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.
$\square$

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