The Stacks project

Lemma 63.8.5. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$. Let $F$ be a presheaf of sets on $(\mathit{Sch}/S)_{fppf}$. Assume

  1. $F$ is a sheaf,

  2. there exists an index set $I$ and subfunctors $F_ i \subset F$ such that

    1. each $F_ i$ is an algebraic space,

    2. each $F_ i \to F$ is representable,

    3. each $F_ i \to F$ is an open immersion (see Definition 63.5.1),

    4. the map $\coprod F_ i \to F$ is surjective as a map of sheaves, and

    5. $\coprod F_ i$ is an algebraic space (set theoretic condition, see Lemma 63.8.4).

Then $F$ is an algebraic space.

Proof. Let $T$ be an object of $(\mathit{Sch}/S)_{fppf}$. Let $T \to F$ be a morphism. By assumption (2)(b) and (2)(c) the fibre product $F_ i \times _ F T$ is representable by an open subscheme $V_ i \subset T$. It follows that $(\coprod F_ i) \times _ F T$ is represented by the scheme $\coprod V_ i$ over $T$. By assumption (2)(d) there exists an fppf covering $\{ T_ j \to T\} _{j \in J}$ such that $T_ j \to T \to F$ factors through $F_ i$, $i = i(j)$. Hence $T_ j \to T$ factors through the open subscheme $V_{i(j)} \subset T$. Since $\{ T_ j \to T\} $ is jointly surjective, it follows that $T = \bigcup V_ i$ is an open covering. In particular, the transformation of functors $\coprod F_ i \to F$ is representable and surjective in the sense of Definition 63.5.1 (see Remark 63.5.2 for a discussion).

Next, let $T' \to F$ be a second morphism from an object in $(\mathit{Sch}/S)_{fppf}$. Write as above $T' = \bigcup V'_ i$ with $V'_ i = T' \times _ F F_ i$. To show that the diagonal $F \to F \times F$ is representable we have to show that $G = T \times _ F T'$ is representable, see Lemma 63.5.10. Consider the subfunctors $G_ i = G \times _ F F_ i$. Note that $G_ i = V_ i \times _{F_ i} V'_ i$, and hence is representable as $F_ i$ is an algebraic space. By the above the $G_ i$ form a Zariski covering of $G$. Hence by Schemes, Lemma 26.15.4 we see $G$ is representable.

Choose a scheme $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a surjective étale morphism $U \to \coprod F_ i$ (this exists by hypothesis). We may write $U = \coprod U_ i$ with $U_ i$ the inverse image of $F_ i$, see Lemma 63.8.2. We claim that $U \to F$ is surjective and étale. Surjectivity follows as $\coprod F_ i \to F$ is surjective (see first paragraph of the proof) by applying Lemma 63.5.4. Consider the fibre product $U \times _ F T$ where $T \to F$ is as above. We have to show that $U \times _ F T \to T$ is étale. Since $U \times _ F T = \coprod U_ i \times _ F T$ it suffices to show each $U_ i \times _ F T \to T$ is étale. Since $U_ i \times _ F T = U_ i \times _{F_ i} V_ i$ this follows from the fact that $U_ i \to F_ i$ is étale and $V_ i \to T$ is an open immersion (and Morphisms, Lemmas 29.36.9 and 29.36.3). $\square$


Comments (3)

Comment #3521 by Laurent Moret-Bailly on

Typo in (2)(b): "is a representable".

Comment #3590 by Jeroen van der Meer on

Typo in proof: "this exist" -> "this exists" or "these exist".

There are also:

  • 8 comment(s) on Section 63.8: Glueing algebraic spaces

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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02WR. Beware of the difference between the letter 'O' and the digit '0'.