Lemma 76.15.3. Let $S$ be a scheme. Let $Z \to Y \to X$ be morphisms of algebraic spaces over $S$. Assume $Z \to Y$ is étale.

1. If $Y \subset Y'$ is a universal first order thickening of $Y$ over $X$, then the unique étale morphism $Z' \to Y'$ such that $Z = Y \times _{Y'} Z'$ (see Theorem 76.8.1) is a universal first order thickening of $Z$ over $X$.

2. If $Z \to Y$ is surjective and $(Z \subset Z') \to (Y \subset Y')$ is an étale morphism of first order thickenings over $X$ and $Z'$ is a universal first order thickening of $Z$ over $X$, then $Y'$ is a universal first order thickening of $Y$ over $X$.

Proof. Proof of (1). By Lemma 76.15.1 it suffices to consider solid commutative diagrams (76.15.0.1) with $T'$ an affine scheme. The composition $T \to Z \to Y$ lifts uniquely to $T' \to Y'$ as $Y'$ is the universal first order thickening. Then the fact that $Z' \to Y'$ is étale implies (see Lemma 76.13.5) that $T' \to Y'$ lifts to the desired morphism $a' : T' \to Z'$.

Proof of (2). Let $T \subset T'$ be a first order thickening over $X$ and let $a : T \to Y$ be a morphism. Set $W = T \times _ Y Z$ and denote $c : W \to Z$ the projection Let $W' \to T'$ be the unique étale morphism such that $W = T \times _{T'} W'$, see Theorem 76.8.1. Note that $W' \to T'$ is surjective as $Z \to Y$ is surjective. By assumption we obtain a unique morphism $c' : W' \to Z'$ over $X$ restricting to $c$ on $W$. By uniqueness the two restrictions of $c'$ to $W' \times _{T'} W'$ are equal (as the two restrictions of $c$ to $W \times _ T W$ are equal). Hence $c'$ descends to a unique morphism $a' : T' \to Y'$ and we win. $\square$

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