The Stacks project

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 $\mathcal{C}$. $\square$