Lemma 76.15.4. Let $S$ be a scheme. Let $h : Z \to X$ be a formally unramified morphism of algebraic spaces over $S$. There exists a universal first order thickening $Z \subset Z'$ of $Z$ over $X$.

**Proof.**
Choose any commutative diagram

where $V$ and $U$ are schemes and the vertical arrows are étale. Note that $V \to U$ is a formally unramified morphism of schemes, see Lemma 76.14.2. Combining Lemma 76.15.1 and More on Morphisms, Lemma 37.7.1 we see that a universal first order thickening $V \subset V'$ of $V$ over $U$ exists. By Lemma 76.15.2 part (1) $V'$ is a universal first order thickening of $V$ over $X$.

Fix a scheme $U$ and a surjective étale morphism $U \to X$. The argument above shows that for any $V \to Z$ étale with $V$ a scheme such that $V \to Z \to X$ factors through $U$ a universal first order thickening $V \subset V'$ of $V$ over $X$ exists (but does not depend on the chosen factorization of $V \to X$ through $U$). Now we may choose $V$ such that $V \to Z$ is surjective étale (see Spaces, Lemma 65.11.6). Then $R = V \times _ Z V$ a scheme étale over $Z$ such that $R \to X$ factors through $U$ also. Hence we obtain universal first order thickenings $V \subset V'$ and $R \subset R'$ over $X$. As $V \subset V'$ is a universal first order thickening, the two projections $s, t : R \to V$ lift to morphisms $s', t': R' \to V'$. By Lemma 76.15.3 as $R'$ is the universal first order thickening of $R$ over $X$ these morphisms are étale. Then $(t', s') : R' \to V'$ is an étale equivalence relation and we can set $Z' = V'/R'$. Since $V' \to Z'$ is surjective étale and $v'$ is the universal first order thickening of $V$ over $X$ we conclude from Lemma 76.15.2 part (2) that $Z'$ is a universal first order thickening of $Z$ over $X$. $\square$

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