Lemma 65.8.1. 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.

## 65.8 Glueing algebraic spaces

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

**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 65.8.2. 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 65.8.1 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 65.8.3. 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 65.5.5. On the other hand, $U \times _ F F_ i$ is a scheme by Lemma 65.8.1. Thus we have verified all the conditions in Definition 65.6.1 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 65.8.4. 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 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

$F$ is a sheaf,

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

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). $\square$

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