The Stacks project

Proof. The morphism $\mathcal{F}_{x_0} \to \overline{\mathcal{F}_{x_0}}$ is an equivalence if and only if $\mathcal{F}_{x_0}$ is fibered in setoids, cf. Categories, Section 4.39 (a setoid is by definition a groupoid in which the only automorphism of any object is the identity). We prove that $\text{Inf}_{x_0}(\mathcal{F}) = 0$ if and only if this condition holds for $\mathcal{F}_{x_0}$. Obviously if $\mathcal{F}_{x_0}$ is fibered in setoids then $\text{Inf}_{x_0}(\mathcal{F}) = 0$. Conversely assume $\text{Inf}_{x_0}(\mathcal{F}) = 0$. Let $A$ be an object of $\mathcal{C}_\Lambda $. Then by Lemma 90.19.12, $\text{Inf}(x/x_0) = 0$ for any object $x \to x_0$ of $\mathcal{F}_{x_0}(A)$. Since by definition $\text{Inf}(x/x_0)$ equals the group of automorphisms of $x \to x_0$ in $\mathcal{F}_{x_0}(A)$, this proves $\mathcal{F}_{x_0}(A)$ is a setoid. $\square$