The Stacks project

93.5 The 2-Yoneda lemma

Let $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, and let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. We will frequently use the $2$-Yoneda lemma, see Categories, Lemma 4.41.2. Technically it says that there is an equivalence of categories

\[ \mathop{\mathrm{Mor}}\nolimits _{\textit{Cat}/(\mathit{Sch}/S)_{fppf}}( (\mathit{Sch}/U)_{fppf}, \mathcal{X}) \longrightarrow \mathcal{X}_ U, \quad f \longmapsto f(U/U). \]

It says that $1$-morphisms $(\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ correspond to objects $x$ of the fibre category $\mathcal{X}_ U$. Namely, given a $1$-morphism $f : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ we obtain the object $x = f(U/U) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$. Conversely, given a choice of pullbacks for $\mathcal{X}$ as in Categories, Definition 4.33.6, and an object $x$ of $\mathcal{X}_ U$, we obtain a functor $(\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ defined by the rule

\[ (\varphi : V \to U) \longmapsto \varphi ^*x \]

on objects. By abuse of notation we use $x : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ to indicate this functor. It indeed has the property that $x(U/U) = x$ and moreover, given any other functor $f$ with $f(U/U) = x$ there exists a unique $2$-isomorphism $x \to f$. In other words the functor $x$ is well determined by the object $x$ up to unique $2$-isomorphism.

We will use this without further mention in the following.

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Your email address will not be published.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