Example 4.36.1. Let $\mathcal{C}$ be a category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Cat}$ is a functor to the $2$-category of categories, see Definition 4.29.5. For $f : V \to U$ in $\mathcal{C}$ we will suggestively write $F(f) = f^\ast $ for the functor from $F(U)$ to $F(V)$. From this we can construct a fibred category $\mathcal{S}_ F$ 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 $(U, x) \mapsto U$. Let us check that this is indeed a fibred category. Given $f: V \to U$ in $\mathcal{C}$ and $(U, x)$ a lift of $U$, then we claim $(f, \text{id}_{f^\ast x}): (V, {f^\ast x}) \to (U, x)$ is a strongly cartesian lift of $f$. We have to show a $h$ in the diagram on the left determines $(h, \nu )$ on the right:

Just take $\nu = \psi $ which works because $f \circ h = g$ and hence $g^*x = h^*f^*x$. Moreover, this is the only lift making the diagram (on the right) commute.

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