Categories fibred in sets are precisely presheaves.
Lemma 4.38.6. Let \mathcal{C} be a category. The only 2-morphisms between categories fibred in sets are identities. In other words, the 2-category of categories fibred in sets is a category. Moreover, there is an equivalence of categories
\left\{ \begin{matrix} \text{the category of presheaves}
\\ \text{of sets over }\mathcal{C}
\end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} \text{the category of categories}
\\ \text{fibred in sets over }\mathcal{C}
\end{matrix} \right\}
The functor from left to right is the construction F \to \mathcal{S}_ F discussed in Example 4.38.5. The functor from right to left assigns to p : \mathcal{S} \to \mathcal{C} the presheaf of objects U \mapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U).
Proof.
The first assertion is clear, as the only morphisms in the fibre categories are identities.
Suppose that p : \mathcal{S} \to \mathcal{C} is fibred in sets. Let f : V \to U be a morphism in \mathcal{C} and let x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U). Then there is exactly one choice for the object f^\ast x. Thus we see that (f \circ g)^\ast x = g^\ast (f^\ast x) for f, g as in Lemma 4.35.2. It follows that we may think of the assignments U \mapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U) and f \mapsto f^\ast as a presheaf on \mathcal{C}.
\square
Comments (1)
Comment #991 by Johan Commelin on