Lemma 65.8.5. 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 65.5.1),

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 65.8.4).

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 65.5.1 (see Remark 65.5.2 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 65.5.10. 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 65.8.2. 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 65.5.4. 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:

• 10 comment(s) on Section 65.8: Glueing algebraic spaces

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