Lemma 99.13.1. The category \mathcal{S}\! \mathit{paces}'_{ft} is fibred in groupoids over \mathit{Sch}_{fppf}. The same is true for \mathcal{S}\! \mathit{paces}'_{fp, flat, proper}.
Proof. We have seen this in Examples of Stacks, Section 95.12 for the case of \mathcal{S}\! \mathit{paces}'_{ft} and this easily implies the result for the other case. However, let us also prove this directly by checking conditions (1) and (2) of Categories, Definition 4.35.1.
Condition (1). Let X \to S be an object of \mathcal{S}\! \mathit{paces}'_{ft} and let S' \to S be a morphism of schemes. Then we set X' = S' \times _ S X. Note that X' \to S' is of finite type by Morphisms of Spaces, Lemma 67.23.3. to obtain a morphism (X' \to S') \to (X \to S) lying over S' \to S. Argue similarly for the other case using Morphisms of Spaces, Lemmas 67.28.3, 67.30.4, and 67.40.3.
Condition (2). Consider morphisms (f, g) : (X' \to S') \to (X \to S) and (a, b) : (Y \to T) \to (X \to S) of \mathcal{S}\! \mathit{paces}'_{ft}. Given a morphism h : T \to S' with g \circ h = b we have to show there is a unique morphism (k, h) : (Y \to T) \to (X' \to S') of \mathcal{S}\! \mathit{paces}'_{ft} such that (f, g) \circ (k, h) = (a, b). This is clear from the fact that X' = S' \times _ S X. The same therefore works for any full subcategory of \mathcal{S}\! \mathit{paces}'_{ft} satisfying (1). \square
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