The Stacks project

Proof. Choose a scheme $U'$ and a surjective smooth morphism $U' \to \mathcal{X}'$. As usual we set $U = \mathcal{X} \times _{\mathcal{X}'} U'$. Then $U \to \mathcal{X}$ is a surjective smooth morphism. Therefore the base change

is a surjective smooth morphism of algebraic spaces. By Topologies on Spaces, Lemma 73.4.4 we can find an étale covering $\{ W_ i \to W\} $ such that $W_ i \to W$ factors through $V \to W$. After covering $W_ i$ by affines (Properties of Spaces, Lemma 66.6.1) we may assume each $W_ i$ is affine. We may and do replace $W$ by $W_ i$ which reduces us to the situation discussed in the next paragraph.

Assume $W$ is affine and the given morphism $W \to \mathcal{X}$ factors through $U$. Picture

Since $W$ and $U$ are smooth over $\mathcal{X}$ we see that $i$ is locally of finite type (Morphisms of Stacks, Lemma 101.17.8). After replacing $U$ by $\mathbf{A}^ n_ U$ we may assume that $i$ is an immersion, see Morphisms, Lemma 29.39.2. By Morphisms of Stacks, Lemma 101.44.4 the morphism $i$ is a local complete intersection. Hence $i$ is a Koszul-regular immersion (as defined in Divisors, Definition 31.21.1) by More on Morphisms, Lemma 37.62.3.

We may still replace $W$ by an affine open covering. For every point $w \in W$ we can choose an affine open $U'_ w \subset U'$ such that if $U_ w \subset U$ is the corresponding affine open, then $w \in i^{-1}(U_ w)$ and $i^{-1}(U_ w) \to U_ w$ is a closed immersion cut out by a Koszul-regular sequence $f_1, \ldots , f_ r \in \Gamma (U_ w, \mathcal{O}_{U_ w})$. This follows from the definition of Koszul-regular immersions and Divisors, Lemma 31.20.7. Set $W_ w = i^{-1}(U_ w)$; this is an affine open neighbourhood of $w \in W$. Choose lifts $f'_1, \ldots , f'_ r \in \Gamma (U'_ w, \mathcal{O}_{U'_ w})$ of $f_1, \ldots , f_ r$. This is possible as $U_ w \to U'_ w$ is a closed immersion of affine schemes. Let $W'_ w \subset U'_ w$ be the closed subscheme cut out by $f'_1, \ldots , f'_ r$. We claim that $W'_ w \to \mathcal{X}'$ is smooth. The claim finishes the proof as $W_ w = \mathcal{X} \times _{\mathcal{X}'} W'_ w$ by construction.

To check the claim it suffices to check that the base change $W'_ w \times _{\mathcal{X}'} X' \to X'$ is smooth for every affine scheme $X'$ smooth over $\mathcal{X}'$. Choose an étale morphism

with $Y'$ affine. Because $U'_ w \times _{\mathcal{X}'} X'$ is covered by the images of such morphisms, it is enough to show that the closed subscheme $Z'$ of $Y'$ cut out by $f'_1, \ldots , f'_ r$ is smooth over $X'$. Picture

Set $X = \mathcal{X} \times _{\mathcal{X}'} X'$, $Y = X \times _{X'} Y' = \mathcal{X} \times _{\mathcal{X}'} Y'$, and $Z = Y \times _{Y'} Z' = X \times _{X'} Z' = \mathcal{X} \times _{\mathcal{X}'} Z'$. Then $(Z \subset Z') \to (Y \subset Y') \subset (X \subset X')$ are (cartesian) morphisms of thickenings of affine schemes and we are given that $Z \to X$ and $Y' \to X'$ are smooth. Finally, the sequence of functions $f'_1, \ldots , f'_ r$ map to a Koszul-regular sequence in $\Gamma (Y', \mathcal{O}_{Y'})$ by More on Algebra, Lemma 15.30.5 because $Y' \to U'_ w$ is smooth and hence flat. By More on Algebra, Lemma 15.31.6 (and the fact that Koszul-regular sequences are quasi-regular sequences by More on Algebra, Lemmas 15.30.2, 15.30.3, and 15.30.6) we conclude that $Z' \to X'$ is smooth as desired. $\square$