The Stacks project

Proof. We map the object $(U, Z, y', x', \alpha ')$ to the object $(U, Z, H(y'), G(x'), \gamma \star \text{id}_ H \star \alpha ')$ where $\star $ denotes horizontal composition of $2$-morphisms, see Categories, Definition 4.28.1. To a morphism $(f, g, b, a) : (U_1, Z_1, y_1', x_1', \alpha _1') \to (U_2, Z_2, y_2', x_2', \alpha _2')$ we assign $(f, g, H(b), G(a))$. We omit the verification that this defines a functor between categories over $(\mathit{Sch}/S)_{fppf}$. $\square$

Proof. We get a $2$-commutative diagram by Lemma 97.12.1 and hence we get a $1$-morphism (i.e., a functor)

We indicate why this functor is essentially surjective. Namely, an object of the category on the right hand side is given by a scheme $U$ over $S$, an object $y'$ of $\mathcal{Y}'_ U$, an object $(U, Z, y, x, \alpha )$ of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U$ and an isomorphism $H(y') \to y$ in $\mathcal{Y}_ U$. The assumption means exactly that there exists an object $x'$ of $\mathcal{X}'_ Z$ such that there exist isomorphisms $G(x') \cong x$ and $\alpha ' : y'|_ Z \to F'(x')$ compatible with $\alpha $. Then we see that $(U, Z, y', x', \alpha ')$ is an object of $\mathcal{H}_ d(\mathcal{X}'/\mathcal{Y}')$ over $U$. Details omitted. $\square$

Proof. Let $U$ be a scheme and let $\xi = (U, Z, y, x, \alpha )$ be an object of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U$. We have to prove that the $2$-fibre product

is representable by an algebraic space étale over $U$. An object of this over $U'$ corresponds to an object $x'$ in the fibre category of $\mathcal{X}'$ over $Z_{U'}$ such that $G(x') \cong x|_{Z_{U'}}$. By assumption the $2$-fibre product

is representable by an algebraic space $W$ such that the projection $W \to Z$ is étale. Then (97.12.3.1) is representable by the algebraic space $F$ parametrizing sections of $W \to Z$ over $U$ introduced in Lemma 97.9.2. Since $F \to U$ is étale we conclude that $\mathcal{H}_ d(\mathcal{X}'/\mathcal{Y}) \to \mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ is representable by algebraic spaces and étale. Finally, if $\mathcal{X}' \to \mathcal{X}$ is surjective also, then $W \to Z$ is surjective, and hence $F \to U$ is surjective by Lemma 97.9.1. Thus in this case $\mathcal{H}_ d(\mathcal{X}'/\mathcal{Y}) \to \mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ is also surjective. $\square$

Proof. Set $\mathcal{X}'' = \mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$. By Lemma 97.4.1 the $1$-morphism $\mathcal{X}' \to \mathcal{X}''$ is representable by algebraic spaces and étale (in particular the condition in the second statement of the lemma that $\mathcal{X}' \to \mathcal{X}''$ be surjective makes sense). We obtain a $2$-commutative diagram

It follows from Lemma 97.12.2 that $\mathcal{H}_ d(\mathcal{X}''/\mathcal{Y}')$ is the base change of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ by $\mathcal{Y}' \to \mathcal{Y}$. In particular we see that $\mathcal{H}_ d(\mathcal{X}''/\mathcal{Y}') \to \mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ is representable by algebraic spaces and étale, see Algebraic Stacks, Lemma 94.10.6. Moreover, it is also surjective if $H$ is. Hence if we can show that the result holds for the left square in the diagram, then we're done. In this way we reduce to the case where $\mathcal{Y}' = \mathcal{Y}$ which is the content of Lemma 97.12.3. $\square$

Proof. Let $U$ be a scheme and let $\xi = (U, Z, p, x, 1)$ be an object of $\mathcal{H}_ d(\mathcal{X}) = \mathcal{H}_ d(\mathcal{X}/S)$ over $U$. Here $p$ is just the structure morphism of $U$. The fifth component $1$ exists and is unique since everything is over $S$. Also, let $y$ be an object of $\mathcal{Y}$ over $U$. We have to show the $2$-fibre product

is representable by an algebraic space. To explain why this is so we introduce

which is an algebraic space over $Z$ by assumption. Let $a : U' \to U$ be a scheme over $U$. What does it mean to give an object of the fibre category of (97.12.5.1) over $U'$? Well, it means that we have an object $\xi ' = (U', Z', y', x', \alpha ')$ of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U'$ and isomorphisms $(U', Z', p', x', 1) \cong (U, Z, p, x, 1)|_{U'}$ and $y' \cong y|_{U'}$. Thus $\xi '$ is isomorphic to $(U', U' \times _{a, U} Z, a^*y, x|_{U' \times _{a, U} Z}, \alpha )$ for some morphism

in the fibre category of $\mathcal{Y}$ over $U' \times _{a, U} Z$. Hence we can view $\alpha $ as a morphism $b : U' \times _{a, U} Z \to I$. In this way we see that (97.12.5.1) is representable by $\text{Res}_{Z/U}(I)$ which is an algebraic space by Proposition 97.11.5. $\square$

Proof. Let $U$ be a scheme and let $\xi = (U, Z, y, x, \alpha )$ be an object of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U$. We have to prove that the $2$-fibre product

is representable by an algebraic space étale over $U$. An object of this over $a : U' \to U$ corresponds to an object $x'$ of $\mathcal{X}'$ over $U' \times _{a, U} Z$ such that $G(x') \cong x|_{U' \times _{a, U} Z}$. By assumption the $2$-fibre product

is representable by an algebraic space $X$ over $Z$. It follows that (97.12.6.1) is representable by $\text{Res}_{Z/U}(X)$, which is an algebraic space by Proposition 97.11.5. $\square$

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. $\square$