## 96.4 Morphisms of stacks in groupoids

This section is preliminary and should be skipped on a first reading.

Lemma 96.4.1. Let $\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $\mathcal{X} \to \mathcal{Z}$ and $\mathcal{Y} \to \mathcal{Z}$ are representable by algebraic spaces and étale so is $\mathcal{X} \to \mathcal{Y}$.

Proof. Let $\mathcal{U}$ be a representable category fibred in groupoids over $S$. Let $f : \mathcal{U} \to \mathcal{Y}$ be a $1$-morphism. We have to show that $\mathcal{X} \times _\mathcal {Y} \mathcal{U}$ is representable by an algebraic space and étale over $\mathcal{U}$. Consider the composition $h : \mathcal{U} \to \mathcal{Z}$. Then

$\mathcal{X} \times _\mathcal {Z} \mathcal{U} \longrightarrow \mathcal{Y} \times _\mathcal {Z} \mathcal{U}$

is a $1$-morphism between categories fibres in groupoids which are both representable by algebraic spaces and both étale over $\mathcal{U}$. Hence by Properties of Spaces, Lemma 65.16.6 this is represented by an étale morphism of algebraic spaces. Finally, we obtain the result we want as the morphism $f$ induces a morphism $\mathcal{U} \to \mathcal{Y} \times _\mathcal {Z} \mathcal{U}$ and we have

$\mathcal{X} \times _\mathcal {Y} \mathcal{U} = (\mathcal{X} \times _\mathcal {Z} \mathcal{U}) \times _{(\mathcal{Y} \times _\mathcal {Z} \mathcal{U})} \mathcal{U}.$
$\square$

Lemma 96.4.2. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. Suppose that $\mathcal{X} \to \mathcal{Y}$ and $\mathcal{Z} \to \mathcal{Y}$ are $1$-morphisms. If

1. $\mathcal{Y}$, $\mathcal{Z}$ are representable by algebraic spaces $Y$, $Z$ over $S$,

2. the associated morphism of algebraic spaces $Y \to Z$ is surjective, flat and locally of finite presentation, and

3. $\mathcal{Y} \times _\mathcal {Z} \mathcal{X}$ is a stack in setoids,

then $\mathcal{X}$ is a stack in setoids.

Proof. This is a special case of Stacks, Lemma 8.6.10. $\square$

The following lemma is the analogue of Algebraic Stacks, Lemma 93.15.3 and will be superseded by the stronger Theorem 96.16.1.

Lemma 96.4.3. Let $S$ be a scheme. Let $u : \mathcal{U} \to \mathcal{X}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If

1. $\mathcal{U}$ is representable by an algebraic space, and

2. $u$ is representable by algebraic spaces, surjective, flat and locally of finite presentation,

then $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ representable by algebraic spaces.

Proof. Given two schemes $T_1$, $T_2$ over $S$ denote $\mathcal{T}_ i = (\mathit{Sch}/T_ i)_{fppf}$ the associated representable fibre categories. Suppose given $1$-morphisms $f_ i : \mathcal{T}_ i \to \mathcal{X}$. According to Algebraic Stacks, Lemma 93.10.11 it suffices to prove that the $2$-fibered product $\mathcal{T}_1 \times _\mathcal {X} \mathcal{T}_2$ is representable by an algebraic space. By Stacks, Lemma 8.6.8 this is in any case a stack in setoids. Thus $\mathcal{T}_1 \times _\mathcal {X} \mathcal{T}_2$ corresponds to some sheaf $F$ on $(\mathit{Sch}/S)_{fppf}$, see Stacks, Lemma 8.6.3. Let $U$ be the algebraic space which represents $\mathcal{U}$. By assumption

$\mathcal{T}_ i' = \mathcal{U} \times _{u, \mathcal{X}, f_ i} \mathcal{T}_ i$

is representable by an algebraic space $T'_ i$ over $S$. Hence $\mathcal{T}_1' \times _\mathcal {U} \mathcal{T}_2'$ is representable by the algebraic space $T'_1 \times _ U T'_2$. Consider the commutative diagram

