Lemma 91.15.3. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $u : \mathcal{U} \to \mathcal{X}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If

$\mathcal{U}$ is representable by an algebraic space, and

$u$ is representable by algebraic spaces, surjective and smooth,

then $\mathcal X$ is an algebraic stack over $S$.

**Proof.**
We have to show that $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces, see Definition 91.12.1. Given two schemes $T_1$, $T_2$ over $S$ denote $\mathcal{T}_ i = (\mathit{Sch}/T_ i)_{fppf}$ the associated representable fibre categories. Suppose given $1$-morphisms $f_ i : \mathcal{T}_ i \to \mathcal{X}$. According to Lemma 91.10.11 it suffices to prove that the $2$-fibered product $\mathcal{T}_1 \times _\mathcal {X} \mathcal{T}_2$ is representable by an algebraic space. By Stacks, Lemma 8.6.8 this is in any case a stack in setoids. Thus $\mathcal{T}_1 \times _\mathcal {X} \mathcal{T}_2$ corresponds to some sheaf $F$ on $(\mathit{Sch}/S)_{fppf}$, see Stacks, Lemma 8.6.3. Let $U$ be the algebraic space which represents $\mathcal{U}$. By assumption

\[ \mathcal{T}_ i' = \mathcal{U} \times _{u, \mathcal{X}, f_ i} \mathcal{T}_ i \]

is representable by an algebraic space $T'_ i$ over $S$. Hence $\mathcal{T}_1' \times _\mathcal {U} \mathcal{T}_2'$ is representable by the algebraic space $T'_1 \times _ U T'_2$. Consider the commutative diagram

\[ \xymatrix{ & \mathcal{T}_1 \times _{\mathcal X} \mathcal{T}_2 \ar[rr]\ar '[d][dd] & & \mathcal{T}_1 \ar[dd] \\ \mathcal{T}_1' \times _\mathcal {U} \mathcal{T}_2' \ar[ur]\ar[rr]\ar[dd] & & \mathcal{T}_1' \ar[ur]\ar[dd] \\ & \mathcal{T}_2 \ar '[r][rr] & & \mathcal X \\ \mathcal{T}_2' \ar[rr]\ar[ur] & & \mathcal{U} \ar[ur] } \]

In this diagram the bottom square, the right square, the back square, and the front square are $2$-fibre products. A formal argument then shows that $\mathcal{T}_1' \times _\mathcal {U} \mathcal{T}_2' \to \mathcal{T}_1 \times _{\mathcal X} \mathcal{T}_2$ is the “base change” of $\mathcal{U} \to \mathcal{X}$, more precisely the diagram

\[ \xymatrix{ \mathcal{T}_1' \times _\mathcal {U} \mathcal{T}_2' \ar[d] \ar[r] & \mathcal{U} \ar[d] \\ \mathcal{T}_1 \times _{\mathcal X} \mathcal{T}_2 \ar[r] & \mathcal{X} } \]

is a $2$-fibre square. Hence $T'_1 \times _ U T'_2 \to F$ is representable by algebraic spaces, smooth, and surjective, see Lemmas 91.9.6, 91.9.7, 91.10.4, and 91.10.6. Therefore $F$ is an algebraic space by Bootstrap, Theorem 77.10.1 and we win.
$\square$

## Comments (0)

There are also: