The Stacks project

Proof. This means we have to show the following: Given

1. an affine scheme $U = \mathop{\mathrm{lim}}\nolimits _ i U_ i$ which is written as the directed limit of affine schemes $U_ i$ over $S$,

2. an object $y_ i$ of $\mathcal{Y}$ over $U_ i$ for some $i$, and

3. an object $\Xi = (U, Z, y, x, \alpha )$ of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U$ such that $y = y_ i|_ U$,

then there exists an $i' \geq i$ and an object $\Xi _{i'} = (U_{i'}, Z_{i'}, y_{i'}, x_{i'}, \alpha _{i'})$ of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U_{i'}$ with $\Xi _{i'}|_ U = \Xi$ and $y_{i'} = y_ i|_{U_{i'}}$. Namely, the last two equalities will take care of the commutativity of (97.5.0.1).

Let $X_{y_ i} \to U_ i$ be an algebraic space representing the $2$-fibre product

Note that $X_{y_ i} \to U_ i$ is locally of finite presentation by our assumption on $F$. Write $\Xi$. It is clear that $\xi = (Z, Z \to U_ i, x, \alpha )$ is an object of the $2$-fibre product displayed above, hence $\xi$ gives rise to a morphism $f_\xi : Z \to X_{y_ i}$ of algebraic spaces over $U_ i$ (since $X_{y_ i}$ is the functor of isomorphisms classes of objects of $(\mathit{Sch}/U_ i)_{fppf} \times _{y, \mathcal{Y}, F} \mathcal{X}$, see Algebraic Stacks, Lemma 94.8.2). By Limits, Lemmas 32.10.1 and 32.8.8 there exists an $i' \geq i$ and a finite locally free morphism $Z_{i'} \to U_{i'}$ of degree $d$ whose base change to $U$ is $Z$. By Limits of Spaces, Proposition 70.3.10 we may, after replacing $i'$ by a bigger index, assume there exists a morphism $f_{i'} : Z_{i'} \to X_{y_ i}$ such that

is commutative. We set $\Xi _{i'} = (U_{i'}, Z_{i'}, y_{i'}, x_{i'}, \alpha _{i'})$ where

1. $y_{i'}$ is the object of $\mathcal{Y}$ over $U_{i'}$ which is the pullback of $y_ i$ to $U_{i'}$,

2. $x_{i'}$ is the object of $\mathcal{X}$ over $Z_{i'}$ corresponding via the $2$-Yoneda lemma to the $1$-morphism

   $$(\mathit{Sch}/Z_{i'})_{fppf} \to \mathcal{S}_{X_{y_ i}} \to (\mathit{Sch}/U_ i)_{fppf} \times _{y_ i, \mathcal{Y}, F} \mathcal{X} \to \mathcal{X}$$

   where the middle arrow is the equivalence which defines $X_{y_ i}$ (notation as in Algebraic Stacks, Sections 94.8 and 94.7).

3. $\alpha _{i'} : y_{i'}|_{Z_{i'}} \to F(x_{i'})$ is the isomorphism coming from the $2$-commutativity of the diagram

   $$\xymatrix{ (\mathit{Sch}/Z_{i'})_{fppf} \ar[r] \ar[rd] & (\mathit{Sch}/U_ i)_{fppf} \times _{y_ i, \mathcal{Y}, F} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d]^ F \\ & (\mathit{Sch}/U_{i'})_{fppf} \ar[r] & \mathcal{Y} }$$

Recall that $f_\xi : Z \to X_{y_ i}$ was the morphism corresponding to the object $\xi = (Z, Z \to U_ i, x, \alpha )$ of $(\mathit{Sch}/U_ i)_{fppf} \times _{y_ i, \mathcal{Y}, F} \mathcal{X}$ over $Z$. By construction $f_{i'}$ is the morphism corresponding to the object $\xi _{i'} = (Z_{i'}, Z_{i'} \to U_ i, x_{i'}, \alpha _{i'})$. As $f_\xi = f_{i'} \circ (Z \to Z_{i'})$ we see that the object $\xi _{i'} = (Z_{i'}, Z_{i'} \to U_ i, x_{i'}, \alpha _{i'})$ pulls back to $\xi$ over $Z$. Thus $x_{i'}$ pulls back to $x$ and $\alpha _{i'}$ pulls back to $\alpha$. This means that $\Xi _{i'}$ pulls back to $\Xi$ over $U$ and we win. $\square$