Lemma 94.9.8. 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
$f$ is representable by algebraic spaces, and
$\mathcal{Z}$ 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 80.3.6. 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 94.9.7. 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 94.9.6. Hence Bootstrap, Lemma 80.3.6 implies that $F$ is an algebraic space as desired.
$\square$
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