Let $\mathcal{C}$ be a category. The $2$-category of fibred categories over $\mathcal{C}$ was constructed/defined in Definition 4.33.9. If $\mathcal{S}$, $\mathcal{S}'$ are fibred categories over $\mathcal{C}$ then
denotes the category of $1$-morphisms in this $2$-category. Here is the $2$-category analogue of the Yoneda lemma in the setting of fibred categories.
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 given by $G \mapsto G(\text{id}_ U)$ is an equivalence.
\[ \mathop{\mathrm{Mor}}\nolimits _{\textit{Fib}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}) \longrightarrow \mathcal{S}_ U \]
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 4.33.7.
We omit the verification that this is an inverse to the functor of the lemma. $\square$
Let $\mathcal{C}$ be a category. The $2$-category of categories fibred in groupoids over $\mathcal{C}$ is a “full” sub $2$-category of the $2$-category of categories over $\mathcal{C}$ (see Definition 4.35.6). Hence if $\mathcal{S}$, $\mathcal{S}'$ are fibred in groupoids over $\mathcal{C}$ then
denotes the category of $1$-morphisms in this $2$-category (see Definition 4.32.1). These are all groupoids, see remarks following Definition 4.35.6. Here is the $2$-category analogue of the Yoneda lemma.
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.
\[ \mathop{\mathrm{Mor}}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}) \longrightarrow \mathcal{S}_ U \]
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. $\square$
Remark 4.41.3. We can use the $2$-Yoneda lemma to give an alternative proof of Lemma 4.37.3. Let $p : \mathcal{S} \to \mathcal{C}$ be a category fibred in groupoids. We define a contravariant functor $F$ from $\mathcal{C}$ to the category of groupoids as follows: for $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ let If $f : U \to V$ the induced functor $\mathcal{C}/U \to \mathcal{C}/V$ induces the morphism $F(f) : F(V) \to F(U)$. Clearly $F$ is a functor. Let $\mathcal{S}'$ be the associated category fibred in groupoids from Example 4.37.1. There is an obvious functor $G : \mathcal{S}' \to \mathcal{S}$ over $\mathcal{C}$ given by taking the pair $(U, x)$, where $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x \in F(U)$, to $x(\text{id}_ U) \in \mathcal{S}$. Now Lemma 4.41.2 implies that for each $U$, is an equivalence, and thus $G$ is an equivalence between $\mathcal{S}$ and $\mathcal{S}'$ by Lemma 4.35.9.
\[ F(U) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}). \]
\[ G_ U : \mathcal{S}'_ U = F(U)= \mathop{\mathrm{Mor}}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}) \to \mathcal{S}_ U \]
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