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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.


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