The Stacks project

Lemma 92.19.1. 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 on

The "ismorphisms" in the statement should be "isomorphisms". The sentence starting with "But for example" seems a bit off. And in the sentence after that, "the the" should be "then the".

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04X5. Beware of the difference between the letter 'O' and the digit '0'.