94.5 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
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
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.
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)