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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