## 64.8 Glueing algebraic spaces

In this section we really start abusing notation and not distinguish between schemes and the spaces they represent.

Lemma 64.8.1. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$. Let $F$ and $G$ be sheaves on $(\mathit{Sch}/S)_{fppf}^{opp}$ and denote $F \amalg G$ the coproduct in the category of sheaves. The map $F \to F \amalg G$ is representable by open and closed immersions.

Proof. Let $U$ be a scheme and let $\xi \in (F \amalg G)(U)$. Recall the coproduct in the category of sheaves is the sheafification of the coproduct presheaf (Sites, Lemma 7.10.13). Thus there exists an fppf covering $\{ g_ i : U_ i \to U\} _{i \in I}$ and a disjoint union decomposition $I = I' \amalg I''$ such that $U_ i \to U \to F \amalg G$ factors through $F$, resp. $G$ if and only if $i \in I'$, resp. $i \in I''$. Since $F$ and $G$ have empty intersection in $F \amalg G$ we conclude that $U_ i \times _ U U_ j$ is empty if $i \in I'$ and $j \in I''$. Hence $U' = \bigcup _{i \in I'} g_ i(U_ i)$ and $U'' = \bigcup _{i \in I''} g_ i(U_ i)$ are disjoint open (Morphisms, Lemma 29.25.10) subschemes of $U$ with $U = U' \amalg U''$. We omit the verification that $U' = U \times _{F \amalg G} F$. $\square$

Lemma 64.8.2. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$. Let $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. Given a set $I$ and sheaves $F_ i$ on $\mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, if $U \cong \coprod _{i\in I} F_ i$ as sheaves, then each $F_ i$ is representable by an open and closed subscheme $U_ i$ and $U \cong \coprod U_ i$ as schemes.

Proof. By Lemma 64.8.1 the map $F_ i \to U$ is representable by open and closed immersions. Hence $F_ i$ is representable by an open and closed subscheme $U_ i$ of $U$. We have $U = \coprod U_ i$ because we have $U \cong \coprod F_ i$ as sheaves and we can test the equality on points. $\square$

Lemma 64.8.3. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$. Let $F$ be an algebraic space over $S$. Given a set $I$ and sheaves $F_ i$ on $\mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, if $F \cong \coprod _{i\in I} F_ i$ as sheaves, then each $F_ i$ is an algebraic space over $S$.

Proof. The representability of $F \to F \times F$ implies that each diagonal morphism $F_ i \to F_ i \times F_ i$ is representable (immediate from the definitions and the fact that $F \times _{(F \times F)} (F_ i \times F_ i) = F_ i$). Choose a scheme $U$ in $(\mathit{Sch}/S)_{fppf}$ and a surjective étale morphism $U \to F$ (this exist by hypothesis). The base change $U \times _ F F_ i \to F_ i$ is surjective and étale by Lemma 64.5.5. On the other hand, $U \times _ F F_ i$ is a scheme by Lemma 64.8.1. Thus we have verified all the conditions in Definition 64.6.1 and $F_ i$ is an algebraic space. $\square$

The condition on the size of $I$ and the $F_ i$ in the following lemma may be ignored by those not worried about set theoretic questions.

Lemma 64.8.4. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$. Suppose given a set $I$ and algebraic spaces $F_ i$, $i \in I$. Then $F = \coprod _{i \in I} F_ i$ is an algebraic space provided $I$, and the $F_ i$ are not too “large”: for example if we can choose surjective étale morphisms $U_ i \to F_ i$ such that $\coprod _{i \in I} U_ i$ is isomorphic to an object of $(\mathit{Sch}/S)_{fppf}$, then $F$ is an algebraic space.

Proof. By construction $F$ is a sheaf. We omit the verification that the diagonal morphism of $F$ is representable. Finally, if $U$ is an object of $(\mathit{Sch}/S)_{fppf}$ isomorphic to $\coprod _{i \in I} U_ i$ then it is straightforward to verify that the resulting map $U \to \coprod F_ i$ is surjective and étale. $\square$

Here is the analogue of Schemes, Lemma 26.15.4.

Lemma 64.8.5. 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 64.5.1),

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 64.8.4).

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 64.5.1 (see Remark 64.5.2 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 64.5.10. 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 64.8.2. 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 64.5.4. 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). $\square$

Comment #599 by yogesh on

In lemma 47.8.4 condition 2.4, the way it's written sounds like the map $\coprod F_i \to F$ is surjective as a map of fppf sheaves. I'm wondering if that's what you meant, or if you mean surjective in the sense of Definition 47.5.1. I'm suspecting the later because in the proof the latter would imply the $V_i$ cover $T$.

A separate comment: in the paragraph in the proof beginning "To show diagonal is representable ..." it might be helpful to the novice if you to cite lemma-representable-diagonal showing why it is equivalent to show $G$ is representable.

Comment #602 by on

For your first comment/question: It seems to me that here we do mean surjective as a map of sheaves, exactly because we say "map of sheaves $\ldots$ is surjective". Namely, at the exact moment we introduce this condition it hasn't yet been stated that $\coprod F_i \to F$ is representable (of course I do realize this is a trivial consequence of 2, but still) and hence it does not yet make sense to use the notion as in Definition 64.5.1. OK?

I made the change you suggested for the proof, see here.

Comment #611 by yogesh on

Ok, thanks for the revision, it makes sense to me now and I see what I didn't know - it seems a representable morphism of sheaves $F \to G$ that is surjective as a map of sheaves is then surjective as a representable morphism i.e. in the sense of 025V. If so, it would be good to mention it in tag 02YN, although it now seems obvious. What's an example of a representable morphism $f:F \to G$ where $F, G$ are sheaves, and $f$ is surjective as a representable morphism but not surjective as a morphisms of sheaves?

In the case at hand I think the two types of surjectivity conditions on $\coprod F_i \to F$ are equivalent because you can apply Lemma \cite[Tag 05VM]{stacks-project} - correct me if I am wrong.

Incidentally in the proof you refer to 2b and 2c etc instead of 2(2) etc but it's clear what you meant.

Comment #616 by on

OK the problem with 2(2) and 2(b) etc comes from the web-interface and will be fixed in the future. In the pdf the numbering should be correct.

Comment #3519 by Laurent Moret-Bailly on

For both 02WO and 02WP, one could just observe that (without any assumption on $U$) the canonical $F_i\to\coprod_j F_j$ is representable by an open and closed immersion. In fact, it is the pullback of the section $1:S\to(\mathbb{Z}/2\mathbb{Z})_S$ via the "characteristic function of $F_i$" $\chi_i:F\to (\mathbb{Z}/2\mathbb{Z})_S$ deduced from the universal property of the sum.

Comment #3659 by on

OK, yes, this is true. I added a lemma stating the fact you are using, namely, that the inclusion of a summand in a coproduct of sheaves is representable and an open and closed immersion. Then I was able to use that directly in the proof of 64.8.2. But in the proof of 64.8.3 I was not able to use it directly, because the lemma stating that something representable over an algebraic space is an algebraic space comes later in the Stacks project. See changes here.

Comment #4355 by oregontrailmixtape on

A small typo: in the second paragraph of the proof of 02WR, it says "By the above the $G_i$ form a Zariski covering of $F$." Shouldn't it say ""By the above the $G_i$ form a Zariski covering of $G$."?

Comment #5994 by Hanlin on

Minor typo: In the first line of the proof of Lemma 63.8.1, it should be $\xi \in (F \coprod G)(U)$ instead of $\xi \in (F \coprod G)(\xi)$.

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