The Stacks project

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

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 66.16.6 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

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

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 94.10.11 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

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

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

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 94.9.6, 94.9.7, 94.10.4, and 94.10.6. Therefore $F$ is an algebraic space by Bootstrap, Theorem 80.10.1 and we win. $\square$

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 $\beta = \text{id}_ x$. $\square$