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The Stacks project

Lemma 4.39.5. Let \mathcal{C} be a category. Let \mathcal{S} be a category over \mathcal{C}.

  1. If \mathcal{S} \to \mathcal{S}' is an equivalence over \mathcal{C} with \mathcal{S}' fibred in sets over \mathcal{C}, then

    1. \mathcal{S} is fibred in setoids over \mathcal{C}, and

    2. for each U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) the map \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U) \to \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}'_ U) identifies the target as the set of isomorphism classes of the source.

  2. If p : \mathcal{S} \to \mathcal{C} is a category fibred in setoids, then there exists a category fibred in sets p' : \mathcal{S}' \to \mathcal{C} and an equivalence \text{can} : \mathcal{S} \to \mathcal{S}' over \mathcal{C}.

Proof. Let us prove (2). An object of the category \mathcal{S}' will be a pair (U, \xi ), where U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) and \xi is an isomorphism class of objects of \mathcal{S}_ U. A morphism (U, \xi ) \to (V , \psi ) is given by a morphism x \to y, where x \in \xi and y \in \psi . Here we identify two morphisms x \to y and x' \to y' if they induce the same morphism U \to V, and if for some choices of isomorphisms x \to x' in \mathcal{S}_ U and y \to y' in \mathcal{S}_ V the compositions x \to x' \to y' and x \to y \to y' agree. By construction there are surjective maps on objects and morphisms from \mathcal{S} \to \mathcal{S}'. We define composition of morphisms in \mathcal{S}' to be the unique law that turns \mathcal{S} \to \mathcal{S}' into a functor. Some details omitted. \square


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