Lemma 36.66.2. 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 36.66.2. 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 36.33.4. 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 36.66.1 it suffices to prove $f$ is open. By Morphisms, Lemma 28.22.2 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 36.36.1 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 étale neighbourhood, $U \to V$ is finite, $v$ is the unique point of $V$ mapping to $u$, $U \subset V \times _ Y X$ 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 36.33.2). 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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (1)

Comment #4649 by Noah Olander on