Lemma 97.4.3. Let S be a scheme. 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, flat and locally of finite presentation,
then \Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X} representable by algebraic spaces.
Proof.
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 Algebraic Stacks, Lemma 94.10.11 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, flat, locally of finite presentation and surjective, see Algebraic Stacks, Lemmas 94.9.6, 94.9.7, 94.10.4, and 94.10.6. Therefore F is an algebraic space by Bootstrap, Theorem 80.10.1 and we win.
\square
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