Definition 29.23.1. Let $f : X \to S$ be a morphism.

We say $f$ is

*open*if the map on underlying topological spaces is open.We say $f$ is

*universally open*if for any morphism of schemes $S' \to S$ the base change $f' : X_{S'} \to S'$ is open.

Definition 29.23.1. Let $f : X \to S$ be a morphism.

We say $f$ is

*open*if the map on underlying topological spaces is open.We say $f$ is

*universally open*if for any morphism of schemes $S' \to S$ the base change $f' : X_{S'} \to S'$ is open.

According to Topology, Lemma 5.19.7 generalizations lift along certain types of open maps of topological spaces. In fact generalizations lift along any open morphism of schemes (see Lemma 29.23.5). Also, we will see that generalizations lift along flat morphisms of schemes (Lemma 29.25.9). This sometimes in turn implies that the morphism is open.

Lemma 29.23.2. Let $f : X \to S$ be a morphism.

If $f$ is locally of finite presentation and generalizations lift along $f$, then $f$ is open.

If $f$ is locally of finite presentation and generalizations lift along every base change of $f$, then $f$ is universally open.

**Proof.**
It suffices to prove the first assertion. This reduces to the case where both $X$ and $S$ are affine. In this case the result follows from Algebra, Lemma 10.40.3 and Proposition 10.40.8.
$\square$

See also Lemma 29.25.10 for the case of a morphism flat of finite presentation.

Lemma 29.23.3. A composition of (universally) open morphisms is (universally) open.

**Proof.**
Omitted.
$\square$

Lemma 29.23.4. Let $k$ be a field. Let $X$ be a scheme over $k$. The structure morphism $X \to \mathop{\mathrm{Spec}}(k)$ is universally open.

**Proof.**
Let $S \to \mathop{\mathrm{Spec}}(k)$ be a morphism. We have to show that the base change $X_ S \to S$ is open. The question is local on $S$ and $X$, hence we may assume that $S$ and $X$ are affine. In this case the result is Algebra, Lemma 10.40.10.
$\square$

Lemma 29.23.5. Let $\varphi : X \to Y$ be a morphism of schemes. If $\varphi $ is open, then $\varphi $ is generizing (i.e., generalizations lift along $\varphi $). If $\varphi $ is universally open, then $\varphi $ is universally generizing.

**Proof.**
Assume $\varphi $ is open. Let $y' \leadsto y$ be a specialization of points of $Y$. Let $x \in X$ with $\varphi (x) = y$. Choose affine opens $U \subset X$ and $V \subset Y$ such that $\varphi (U) \subset V$ and $x \in U$. Then also $y' \in V$. Hence we may replace $X$ by $U$ and $Y$ by $V$ and assume $X$, $Y$ affine. The affine case is Algebra, Lemma 10.40.2 (combined with Algebra, Lemma 10.40.3).
$\square$

Lemma 29.23.6. Let $f : X \to Y$ be a morphism of schemes. Let $g : Y' \to Y$ be open and surjective such that the base change $f' : X' \to Y'$ is quasi-compact. Then $f$ is quasi-compact.

**Proof.**
Let $V \subset Y$ be a quasi-compact open. As $g$ is open and surjective we can find a quasi-compact open $W' \subset W$ such that $g(W') = V$. By assumption $(f')^{-1}(W')$ is quasi-compact. The image of $(f')^{-1}(W')$ in $X$ is equal to $f^{-1}(V)$, see Lemma 29.9.3. Hence $f^{-1}(V)$ is quasi-compact as the image of a quasi-compact space, see Topology, Lemma 5.12.7. Thus $f$ is quasi-compact.
$\square$

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