## 92.9 Morphisms representable by algebraic spaces

In analogy with Categories, Definition 4.41.5 we make the following definition.

Definition 92.9.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. A $1$-morphism $f : \mathcal{X} \to \mathcal{Y}$ of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$ is called representable by algebraic spaces if for any $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and any $y : (\mathit{Sch}/U)_{fppf} \to \mathcal{Y}$ the category fibred in groupoids

$(\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$

over $(\mathit{Sch}/U)_{fppf}$ is representable by an algebraic space over $U$.

Choose an algebraic space $F_ y$ over $U$ which represents $(\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$. We may think of $F_ y$ as an algebraic space over $S$ which comes equipped with a canonical morphism $f_ y : F_ y \to U$ over $S$, see Spaces, Section 63.16. Here is the diagram

92.9.1.1
$$\label{algebraic-equation-representable-by-algebraic-spaces} \vcenter { \xymatrix{ F_ y \ar[d]_{f_ y} & (\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X} \ar@{~>}[l] \ar[d]_{\text{pr}_0} \ar[r]_-{\text{pr}_1} & \mathcal{X} \ar[d]^ f \\ U & (\mathit{Sch}/U)_{fppf} \ar@{~>}[l] \ar[r]^-y & \mathcal{Y} } }$$

where the squiggly arrows represent the construction which associates to a stack fibred in setoids its associated sheaf of isomorphism classes of objects. The right square is $2$-commutative, and is a $2$-fibre product square.

Here is the analogue of Categories, Lemma 4.41.7.

Lemma 92.9.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. The following are necessary and sufficient conditions for $f$ to be representable by algebraic spaces:

1. for each scheme $U/S$ the functor $f_ U : \mathcal{X}_ U \longrightarrow \mathcal{Y}_ U$ between fibre categories is faithful, and

2. for each $U$ and each $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$ the presheaf

$(h : V \to U) \longmapsto \{ (x, \phi ) \mid x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ V), \phi : h^*y \to f(x)\} /\cong$

is an algebraic space over $U$.

Here we have made a choice of pullbacks for $\mathcal{Y}$.

Proof. This follows from the description of fibre categories of the $2$-fibre products $(\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$ in Categories, Lemma 4.41.3 combined with Lemma 92.8.2. $\square$

Lemma 92.9.3. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Consider a $2$-commutative diagram

$\xymatrix{ \mathcal{X}' \ar[r] \ar[d]_{f'} & \mathcal{X} \ar[d]^ f \\ \mathcal{Y}' \ar[r] & \mathcal{Y} }$

of $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume the horizontal arrows are equivalences. Then $f$ is representable by algebraic spaces if and only if $f'$ is representable by algebraic spaces.

Proof. Omitted. $\square$

Lemma 92.9.4. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $S$. If $\mathcal{X}$ and $\mathcal{Y}$ are representable by algebraic spaces over $S$, then the $1$-morphism $f$ is representable by algebraic spaces.

Proof. Omitted. This relies only on the fact that the category of algebraic spaces over $S$ has fibre products, see Spaces, Lemma 63.7.3. $\square$

Lemma 92.9.5. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Let $a : F \to G$ be a map of presheaves of sets on $(\mathit{Sch}/S)_{fppf}$. Denote $a' : \mathcal{S}_ F \to \mathcal{S}_ G$ the associated map of categories fibred in sets. Then $a$ is representable by algebraic spaces (see Bootstrap, Definition 78.3.1) if and only if $a'$ is representable by algebraic spaces.

Proof. Omitted. $\square$

Lemma 92.9.6. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in setoids over $(\mathit{Sch}/S)_{fppf}$. Let $F$, resp. $G$ be the presheaf which to $T$ associates the set of isomorphism classes of objects of $\mathcal{X}_ T$, resp. $\mathcal{Y}_ T$. Let $a : F \to G$ be the map of presheaves corresponding to $f$. Then $a$ is representable by algebraic spaces (see Bootstrap, Definition 78.3.1) if and only if $f$ is representable by algebraic spaces.

Proof. Omitted. Hint: Combine Lemmas 92.9.3 and 92.9.5. $\square$

Lemma 92.9.7. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism representable by algebraic spaces. Let $g : \mathcal{Z} \to \mathcal{Y}$ be any $1$-morphism. Consider the fibre product diagram

$\xymatrix{ \mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X} \ar[r]_-{g'} \ar[d]_{f'} & \mathcal{X} \ar[d]^ f \\ \mathcal{Z} \ar[r]^ g & \mathcal{Y} }$

Then the base change $f'$ is a $1$-morphism representable by algebraic spaces.

