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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