Lemma 4.40.1 (2-Yoneda lemma). Let $\mathcal{S}\to \mathcal{C}$ be fibred in groupoids. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. The functor

given by $G \mapsto G(\text{id}_ U)$ is an equivalence.

Lemma 4.40.1 (2-Yoneda lemma). Let $\mathcal{S}\to \mathcal{C}$ be fibred in groupoids. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. The functor

\[ \mathop{Mor}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}) \longrightarrow \mathcal{S}_ U \]

given by $G \mapsto G(\text{id}_ U)$ is an equivalence.

**Proof.**
Make a choice of pullbacks for $\mathcal{S}$ (see Definition 4.32.6). We define a functor

\[ \mathcal{S}_ U \longrightarrow \mathop{Mor}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}) \]

as follows. Given $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ the associated functor is

on objects: $(f : V \to U) \mapsto f^*x$, and

on morphisms: the arrow $(g : V'/U \to V/U)$ maps to the composition

\[ (f \circ g)^*x \xrightarrow {(\alpha _{g, f})_ x} g^*f^*x \rightarrow f^*x \]where $\alpha _{g, f}$ is as in Lemma 4.34.2.

We omit the verification that this is an inverse to the functor of the lemma. $\square$

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