Lemma 86.11.1. Let $S$ be a scheme. Let $\{ X_ i \to X\} _{i \in I}$ be a family of maps of sheaves on $(\mathit{Sch}/S)_{fppf}$. Assume (a) $X_ i$ is a formal algebraic space over $S$, (b) $X_ i \to X$ is representable by algebraic spaces and étale, and (c) $\coprod X_ i \to X$ is a surjection of sheaves. Then $X$ is a formal algebraic space over $S$.

## 86.11 Fibre products

Obligatory section about fibre products of formal algebraic spaces.

**Proof.**
For each $i$ pick $\{ X_{ij} \to X_ i\} _{j \in J_ i}$ as in Definition 86.7.1. Then $\{ X_{ij} \to X\} _{i \in I, j \in J_ i}$ is a family as in Definition 86.7.1 for $X$.
$\square$

Lemma 86.11.2. Let $S$ be a scheme. Let $X, Y$ be formal algebraic spaces over $S$ and let $Z$ be a sheaf whose diagonal is representable by algebraic spaces. Let $X \to Z$ and $Y \to Z$ be maps of sheaves. Then $X \times _ Z Y$ is a formal algebraic space.

**Proof.**
Choose $\{ X_ i \to X\} $ and $\{ Y_ j \to Y\} $ as in Definition 86.7.1. Then $\{ X_ i \times _ Z Y_ j \to X \times _ Z Y\} $ is a family of maps which are representable by algebraic spaces and étale. Thus Lemma 86.11.1 tells us it suffices to show that $X \times _ Z Y$ is a formal algebraic space when $X$ and $Y$ are affine formal algebraic spaces.

Assume $X$ and $Y$ are affine formal algebraic spaces. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ and $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu $ as in Definition 86.5.1. Then $X \times _ Z Y = \mathop{\mathrm{colim}}\nolimits X_\lambda \times _ Z Y_\mu $. Each $X_\lambda \times _ Z Y_\mu $ is an algebraic space. For $\lambda \leq \lambda '$ and $\mu \leq \mu '$ the morphism

is a thickening as a composition of base changes of thickenings. Thus we conclude by applying Lemma 86.9.1. $\square$

Lemma 86.11.3. Let $S$ be a scheme. The category of formal algebraic spaces over $S$ has fibre products.

**Proof.**
Special case of Lemma 86.11.2 because formal algebraic spaces have representable diagonals, see Lemma 86.7.2.
$\square$

Lemma 86.11.4. Let $S$ be a scheme. Let $X \to Z$ and $Y \to Z$ be morphisms of formal algebraic spaces over $S$. Then $(X \times _ Z Y)_{red} = (X_{red} \times _{Z_{red}} Y_{red})_{red}$.

**Proof.**
This follows from the universal property of the reduction in Lemma 86.8.1.
$\square$

We have already proved the following lemma (without knowing that fibre products exist).

Lemma 86.11.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$. The diagonal morphism $\Delta : X \to X \times _ Y X$ is representable (by schemes), a monomorphism, locally quasi-finite, locally of finite type, and separated.

**Proof.**
Let $T$ be a scheme and let $T \to X \times _ Y X$ be a morphism. Then

Hence the result follows immediately from Lemma 86.7.2. $\square$

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