
Lemma 86.10.11. 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:

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

2. 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 $U$,

3. 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,

4. 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,

5. 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,

6. 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,

7. 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,

8. 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 4.40.8. 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 86.9.3. 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 86.9.8 implies what we want. Finally, (7) directly implies (3). $\square$

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