Lemma 67.6.4 (03Z4)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.6: Open morphisms
Lemma 67.6.4 (cite)

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Lemma 67.6.4. The composition of a pair of (universally) open morphisms of algebraic spaces is (universally) open.

Proof. Omitted. $\square$

Comments (2)

Comment #501 by Kestutis Cesnavicius on April 01, 2014 at 19:04

Proof: This is immediate from the definition.

Also, I would change '(universally) open' to 'open (resp., universally open)' for clarity.

Comment #508 by Johan on April 02, 2014 at 19:48

OK, in a way it is and in a way it isn't. I guess you use also something about associativity of fibre product. Maybe we just leave it as omitted till somebody writes out a proof.

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