Lemma 95.9.6. Let $S$ be a scheme. Let $p : \mathcal{X} \to \mathcal{Y}$ and $q : \mathcal{Z} \to \mathcal{Y}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If the functor (95.9.3.1) is an equivalence for $\mathcal{X}$, $\mathcal{Y}$, and $\mathcal{Z}$, then it is an equivalence for $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$.

Proof. The left and the right hand side of (95.9.3.1) for $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ are simply the $2$-fibre products of the left and the right hand side of (95.9.3.1) for $\mathcal{X}$, $\mathcal{Z}$ over $\mathcal{Y}$. Hence the result follows as taking $2$-fibre products is compatible with equivalences of categories, see Categories, Lemma 4.30.7. $\square$

Comment #2984 by Tanya Kaushal Srivastava on

Small typo in the statement of lemma: then it is and equivalence ---> then it is an equivalence.

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