ScalingStacks

Proof. Let us show the first statement. We can assume $\theta_{\zeta(0)}^{\zeta}\neq\operatorname{id}\nolimits$.

Assume $\theta_{\zeta(0)}^{\zeta}$ has the same orientation as $\zeta^{-1}$. We have $\theta_{\zeta(1)}=\theta_{\zeta(0)}^{\zeta}\circ\zeta^{-1}$. If $\gamma$ and $\gamma^{\prime}$ are minimal paths in $\theta_{\zeta(0)}^{\zeta}$ and $\zeta^{-1}$, then $\gamma\circ\gamma^{\prime}$ is a minimal path in $\theta_{\zeta(1)}$. Since $\gamma\circ\gamma^{\prime}$ is admissible, it follows that $\gamma$ is admissible, hence $\theta_{\zeta(0)}^{\zeta}$ is admissible.

Otherwise, $\theta_{\zeta(0)}^{\zeta}$ has the same orientation as $\bar{\zeta}^{-1}$ and $\theta_{\zeta(0)}=\bar{\zeta}^{-1}\circ\theta_{\zeta(0)}^{\zeta}$, hence we deduce as above that $\theta_{\zeta(0)}^{\zeta}$ is admissible.

Similarly, $\theta_{\zeta(1)}^{\zeta}$ is admissible and we deduce that $\theta^{\zeta}$ is a braid.

Let us prove the second part of the lemma. When $Z=S^{1}$ unoriented, this holds by Lemmas 7.4.20, 6.2.9 and 7.4.19 and Proposition 7.4.18. We deduce that the lemma holds when $Z$ is a connected non-singular curve, by embedding $Z$ in $S^{1}$. So, it holds when $Z$ is a non-singular curve (since $\operatorname{supp}\nolimits(\zeta)$ is contained in a connected component of $Z$).

Consider now a general $Z$ and the non-singular cover $q:\hat{Z}\to Z$. There is a braid $\hat{\theta}$ in $\hat{Z}$ with $q(\hat{\theta})=\theta$ (Lemma 7.4.5) and there is $\hat{\zeta}\in L(\hat{\theta})$ such that $\zeta=q(\hat{\zeta})$ (Lemma 7.4.28). The considerations above show that $\hat{\theta}^{\hat{\zeta}}$ is a braid in $\hat{Z}$, hence $\theta^{\zeta}=q(\hat{\theta}^{\hat{\zeta}})$ is a braid in $Z$. The statement on degrees follows from Lemmas 7.4.28 and 7.4.12. ∎