Proof.
Assume (2). The category \mathcal{I}_\mathcal {S} has objects (x, \alpha ) where x \in \mathcal{S}, say with p(x) = U, and \alpha : x \to x is a morphism in \mathcal{S}_ U. Hence if \mathcal{I}_\mathcal {S} \to \mathcal{S} is an equivalence over \mathcal{C} then every pair of objects (x, \alpha ), (x, \alpha ') are isomorphic in the fibre category of \mathcal{I}_\mathcal {S} over U. Looking at the definition of morphisms in \mathcal{I}_\mathcal {S} we conclude that \alpha , \alpha ' are conjugate in the group of automorphisms of x. Hence taking \alpha ' = \text{id}_ x we conclude that every automorphism of x is equal to the identity. Since \mathcal{S} \to \mathcal{C} is fibred in groupoids this implies that \mathcal{S} \to \mathcal{C} is fibred in setoids. We omit the proof of (1) \Rightarrow (2).
\square
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