Lemma 4.41.2 (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 4.41.2 (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.
Proof. Make a choice of pullbacks for $\mathcal{S}$ (see Definition 4.33.6). We define a functor
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
where $\alpha _{g, f}$ is as in Lemma 4.35.2.
We omit the verification that this is an inverse to the functor of the lemma. $\square$
Comments (2)
Comment #7202 by Anonymous on
Comment #7323 by Johan on