Lemma 37.74.2. Let f : X \to Y be a morphism of schemes. If
f is locally quasi-finite,
Y is geometrically unibranch and locally Noetherian, and
every irreducible component of X dominates an irreducible component of Y,
then f is universally open.
Lemma 37.74.2. Let f : X \to Y be a morphism of schemes. If
f is locally quasi-finite,
Y is geometrically unibranch and locally Noetherian, and
every irreducible component of X dominates an irreducible component of Y,
then f is universally open.
Proof. For any n the scheme \mathbf{A}^ n \times Y is geometrically unibranch by Lemma 37.36.4 and Properties, Lemma 28.15.6. Hence the hypotheses of the lemma hold for the morphisms \mathbf{A}^ n \times X \to \mathbf{A}^ n \times Y for all n. By Lemma 37.74.1 it suffices to prove f is open. By Morphisms, Lemma 29.23.2 it suffices to show that generalizations lift along f. Suppose that y' \leadsto y is a specialization of points in Y and x \in X is a point mapping to y. As in Lemma 37.41.1 choose a diagram
where (V, v) \to (Y, y) is an elementary étale neighbourhood, U \to V is finite, u is the unique point of U mapping to v, U \subset V \times _ Y X is open, and v \mapsto y and u \mapsto x. Let E be an irreducible component of U passing through u (there is at least one of these). Since U \to X is étale, E maps to an irreducible component of X, which in turn dominates an irreducible component of Y (by assumption). Since U \to V is finite hence closed, we conclude that the image E' \subset V of E is an irreducible closed subset passing through v which dominates an irreducible component of Y. Since V \to Y is étale E' must be an irreducible component of V passing through v. Since Y is geometrically unibranch we see that E' is the unique irreducible component of V passing through v (Lemma 37.36.2). Since V is locally Noetherian we may after shrinking V assume that E' = V (equality of sets).
Since V \to Y is étale we can find a specialization v' \leadsto v whose image is y' \leadsto y. By the above we can find u' \in U mapping to v'. Then u' \leadsto u because u is the only point of U mapping to v and U \to V is closed. Then finally the image x' \in X of u' is a point specializing to x and mapping to y' and the proof is complete. \square
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