Lemma 66.6.4. The composition of a pair of (universally) open morphisms of algebraic spaces is (universally) open.

Proof. Omitted. $\square$

Comment #501 by Kestutis Cesnavicius on

Proof: This is immediate from the definition.

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

Comment #508 by on

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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