Example 4.36.1. This example is the analogue of Example 4.35.1, for “presheaves of groupoids” instead of “presheaves of categories”. The output will be a category fibred in groupoids instead of a fibred category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Groupoids}$ is a functor to the category of groupoids, see Definition 4.28.5. For $f : V \to U$ in $\mathcal{C}$ we will suggestively write $F(f) = f^\ast $ for the functor from $F(U)$ to $F(V)$. We construct a category $\mathcal{S}_ F$ fibred in groupoids over $\mathcal{C}$ as follows. Define

For $(U, x), (V, y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F)$ we define

In order to define composition we use that $g^\ast \circ f^\ast = (f \circ g)^\ast $ for a pair of composable morphisms of $\mathcal{C}$ (by definition of a functor into a $2$-category). Namely, we define the composition of $\psi : z \to g^\ast y$ and $ \phi : y \to f^\ast x$ to be $ g^\ast (\phi ) \circ \psi $. The functor $p_ F : \mathcal{S}_ F \to \mathcal{C}$ is given by the rule $(U, x) \mapsto U$. The condition that $F(U)$ is a groupoid for every $U$ guarantees that $\mathcal{S}_ F$ is fibred in groupoids over $\mathcal{C}$, as we have already seen in Example 4.35.1 that $\mathcal{S}_ F$ is a fibred category, see Lemma 4.34.2. But we can also prove conditions (1), (2) of Definition 4.34.1 directly as follows: (1) Lifts of morphisms exist since given $f: V \to U$ in $\mathcal{C}$ and $(U, x)$ an object of $\mathcal{S}_ F$ over $U$, then $(f, \text{id}_{f^\ast x}): (V, {f^\ast x}) \to (U, x)$ is a lift of $f$. (2) Suppose given solid diagrams as follows

Then for the dotted arrows we have $\nu = (h^\ast \phi )^{-1} \circ \psi $ so given $h$ there exists a $\nu $ which is unique by uniqueness of inverses.

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