Proof. This is formal. $\square$

Lemma 92.9.8. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$ Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Z} \to \mathcal{Y}$ be $1$-morphisms. Assume

1. $f$ is representable by algebraic spaces, and

2. $\mathcal{Z}$ is representable by an algebraic space over $S$.

Then the $2$-fibre product $\mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X}$ is representable by an algebraic space.

Proof. This is a reformulation of Bootstrap, Lemma 78.3.6. First note that $\mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X}$ is fibred in setoids over $(\mathit{Sch}/S)_{fppf}$. Hence it is equivalent to $\mathcal{S}_ F$ for some presheaf $F$ on $(\mathit{Sch}/S)_{fppf}$, see Categories, Lemma 4.39.5. Moreover, let $G$ be an algebraic space which represents $\mathcal{Z}$. The $1$-morphism $\mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$ is representable by algebraic spaces by Lemma 92.9.7. And $\mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$ corresponds to a morphism $F \to G$ by Categories, Lemma 4.39.6. Then $F \to G$ is representable by algebraic spaces by Lemma 92.9.6. Hence Bootstrap, Lemma 78.3.6 implies that $F$ is an algebraic space as desired. $\square$

Let $S$, $\mathcal{X}$, $\mathcal{Y}$, $\mathcal{Z}$, $f$, $g$ be as in Lemma 92.9.8. Let $F$ and $G$ be algebraic spaces over $S$ such that $F$ represents $\mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X}$ and $G$ represents $\mathcal{Z}$. The $1$-morphism $f' : \mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X} \to \mathcal{Z}$ corresponds to a morphism $f' : F \to G$ of algebraic spaces by (92.8.2.1). Thus we have the following diagram

92.9.8.1
$$\label{algebraic-equation-representable-by-algebraic-spaces-on-space} \vcenter { \xymatrix{ F \ar[d]_{f'} & \mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X} \ar@{~>}[l] \ar[d] \ar[r] & \mathcal{X} \ar[d]^ f \\ G & \mathcal{Z} \ar@{~>}[l] \ar[r]^-g & \mathcal{Y} } }$$

where the squiggly arrows represent the construction which associates to a stack fibred in setoids its associated sheaf of isomorphism classes of objects.

Lemma 92.9.9. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ are $1$-morphisms representable by algebraic spaces, then

$g \circ f : \mathcal{X} \longrightarrow \mathcal{Z}$

is a $1$-morphism representable by algebraic spaces.

Proof. This follows from Lemma 92.9.8. Details omitted. $\square$

Lemma 92.9.10. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}_ i, \mathcal{Y}_ i$ be categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$, $i = 1, 2$. Let $f_ i : \mathcal{X}_ i \to \mathcal{Y}_ i$, $i = 1, 2$ be $1$-morphisms representable by algebraic spaces. Then

$f_1 \times f_2 : \mathcal{X}_1 \times \mathcal{X}_2 \longrightarrow \mathcal{Y}_1 \times \mathcal{Y}_2$

is a $1$-morphism representable by algebraic spaces.

Proof. Write $f_1 \times f_2$ as the composition $\mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{Y}_2$. The first arrow is the base change of $f_1$ by the map $\mathcal{Y}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1$, and the second arrow is the base change of $f_2$ by the map $\mathcal{Y}_1 \times \mathcal{Y}_2 \to \mathcal{Y}_2$. Hence this lemma is a formal consequence of Lemmas 92.9.9 and 92.9.7. $\square$

Lemma 92.9.11. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X} \to \mathcal{Z}$ and $\mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $\mathcal{X} \to \mathcal{Z}$ is representable by algebraic spaces and $\mathcal{Y}$ is a stack in groupoids, then $\mathcal{X} \times _\mathcal {Z} \mathcal{Y}$ is a stack in groupoids.

Proof. The property of a morphism being representable by algebraic spaces is preserved under base-change (Lemma 92.9.8), and so, passing to the base-change $\mathcal{X} \times _\mathcal {Z} \mathcal{Y}$ over $\mathcal{Y}$, we may reduce to the case of a morphism of categories fibred in groupoids $\mathcal{X} \to \mathcal{Y}$ which is representable by algebraic spaces, and whose target is a stack in groupoids; our goal is then to prove that $\mathcal{X}$ is also a stack in groupoids. This follows from Stacks, Lemma 8.6.11 whose assumptions are satisfied as a result of Lemma 92.9.2. $\square$

Comment #2248 by David Hansen on

The use of "middle square" in reference to part of diagram 77.9.8.1 is confusing (cf. second line below the diagram) - I only see left and right squares. :)

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).