Lemma 8.12.7. With notation and assumptions as in Lemma 8.12.6. If $\mathcal{S}$ is fibred in groupoids, then $u_ p\mathcal{S}$ is fibred in groupoids.

Proof. By Lemma 8.12.6 we know that $u_ p\mathcal{S}$ is a fibred category. Let $f : X \to Y$ be a morphism of $u_ p\mathcal{S}$ with $p_ p(f) = \text{id}_ V$. We are done if we can show that $f$ is invertible, see Categories, Lemma 4.35.2. Write $f$ as the equivalence class of a pair $((a, b, \alpha ), r)$ with $r \in R$. Then $p_ p(r) = \text{id}_ V$, hence $p_{pp}((a, b, \alpha )) = \text{id}_ V$. Hence $b = \text{id}_ V$. But any morphism of $\mathcal{S}$ is strongly cartesian, see Categories, Lemma 4.35.2 hence we see that $(a, b, \alpha ) \in R$ is invertible in $u_ p\mathcal{S}$ as desired. $\square$

There are also:

• 2 comment(s) on Section 8.12: Functoriality for stacks

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).