Lemma 97.8.4. Let S be a scheme. Let F : \mathcal{X} \to \mathcal{Y} be a 1-morphism of stacks in groupoids over (\mathit{Sch}/S)_{fppf}. If F is algebraic and \Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y} is representable by algebraic spaces, then \Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X} is representable by algebraic spaces.
Proof. Assume F is algebraic and \Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y} is representable by algebraic spaces. Take a scheme U over S and two objects x_1, x_2 of \mathcal{X} over U. We have to show that \mathit{Isom}(x_1, x_2) is an algebraic space over U, see Algebraic Stacks, Lemma 94.10.11. Set y_ i = F(x_ i). We have a morphism of sheaves of sets
f : \mathit{Isom}(x_1, x_2) \to \mathit{Isom}(y_1, y_2)
and the target is an algebraic space by assumption. Thus it suffices to show that f is representable by algebraic spaces, see Bootstrap, Lemma 80.3.6. Thus we can choose a scheme V over U and an isomorphism \beta : y_{1, V} \to y_{2, V} and we have to show the functor
(\mathit{Sch}/V)_{fppf} \to \textit{Sets},\quad T/V \mapsto \{ \alpha : x_{1, T} \to x_{2, T} \text{ in }\mathcal{X}_ T \mid F(\alpha ) = \beta |_ T\}
is an algebraic space. Consider the objects z_1 = (V, x_{1, V}, \text{id}) and z_2 = (V, x_{2, V}, \beta ) of
\mathcal{Z} = (\mathit{Sch}/V)_{fppf} \times _{y_{1, V}, \mathcal{Y}} \mathcal{X}
Then it is straightforward to verify that the functor above is equal to \mathit{Isom}(z_1, z_2) on (\mathit{Sch}/V)_{fppf}. Hence this is an algebraic space by our assumption that F is algebraic (and the definition of algebraic stacks). \square
Comments (0)