Lemma 4.35.13 (02XT)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 4: Categories
Section 4.35: Categories fibred in groupoids
Lemma 4.35.13 (cite)

Lemma 4.35.13. Let $\mathcal{C}$ be a category. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. If $p : \mathcal{S} \to \mathcal{C}$ is a category fibred in groupoids and $p$ factors through $p' : \mathcal{S} \to \mathcal{C}/U$ then $p' : \mathcal{S} \to \mathcal{C}/U$ is fibred in groupoids.

Proof. We have already seen in Lemma 4.33.11 that $p'$ is a fibred category. Hence it suffices to prove the fibre categories are groupoids, see Lemma 4.35.2. For $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have

\[ \mathcal{S}_ V = \coprod \nolimits _{f : V \to U} \mathcal{S}_{(f : V \to U)} \]

where the left hand side is the fibre category of $p$ and the right hand side is the disjoint union of the fibre categories of $p'$. Hence the result. $\square$

Comments (1)

Comment #11410 by Elías Guisado on May 11, 2026 at 06:43

Here's a generalization:

[Vis2, Proposition 3.28]. Let $p_\mathcal{G}:\mathcal{G}\to\mathcal{C}$ be a fibered category, $\mathcal{F}$ another category, $F:\mathcal{F}\to\mathcal{G}$ a functor. Then $\mathcal{F}$ is fibered in sets (resp., in groupoids) over $\mathcal{G}$ if and only if it is fibered in sets (resp., in groupoids) over $\mathcal{C}$ via the composite $p_\mathcal{G}\circ F:\mathcal{F}\to\mathcal{G}$ .

Proof. Firstly, we have #11409. On the other hand, as the proof of [Vis2, Proposition 3.28] says, “one sees that the fiber of $\mathcal{F}$ over an object $U$ of $\mathcal{C}$ is the disjoint union, as a category, of the fibers of $\mathcal{F}$ over all the objects of $\mathcal{G}$ over $U$ ; these fiber are groupoids, or sets, if and only if their disjoint union is.” $\square$

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