Lemma 98.9.6. 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 (98.9.3.1) 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 (98.9.3.1) for $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ are simply the $2$-fibre products of the left and the right hand side of (98.9.3.1) 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 4.31.7.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2984 by Tanya Kaushal Srivastava on

Comment #3108 by Johan on

There are also: