## 94.10 Properties of morphisms representable by algebraic spaces

Here is the definition that makes this work.

Definition 94.10.1. 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 $(\mathit{Sch}/S)_{fppf}$. Assume $f$ is representable by algebraic spaces. Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which

is preserved under any base change, and

is fppf local on the base, see Descent on Spaces, Definition 74.10.1.

In this case we say that $f$ has *property $\mathcal{P}$* if for every $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and any $y \in \mathcal{Y}_ U$ the resulting morphism of algebraic spaces $f_ y : F_ y \to U$, see diagram (94.9.1.1), has property $\mathcal{P}$.

It is important to note that we will only use this definition for properties of morphisms that are stable under base change, and local in the fppf topology on the target. This is not because the definition doesn't make sense otherwise; rather it is because we may want to give a different definition which is better suited to the property we have in mind.

Lemma 94.10.2. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Let $\mathcal{P}$ be as in Definition 94.10.1. 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 and $f$ (or equivalently $f'$) is representable by algebraic spaces. Then $f$ has $\mathcal{P}$ if and only if $f'$ has $\mathcal{P}$.

**Proof.**
Note that this makes sense by Lemma 94.9.3. Proof omitted.
$\square$

Here is a sanity check.

Lemma 94.10.3. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $a : F \to G$ be a map of presheaves on $(\mathit{Sch}/S)_{fppf}$. Let $\mathcal{P}$ be as in Definition 94.10.1. Assume $a$ is representable by algebraic spaces. Then $a : F \to G$ has property $\mathcal{P}$ (see Bootstrap, Definition 80.4.1) if and only if the corresponding morphism $\mathcal{S}_ F \to \mathcal{S}_ G$ of categories fibred in groupoids has property $\mathcal{P}$.

**Proof.**
Note that the lemma makes sense by Lemma 94.9.5. Proof omitted.
$\square$

Lemma 94.10.4. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Let $\mathcal{P}$ be as in Definition 94.10.1. 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 $a$ has $\mathcal{P}$ if and only if $f$ has $\mathcal{P}$.

**Proof.**
The lemma makes sense by Lemma 94.9.6. The lemma follows on combining Lemmas 94.10.2 and 94.10.3.
$\square$

Lemma 94.10.5. 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 $\mathcal{P}$ be a property as in Definition 94.10.1 which is stable under composition. Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms which are representable by algebraic spaces. If $f$ and $g$ have property $\mathcal{P}$ so does $g \circ f : \mathcal{X} \to \mathcal{Z}$.

**Proof.**
Note that the lemma makes sense by Lemma 94.9.9. Proof omitted.
$\square$

Lemma 94.10.6. 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 $\mathcal{P}$ be a property as in Definition 94.10.1. 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 $2$-fibre product diagram

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

If $f$ has $\mathcal{P}$, then the base change $f'$ has $\mathcal{P}$.

**Proof.**
The lemma makes sense by Lemma 94.9.7. Proof omitted.
$\square$

Lemma 94.10.7. 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 $\mathcal{P}$ be a property as in Definition 94.10.1. 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

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

Assume that for every scheme $U$ and object $x$ of $\mathcal{Y}_ U$, there exists an fppf covering $\{ U_ i \to U\} $ such that $x|_{U_ i}$ is in the essential image of the functor $g : \mathcal{Z}_{U_ i} \to \mathcal{Y}_{U_ i}$. In this case, if $f'$ has $\mathcal{P}$, then $f$ has $\mathcal{P}$.

**Proof.**
Proof omitted. Hint: Compare with the proof of Spaces, Lemma 65.5.6.
$\square$

Lemma 94.10.8. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{P}$ be a property as in Definition 94.10.1 which is stable under composition. 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. If $f_1$ and $f_2$ have property $\mathcal{P}$ so does $ f_1 \times f_2 : \mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{Y}_2 $.

**Proof.**
The lemma makes sense by Lemma 94.9.10. Proof omitted.
$\square$

Lemma 94.10.9. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}$, $\mathcal{Y}$ 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 $\mathcal{P}$, $\mathcal{P}'$ be properties as in Definition 94.10.1. Suppose that for any morphism of algebraic spaces $a : F \to G$ we have $\mathcal{P}(a) \Rightarrow \mathcal{P}'(a)$. If $f$ has property $\mathcal{P}$ then $f$ has property $\mathcal{P}'$.

**Proof.**
Formal.
$\square$

Lemma 94.10.10. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $j : \mathcal X \to \mathcal Y$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume $j$ is representable by algebraic spaces and a monomorphism (see Definition 94.10.1 and Descent on Spaces, Lemma 74.11.30). Then $j$ is fully faithful on fibre categories.

**Proof.**
We have seen in Lemma 94.9.2 that $j$ is faithful on fibre categories. Consider a scheme $U$, two objects $u, v$ of $\mathcal{X}_ U$, and an isomorphism $t : j(u) \to j(v)$ in $\mathcal{Y}_ U$. We have to construct an isomorphism in $\mathcal{X}_ U$ between $u$ and $v$. By the $2$-Yoneda lemma (see Section 94.5) we think of $u$, $v$ as $1$-morphisms $u, v : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ and we consider the $2$-fibre product

\[ (\mathit{Sch}/U)_{fppf} \times _{j \circ v, \mathcal{Y}} \mathcal{X}. \]

By assumption this is representable by an algebraic space $F_{j \circ v}$, over $U$ and the morphism $F_{j \circ v} \to U$ is a monomorphism. But since $(1_ U, v, 1_{j(v)})$ gives a $1$-morphism of $(\mathit{Sch}/U)_{fppf}$ into the displayed $2$-fibre product, we see that $F_{j \circ v} = U$ (here we use that if $V \to U$ is a monomorphism of algebraic spaces which has a section, then $V = U$). Therefore the $1$-morphism projecting to the first coordinate

\[ (\mathit{Sch}/U)_{fppf} \times _{j \circ v, \mathcal{Y}} \mathcal{X} \to (\mathit{Sch}/U)_{fppf} \]

is an equivalence of fibre categories. Since $(1_ U, u, t)$ and $(1_ U, v, 1_{j(v)})$ give two objects in $((\mathit{Sch}/U)_{fppf} \times _{j \circ v, \mathcal{Y}} \mathcal{X})_ U$ which have the same first coordinate, there must be a $2$-morphism between them in the $2$-fibre product. This is by definition a morphism $\tilde t : u \to v$ such that $j(\tilde t) = t$.
$\square$

Here is a characterization of those categories fibred in groupoids for which the diagonal is representable by algebraic spaces.

Lemma 94.10.11. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. The following are equivalent:

the diagonal $\mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces,

for every scheme $U$ over $S$, and any $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$ the sheaf $\mathit{Isom}(x, y)$ is an algebraic space over $U$,

for every scheme $U$ over $S$, and any $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$ the associated $1$-morphism $x : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ is representable by algebraic spaces,

for every pair of schemes $T_1, T_2$ over $S$, and any $x_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_{T_ i})$, $i = 1, 2$ the $2$-fibre product $(\mathit{Sch}/T_1)_{fppf} \times _{x_1, \mathcal{X}, x_2} (\mathit{Sch}/T_2)_{fppf}$ is representable by an algebraic space,

for every representable category fibred in groupoids $\mathcal{U}$ over $(\mathit{Sch}/S)_{fppf}$ every $1$-morphism $\mathcal{U} \to \mathcal{X}$ is representable by algebraic spaces,

for every pair $\mathcal{T}_1, \mathcal{T}_2$ of representable categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$ and any $1$-morphisms $x_ i : \mathcal{T}_ i \to \mathcal{X}$, $i = 1, 2$ the $2$-fibre product $\mathcal{T}_1 \times _{x_1, \mathcal{X}, x_2} \mathcal{T}_2$ is representable by an algebraic space,

