The Stacks project

Proof. We have to show the following: Given

1. an object $(U, Z, y, x, \alpha )$ of $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ over an affine scheme $U$,

2. a first order thickening $U \subset U'$, and

3. an object $y'$ of $\mathcal{Y}$ over $U'$ such that $y'|_ U = y$,

then there exists an object $(U', Z', y', x', \alpha ')$ of $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ over $U'$ with $Z = U \times _{U'} Z'$, with $x = x'|_ Z$, and with $\alpha = \alpha '|_ U$. Namely, the last two equalities will take care of the commutativity of (97.6.0.1).

Consider the morphism $x_\alpha : Z \to X_ y$ constructed in Equation (97.15.0.1). Denote similarly $X'_{y'}$ the algebraic space over $U'$ representing the $2$-fibre product $(\mathit{Sch}/U')_{fppf} \times _{y', \mathcal{Y}, F} \mathcal{X}$. By assumption the morphism $X'_{y'} \to U'$ is flat (and locally of finite presentation). As $y'|_ U = y$ we see that $X_ y = U \times _{U'} X'_{y'}$. Hence we may apply Lemma 97.15.2 to find $Z' \to U'$ finite locally free of degree $d$ with $Z = U \times _{U'} Z'$ and with $Z' \to X'_{y'}$ extending $x_\alpha $. By construction the morphism $Z' \to X'_{y'}$ corresponds to a pair $(x', \alpha ')$. It is clear that $(U', Z', y', x', \alpha ')$ is an object of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U'$ with $Z = U \times _{U'} Z'$, with $x = x'|_ Z$, and with $\alpha = \alpha '|_ U$. As we've seen in Lemma 97.15.1 that $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y}) \subset \mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ is an “open substack” it follows that $(U', Z', y', x', \alpha ')$ is an object of $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ as desired. $\square$