Lemma 76.9.8 (09ZX)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 76: More on Morphisms of Spaces
Section 76.9: Thickenings
Lemma 76.9.8 (cite)

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Lemma 76.9.8. Let $S$ be a scheme. Let $Y \subset Y'$ be a thickening of algebraic spaces over $S$ . Let $X' \to Y'$ be a morphism and set $X = Y \times _{Y'} X'$ . Then $(X \subset X') \to (Y \subset Y')$ is a morphism of thickenings. If $Y \subset Y'$ is a first (resp. finite order) thickening, then $X \subset X'$ is a first (resp. finite order) thickening.

Proof. Omitted. $\square$

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