Lemma 98.5.3. 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 \mathcal{X}, \mathcal{Y}, and \mathcal{Z} satisfy (RS), then so does \mathcal{X} \times _\mathcal {Y} \mathcal{Z}.
Proof. This is formal. Let
\xymatrix{ X \ar[r] \ar[d] & X' \ar[d] \\ Y \ar[r] & Y' = Y \amalg _ X X' }
be a diagram as in Definition 98.5.1. We have to show that
(\mathcal{X} \times _{\mathcal{Y}} \mathcal{Z})_{Y'} \longrightarrow (\mathcal{X} \times _{\mathcal{Y}} \mathcal{Z})_ Y \times _{(\mathcal{X} \times _{\mathcal{Y}} \mathcal{Z})_ X} (\mathcal{X} \times _{\mathcal{Y}} \mathcal{Z})_{X'}
is an equivalence. Using the definition of the 2-fibre product this becomes
98.5.3.1
\begin{equation} \label{artin-equation-RS-fibre-product} \mathcal{X}_{Y'} \times _{\mathcal{Y}_{Y'}} \mathcal{Z}_{Y'} \longrightarrow (\mathcal{X}_ Y \times _{\mathcal{Y}_ Y} \mathcal{Z}_ Y) \times _{(\mathcal{X}_ X \times _{\mathcal{Y}_ X} \mathcal{Z}_ X)} (\mathcal{X}_{X'} \times _{\mathcal{Y}_{X'}} \mathcal{Z}_{X'}). \end{equation}
We are given that each of the functors
\mathcal{X}_{Y'} \to \mathcal{X}_ Y \times _{\mathcal{Y}_ Y} \mathcal{Z}_ Y, \quad \mathcal{Y}_{Y'} \to \mathcal{X}_ X \times _{\mathcal{Y}_ X} \mathcal{Z}_ X, \quad \mathcal{Z}_{Y'} \to \mathcal{X}_{X'} \times _{\mathcal{Y}_{X'}} \mathcal{Z}_{X'}
are equivalences. An object of the right hand side of (98.5.3.1) is a system
((x_ Y, z_ Y, \phi _ Y), (x_{X'}, z_{X'}, \phi _{X'}), (\alpha , \beta )).
Then (x_ Y, x_{Y'}, \alpha ) is isomorphic to the image of an object x_{Y'} in \mathcal{X}_{Y'} and (z_ Y, z_{Y'}, \beta ) is isomorphic to the image of an object z_{Y'} of \mathcal{Z}_{Y'}. The pair of morphisms (\phi _ Y, \phi _{X'}) corresponds to a morphism \psi between the images of x_{Y'} and z_{Y'} in \mathcal{Y}_{Y'}. Then (x_{Y'}, z_{Y'}, \psi ) is an object of the left hand side of (98.5.3.1) mapping to the given object of the right hand side. This proves that (98.5.3.1) is essentially surjective. We omit the proof that it is fully faithful. \square
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