The Stacks project

Lemma 92.10.10. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $j : \mathcal X \to \mathcal Y$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume $j$ is representable by algebraic spaces and a monomorphism (see Definition 92.10.1 and Descent on Spaces, Lemma 72.10.30). Then $j$ is fully faithful on fibre categories.

Proof. We have seen in Lemma 92.9.2 that $j$ is faithful on fibre categories. Consider a scheme $U$, two objects $u, v$ of $\mathcal{X}_ U$, and an isomorphism $t : j(u) \to j(v)$ in $\mathcal{Y}_ U$. We have to construct an isomorphism in $\mathcal{X}_ U$ between $u$ and $v$. By the $2$-Yoneda lemma (see Section 92.5) we think of $u$, $v$ as $1$-morphisms $u, v : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ and we consider the $2$-fibre product

\[ (\mathit{Sch}/U)_{fppf} \times _{j \circ v, \mathcal{Y}} \mathcal{X}. \]

By assumption this is representable by an algebraic space $F_{j \circ v}$, over $U$ and the morphism $F_{j \circ v} \to U$ is a monomorphism. But since $(1_ U, v, 1_{j(v)})$ gives a $1$-morphism of $(\mathit{Sch}/U)_{fppf}$ into the displayed $2$-fibre product, we see that $F_{j \circ v} = U$ (here we use that if $V \to U$ is a monomorphism of algebraic spaces which has a section, then $V = U$). Therefore the $1$-morphism projecting to the first coordinate

\[ (\mathit{Sch}/U)_{fppf} \times _{j \circ v, \mathcal{Y}} \mathcal{X} \to (\mathit{Sch}/U)_{fppf} \]

is an equivalence of fibre categories. Since $(1_ U, u, t)$ and $(1_ U, v, 1_{j(v)})$ give two objects in $((\mathit{Sch}/U)_{fppf} \times _{j \circ v, \mathcal{Y}} \mathcal{X})_ U$ which have the same first coordinate, there must be a $2$-morphism between them in the $2$-fibre product. This is by definition a morphism $\tilde t : u \to v$ such that $j(\tilde t) = t$. $\square$

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Your email address will not be published.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 05UK. Beware of the difference between the letter 'O' and the digit '0'.