The Stacks project

Example 4.35.5. Let $\mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) = \{ A, B, T\}$ and $\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(A, B) = \{ f\}$, $\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(B, T) = \{ g\}$, $\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(A, T) = \{ h\} = \{ gf\} ,$ plus the identity morphism for each object. See the diagram below for a picture of this category. Now let $\mathop{\mathrm{Ob}}\nolimits (\mathcal{S}) = \{ A', B', T'\}$ and $\mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(A', B') = \emptyset$, $\mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(B', T') = \{ g'\}$, $\mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(A', T') = \{ h'\} ,$ plus the identity morphisms. The functor $p : \mathcal{S} \to \mathcal{C}$ is obvious. Then for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $\mathcal{S}_ U$ is the category with one object and the identity morphism on that object, so a groupoid, but the morphism $f: A \to B$ cannot be lifted. Similarly, if we declare $\mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(A', B') = \{ f'_1, f'_2\}$ and $\mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(A', T') = \{ h'\} = \{ g'f'_1 \} = \{ g'f'_2\}$, then the fibre categories are the same and $f: A \to B$ in the diagram below has two lifts.