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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