps to $v$.

Here is a commutative diagram

After shrinking $V$ we may assume that $W'$ maps into $W$: just remove the image the inverse image of $R$ in $W'$; this is a closed set (as $W' \to V$ is finite) not containing $v$. Then $W' \to W$ is finite because both $W \to V$ and $W' \to V$ are finite. Hence $W' \to W$ is finite étale, and there is exactly one point in the fibre over $w$ with $\kappa (w) = \kappa (w')$. Hence $W' \to W$ is an isomorphism in an open neighbourhood $W^\circ $ of $w$, see Étale Morphisms, Lemma 41.14.2. Since $W \to V$ is finite the image of $W \setminus W^\circ $ is a closed subset $T$ of $V$ not containing $v$. Thus after replacing $V$ by $V \setminus T$ we may assume that $W' \to W$ is an isomorphism. Now the decomposition $X \times _ Y V = W \amalg R$ and the morphism $W \to U$ are as desired and we win. $\square$