Example 4.38.5. This example is the analogue of Examples 4.36.1 and 4.37.1 for presheaves instead of “presheaves of categories”. The output will be a category fibred in sets instead of a fibred category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Sets}$ is a presheaf. For $f : V \to U$ in $\mathcal{C}$ we will suggestively write $F(f) = f^\ast : F(U) \to F(V)$. We construct a category $\mathcal{S}_ F$ fibred in sets over $\mathcal{C}$ as follows. Define

$\mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F) = \{ (U, x) \mid U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), x \in \mathop{\mathrm{Ob}}\nolimits (F(U))\} .$

For $(U, x), (V, y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F)$ we define

\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ F}((V, y), (U, x)) & = \{ f \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U) \mid f^*x = y\} \end{align*}

Composition is inherited from composition in $\mathcal{C}$ which works as $g^\ast \circ f^\ast = (f \circ g)^\ast$ for a pair of composable morphisms of $\mathcal{C}$. The functor $p_ F : \mathcal{S}_ F \to \mathcal{C}$ is given by the rule $(U, x) \mapsto U$. As every fibre category $\mathcal{S}_{F, U}$ is discrete with underlying set $F(U)$ and we have already see in Example 4.37.1 that $\mathcal{S}_ F$ is a category fibred in groupoids, we conclude that $\mathcal{S}_ F$ is fibred in sets.

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