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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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