Lemma 4.41.1 (2-Yoneda lemma for fibred categories). Let $\mathcal{C}$ be a category. Let $\mathcal{S} \to \mathcal{C}$ be a fibred category over $\mathcal{C}$. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. The functor

$\mathop{\mathrm{Mor}}\nolimits _{\textit{Fib}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}) \longrightarrow \mathcal{S}_ U$

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

Proof. Make a choice of pullbacks for $\mathcal{S}$ (see Definition 4.33.6). We define a functor

$\mathcal{S}_ U \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\textit{Fib}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S})$

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

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

2. 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 4.33.7.

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

Comment #7457 by nhw on

In the first line should "categori" be category?

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