for every category fibred in groupoids $\mathcal{U}$ over $(\mathit{Sch}/S)_{fppf}$ which is representable by an algebraic space every $1$-morphism $\mathcal{U} \to \mathcal{X}$ is representable by algebraic spaces,

for every pair $\mathcal{T}_1, \mathcal{T}_2$ of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$ which are representable by algebraic spaces, and any $1$-morphisms $x_ i : \mathcal{T}_ i \to \mathcal{X}$ the $2$-fibre product $\mathcal{T}_1 \times _{x_1, \mathcal{X}, x_2} \mathcal{T}_2$ is representable by an algebraic space.

**Proof.**
The equivalence of (1) and (2) follows from Stacks, Lemma 8.2.5 and the definitions. Let us prove the equivalence of (1) and (3). Write $\mathcal{C} = (\mathit{Sch}/S)_{fppf}$ for the base category. We will use some of the observations of the proof of the similar Categories, Lemma 4.42.6. We will use the symbol $\cong $ to mean “equivalence of categories fibred in groupoids over $\mathcal{C} = (\mathit{Sch}/S)_{fppf}$”. Assume (1). Suppose given $U$ and $x$ as in (3). For any scheme $V$ and $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ V)$ we see (compare reference above) that

\[ \mathcal{C}/U \times _{x, \mathcal{X}, y} \mathcal{C}/V \cong (\mathcal{C}/U \times _ S V) \times _{(x, y), \mathcal{X} \times \mathcal{X}, \Delta } \mathcal{X} \]

which is representable by an algebraic space by assumption. Conversely, assume (3). Consider any scheme $U$ over $S$ and a pair $(x, x')$ of objects of $\mathcal{X}$ over $U$. We have to show that $\mathcal{X} \times _{\Delta , \mathcal{X} \times \mathcal{X}, (x, x')} U$ is representable by an algebraic space. This is clear because (compare reference above)

\[ \mathcal{X} \times _{\Delta , \mathcal{X} \times \mathcal{X}, (x, x')} \mathcal{C}/U \cong (\mathcal{C}/U \times _{x, \mathcal{X}, x'} \mathcal{C}/U) \times _{\mathcal{C}/U \times _ S U, \Delta } \mathcal{C}/U \]

and the right hand side is representable by an algebraic space by assumption and the fact that the category of algebraic spaces over $S$ has fibre products and contains $U$ and $S$.

The equivalences (3) $\Leftrightarrow $ (4), (5) $\Leftrightarrow $ (6), and (7) $\Leftrightarrow $ (8) are formal. The equivalences (3) $\Leftrightarrow $ (5) and (4) $\Leftrightarrow $ (6) follow from Lemma 94.9.3. Assume (3), and let $\mathcal{U} \to \mathcal{X}$ be as in (7). To prove (7) we have to show that for every scheme $V$ and $1$-morphism $y : (\mathit{Sch}/V)_{fppf} \to \mathcal{X}$ the $2$-fibre product $(\mathit{Sch}/V)_{fppf} \times _{y, \mathcal{X}} \mathcal{U}$ is representable by an algebraic space. Property (3) tells us that $y$ is representable by algebraic spaces hence Lemma 94.9.8 implies what we want. Finally, (7) directly implies (3).
$\square$

In the situation of the lemma, for any $1$-morphism $x : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ as in the lemma, it makes sense to say that $x$ has property $\mathcal{P}$, for any property as in Definition 94.10.1. In particular this holds for $\mathcal{P} = $ “surjective”, $\mathcal{P} = $ “smooth”, and $\mathcal{P} = $ “étale”, see Descent on Spaces, Lemmas 74.11.6, 74.11.26, and 74.11.28. We will use these three cases in the definitions of algebraic stacks below.

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