We take a break from our habit of introducing new concepts first for rings, then for schemes, and then for algebraic spaces, by introducing formal algebraic spaces without first introducing formal schemes. The general idea will be that a formal algebraic space is a sheaf in the fppf topology which étale locally is an affine formal scheme in the sense of [BVGD]. Related material can be found in [Yasuda].

In the definition of a formal algebraic space we are going to borrow some terminology from Bootstrap, Sections 80.3 and 80.4.

Definition 87.11.1. Let $S$ be a scheme. We say a sheaf $X$ on $(\mathit{Sch}/S)_{fppf}$ is a *formal algebraic space* if there exist a family of maps $\{ X_ i \to X\} _{i \in I}$ of sheaves such that

$X_ i$ is an affine formal algebraic space,

$X_ i \to X$ is representable by algebraic spaces and étale,

$\coprod X_ i \to X$ is surjective as a map of sheaves

and $X$ satisfies a set theoretic condition (see Remark 87.11.5). A *morphism of formal algebraic spaces* over $S$ is a map of sheaves.

Discussion. Sanity check: an affine formal algebraic space is a formal algebraic space. In the situation of the definition the morphisms $X_ i \to X$ are representable (by schemes), see Lemma 87.9.11. By Bootstrap, Lemma 80.4.6 we could instead of asking $\coprod X_ i \to X$ to be surjective as a map of sheaves, require that it be surjective (which makes sense because it is representable).

Our notion of a formal algebraic space is **very general**. In fact, even affine formal algebraic spaces as defined above are very nasty objects.

Lemma 87.11.2. Let $S$ be a scheme. If $X$ is a formal algebraic space over $S$, then the diagonal morphism $\Delta : X \to X \times _ S X$ is representable, a monomorphism, locally quasi-finite, locally of finite type, and separated.

**Proof.**
Suppose given $U \to X$ and $V \to X$ with $U, V$ schemes over $S$. Then $U \times _ X V$ is a sheaf. Choose $\{ X_ i \to X\} $ as in Definition 87.11.1. For every $i$ the morphism

\[ (U \times _ X X_ i) \times _{X_ i} (V \times _ X X_ i) = (U \times _ X V) \times _ X X_ i \to U \times _ X V \]

is representable and étale as a base change of $X_ i \to X$ and its source is a scheme (use Lemmas 87.9.2 and 87.9.11). These maps are jointly surjective hence $U \times _ X V$ is an algebraic space by Bootstrap, Theorem 80.10.1. The morphism $U \times _ X V \to U \times _ S V$ is a monomorphism. It is also locally quasi-finite, because on precomposing with the morphism displayed above we obtain the composition

\[ (U \times _ X X_ i) \times _{X_ i} (V \times _ X X_ i) \to (U \times _ X X_ i) \times _ S (V \times _ X X_ i) \to U \times _ S V \]

which is locally quasi-finite as a composition of a closed immersion (Lemma 87.9.2) and an étale morphism, see Descent on Spaces, Lemma 74.19.2. Hence we conclude that $U \times _ X V$ is a scheme by Morphisms of Spaces, Proposition 67.50.2. Thus $\Delta $ is representable, see Spaces, Lemma 65.5.10.

In fact, since we've shown above that the morphisms of schemes $U \times _ X V \to U \times _ S V$ are aways monomorphisms and locally quasi-finite we conclude that $\Delta : X \to X \times _ S X$ is a monomorphism and locally quasi-finite, see Spaces, Lemma 65.5.11. Then we can use the principle of Spaces, Lemma 65.5.8 to see that $\Delta $ is separated and locally of finite type. Namely, a monomorphism of schemes is separated (Schemes, Lemma 26.23.3) and a locally quasi-finite morphism of schemes is locally of finite type (follows from the definition in Morphisms, Section 29.20).
$\square$

Lemma 87.11.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism from an algebraic space over $S$ to a formal algebraic space over $S$. Then $f$ is representable by algebraic spaces.

**Proof.**
Let $Z \to Y$ be a morphism where $Z$ is a scheme over $S$. We have to show that $X \times _ Y Z$ is an algebraic space. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Then $U \times _ Y Z \to X \times _ Y Z$ is representable surjective étale (Spaces, Lemma 65.5.5) and $U \times _ Y Z$ is a scheme by Lemma 87.11.2. Hence the result by Bootstrap, Theorem 80.10.1.
$\square$

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