The Stacks project

Follows from the implication (a) $\Rightarrow $ (b) in [IV, Corollary 1.10.4, EGA]

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 10.41.2 (combined with Algebra, Lemma 10.41.3). $\square$

Comments (2)

Comment #2740 by Ariyan Javanpeykar on

This follows from the implication "a) implies b)" in EGA IV_1, Corollary 1.10.4

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