$\xymatrix{ & \mathcal{T}_1 \times _{\mathcal X} \mathcal{T}_2 \ar[rr]\ar '[d][dd] & & \mathcal{T}_1 \ar[dd] \\ \mathcal{T}_1' \times _\mathcal {U} \mathcal{T}_2' \ar[ur]\ar[rr]\ar[dd] & & \mathcal{T}_1' \ar[ur]\ar[dd] \\ & \mathcal{T}_2 \ar '[r][rr] & & \mathcal X \\ \mathcal{T}_2' \ar[rr]\ar[ur] & & \mathcal{U} \ar[ur] }$

In this diagram the bottom square, the right square, the back square, and the front square are $2$-fibre products. A formal argument then shows that $\mathcal{T}_1' \times _\mathcal {U} \mathcal{T}_2' \to \mathcal{T}_1 \times _{\mathcal X} \mathcal{T}_2$ is the “base change” of $\mathcal{U} \to \mathcal{X}$, more precisely the diagram

$\xymatrix{ \mathcal{T}_1' \times _\mathcal {U} \mathcal{T}_2' \ar[d] \ar[r] & \mathcal{U} \ar[d] \\ \mathcal{T}_1 \times _{\mathcal X} \mathcal{T}_2 \ar[r] & \mathcal{X} }$

is a $2$-fibre square. Hence $T'_1 \times _ U T'_2 \to F$ is representable by algebraic spaces, flat, locally of finite presentation and surjective, see Algebraic Stacks, Lemmas 93.9.6, 93.9.7, 93.10.4, and 93.10.6. Therefore $F$ is an algebraic space by Bootstrap, Theorem 79.10.1 and we win. $\square$

Lemma 96.4.4. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. The following are equivalent

1. $\Delta _\Delta : \mathcal{X} \to \mathcal{X} \times _{\mathcal{X} \times \mathcal{X}} \mathcal{X}$ is representable by algebraic spaces,

2. for every $1$-morphism $\mathcal{V} \to \mathcal{X} \times \mathcal{X}$ with $\mathcal{V}$ representable (by a scheme) the fibre product $\mathcal{Y} = \mathcal{X} \times _{\Delta , \mathcal{X} \times \mathcal{X}} \mathcal{V}$ has diagonal representable by algebraic spaces.

Proof. Although this is a bit of a brain twister, it is completely formal. Namely, recall that $\mathcal{X} \times _{\mathcal{X} \times \mathcal{X}} \mathcal{X} = \mathcal{I}_\mathcal {X}$ is the inertia of $\mathcal{X}$ and that $\Delta _\Delta$ is the identity section of $\mathcal{I}_\mathcal {X}$, see Categories, Section 4.34. Thus condition (1) says the following: Given a scheme $V$, an object $x$ of $\mathcal{X}$ over $V$, and a morphism $\alpha : x \to x$ of $\mathcal{X}_ V$ the condition “$\alpha = \text{id}_ x$” defines an algebraic space over $V$. (In other words, there exists a monomorphism of algebraic spaces $W \to V$ such that a morphism of schemes $f : T \to V$ factors through $W$ if and only if $f^*\alpha = \text{id}_{f^*x}$.)

On the other hand, let $V$ be a scheme and let $x, y$ be objects of $\mathcal{X}$ over $V$. Then $(x, y)$ define a morphism $\mathcal{V} = (\mathit{Sch}/V)_{fppf} \to \mathcal{X} \times \mathcal{X}$. Next, let $h : V' \to V$ be a morphism of schemes and let $\alpha : h^*x \to h^*y$ and $\beta : h^*x \to h^*y$ be morphisms of $\mathcal{X}_{V'}$. Then $(\alpha , \beta )$ define a morphism $\mathcal{V}' = (\mathit{Sch}/V)_{fppf} \to \mathcal{Y} \times \mathcal{Y}$. Condition (2) now says that (with any choices as above) the condition “$\alpha = \beta$” defines an algebraic space over $V$.

To see the equivalence, given $(\alpha , \beta )$ as in (2) we see that (1) implies that “$\alpha ^{-1} \circ \beta = \text{id}_{h^*x}$” defines an algebraic space. The implication (2) $\Rightarrow$ (1) follows by taking $h = \text{id}_ V$ and $\beta = \text{id}_ x$. $\square$

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