Lemma 95.8.1. The functor $p_{ft} : \mathcal{S}\! \mathit{paces}_{ft} \to (\mathit{Sch}/S)_{fppf}$ satisfies the conditions (1), (2) and (3) of Stacks, Definition 8.4.1.

**Proof.**
We are going to write this out in ridiculous detail (which may make it hard to see what is going on).

We have seen in Lemma 95.7.1 that a morphism $(f, g) : X/U \to Y/V$ of $\mathcal{S}\! \mathit{paces}$ is strongly cartesian if the induced morphism $f : X \to U \times _ V Y$ is an isomorphism. Note that if $Y \to V$ is of finite type then also $U \times _ V Y \to U$ is of finite type, see Morphisms of Spaces, Lemma 67.23.3. So if $(f, g) : X/U \to Y/V$ of $\mathcal{S}\! \mathit{paces}$ is strongly cartesian in $\mathcal{S}\! \mathit{paces}$ and $Y/V$ is an object of $\mathcal{S}\! \mathit{paces}_{ft}$ then automatically also $X/U$ is an object of $\mathcal{S}\! \mathit{paces}_{ft}$, and of course $(f, g)$ is also strongly cartesian in $\mathcal{S}\! \mathit{paces}_{ft}$. In this way we conclude that $\mathcal{S}\! \mathit{paces}_{ft}$ is a fibred category over $(\mathit{Sch}/S)_{fppf}$. This proves (1).

The argument above also shows that the inclusion functor $\mathcal{S}\! \mathit{paces}_{ft} \to \mathcal{S}\! \mathit{paces}$ transforms strongly cartesian morphisms into strongly cartesian morphisms. In other words $\mathcal{S}\! \mathit{paces}_{ft} \to \mathcal{S}\! \mathit{paces}$ is a $1$-morphism of fibred categories over $(\mathit{Sch}/S)_{fppf}$.

Let $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. Let $X, Y$ be algebraic spaces of finite type over $U$. By Stacks, Lemma 8.2.3 we obtain a map of presheaves

which is an isomorphism as $\mathcal{S}\! \mathit{paces}_{ft}$ is a full subcategory of $\mathcal{S}\! \mathit{paces}$. Hence the left hand side is a sheaf, because in Lemma 95.7.2 we showed the right hand side is a sheaf. This proves (2).

To prove condition (3) of Stacks, Definition 8.4.1 we have to show the following: Given

a covering $\{ U_ i \to U\} _{i \in I}$ of $(\mathit{Sch}/S)_{fppf}$,

for each $i \in I$ an algebraic space $X_ i$ of finite type over $U_ i$, and

for each $i, j \in I$ an isomorphism $\varphi _{ij} : X_ i \times _ U U_ j \to U_ i \times _ U X_ j$ of algebraic spaces over $U_ i \times _ U U_ j$ satisfying the cocycle condition over $U_ i \times _ U U_ j \times _ U U_ k$,

there exists an algebraic space $X$ of finite type over $U$ and isomorphisms $X_{U_ i} \cong X_ i$ over $U_ i$ recovering the isomorphisms $\varphi _{ij}$. This follows from Bootstrap, Lemma 80.11.3 part (2). By Descent on Spaces, Lemma 74.11.10 we see that $X \to U$ is of finite type which concludes the proof. $\square$

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