Lemma 76.15.1. Let $S$ be a scheme. Let $h : Z \to X$ be a morphism of algebraic spaces over $S$. Let $Z \subset Z'$ be a first order thickening over $X$. The following are equivalent

**Proof.**
The implications (1) $\Rightarrow $ (2) $\Rightarrow $ (3) are formal. Assume (3) a assume given an arbitrary diagram (76.15.0.1). Choose a presentation $T' = U'/R'$, see Spaces, Definition 65.9.3. We may assume that $U' = \coprod U'_ i$ is a disjoint union of affines, so $R' = U' \times _{T'} U' = \coprod _{i, j} U'_ i \times _ T' U'_ j$. For each pair $(i, j)$ choose an affine open covering $U'_ i \times _ T' U'_ j = \bigcup _ k R'_{ijk}$. Denote $U_ i, R_{ijk}$ the fibre products with $T$ over $T'$. Then each $U_ i \subset U'_ i$ and $R_{ijk} \subset R'_{ijk}$ is a first order thickening of affine schemes. Denote $a_ i : U_ i \to Z$, resp. $a_{ijk} : R_{ijk} \to Z$ the composition of $a : T \to Z$ with the morphism $U_ i \to T$, resp. $R_{ijk} \to T$. By (3) applied to $a_ i : U_ i \to Z$ we obtain unique morphisms $a'_ i : U'_ i \to Z'$. By (3) applied to $a_{ijk}$ we see that the two compositions $R'_{ijk} \to R'_ i \to Z'$ and $R'_{ijk} \to R'_ j \to Z'$ are equal. Hence $a' = \coprod a'_ i : U' = \coprod U'_ i \to Z'$ descends to the quotient sheaf $T' = U'/R'$ and we win.
$\square$

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