The Stacks project

Lemma 95.7.2. The functor $p : \mathcal{S}\! \mathit{paces}\to (\mathit{Sch}/S)_{fppf}$ satisfies conditions (1) and (2) of Stacks, Definition 8.4.1.

Proof. It is follows from Lemma 95.7.1 that $\mathcal{S}\! \mathit{paces}$ is a fibred category over $(\mathit{Sch}/S)_{fppf}$ which proves (1). Suppose that $\{ U_ i \to U\} _{i \in I}$ is a covering of $(\mathit{Sch}/S)_{fppf}$. Suppose that $X, Y$ are algebraic spaces over $U$. Finally, suppose that $\varphi _ i : X_{U_ i} \to Y_{U_ i}$ are morphisms of $\textit{Spaces}/U_ i$ such that $\varphi _ i$ and $\varphi _ j$ restrict to the same morphisms $X_{U_ i \times _ U U_ j} \to Y_{U_ i \times _ U U_ j}$ of algebraic spaces over $U_ i \times _ U U_ j$. To prove (2) we have to show that there exists a unique morphism $\varphi : X \to Y$ over $U$ whose base change to $U_ i$ is equal to $\varphi _ i$. As a morphism from $X$ to $Y$ is the same thing as a map of sheaves this follows directly from Sites, Lemma 7.26.1. $\square$

Comments (0)

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Your email address will not be published.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 04UA. Beware of the difference between the letter 'O' and the digit '0'.