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 79.3 and 79.4.

Definition 86.7.1. 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 86.7.5). 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 86.5.11. By Bootstrap, Lemma 79.4.6 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 86.7.2. 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 86.7.1. 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 86.5.2 and 86.5.11). These maps are jointly surjective hence $U \times _ X V$ is an algebraic space by Bootstrap, Theorem 79.10.1. 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 86.5.2) and an étale morphism, see Descent on Spaces, Lemma 73.18.2. Hence we conclude that $U \times _ X V$ is a scheme by Morphisms of Spaces, Proposition 66.50.2. Thus $\Delta $ is representable, see Spaces, Lemma 64.5.10.

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 64.5.11. Then we can use the principle of Spaces, Lemma 64.5.8 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 86.7.3. 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 64.5.5) and $U \times _ Y Z$ is a scheme by Lemma 86.7.2. Hence the result by Bootstrap, Theorem 79.10.1.
$\square$

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