Lemma 4.38.6 (02Y2)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 4: Categories
Section 4.38: Categories fibred in sets
Lemma 4.38.6 (cite)

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 September 05, 2014 at 13:24

Suggested slogan: Categories fibred in sets are precisely presheaves.

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