Lemma 94.19.1. Let $\mathit{Sch}_{fppf}$ be a big fppf site. Let $S \to S'$ be a morphism of this site. The constructions A and B of Stacks, Section 8.13 above give isomorphisms of $2$-categories
\[ \left\{ \begin{matrix} 2\text{-category of algebraic}
\\ \text{stacks }\mathcal{X}\text{ over }S
\end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} 2\text{-category of pairs }(\mathcal{X}', f)\text{ consisting of an}
\\ \text{algebraic stack }\mathcal{X}'\text{ over }S'\text{ and a morphism}
\\ f : \mathcal{X}' \to (\mathit{Sch}/S)_{fppf}\text{ of algebraic stacks over }S'
\end{matrix} \right\} \]
Proof.
The statement makes sense as the functor $j : (\mathit{Sch}/S)_{fppf} \to (\mathit{Sch}/S')_{fppf}$ is the localization functor associated to the object $S/S'$ of $(\mathit{Sch}/S')_{fppf}$. By Stacks, Lemma 8.13.2 the only thing to show is that the constructions A and B preserve the subcategories of algebraic stacks. For example, if $\mathcal{X} = [U/R]$ then construction A applied to $\mathcal{X}$ just produces $\mathcal{X}' = \mathcal{X}$. Conversely, if $\mathcal{X}' = [U'/R']$ the morphism $p$ induces morphisms of algebraic spaces $U' \to S$ and $R' \to S$, and then $\mathcal{X} = [U'/R']$ but now viewed as a stack over $S$. Hence the lemma is clear.
$\square$
Comments (2)
Comment #193 by Pieter Belmans on
Comment #200 by Johan on