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