Lemma 94.8.1. 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 94.7.1 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 66.23.3. 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 94.7.2 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 79.11.3 part (2). By Descent on Spaces, Lemma 73.11.10 we see that $X \to U$ is of finite type which concludes the proof. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)