Lemma 37.68.2. Let $f : X \to Y$ be a morphism of schemes. If

$f$ is locally quasi-finite,

$Y$ is unibranch and locally Noetherian, and

every irreducible component of $X$ dominates an irreducible component of $Y$,

then $f$ is universally open.

Lemma 37.68.2. Let $f : X \to Y$ be a morphism of schemes. If

$f$ is locally quasi-finite,

$Y$ is 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 unibranch by Lemma 37.33.4. 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 37.68.1 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 37.37.1 choose a diagram

\[ \xymatrix{ u \ar[d] & U \ar[d] \ar[r] & X \ar[d] \\ v & V \ar[r] & Y } \]

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 unibranch we see that $E'$ is the unique irreducible component of $V$ passing through $v$ (Lemma 37.33.2). 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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## Comments (2)

Comment #4649 by Noah Olander on

Comment #4795 by Johan on