ScalingStacks

Proof. We prove the lemma by induction on $|\theta|$. Assume there is a set $I^{\prime}$ satisfying the assumptions of Lemma 7.4.26 and such that $\mu(u)=0$. By induction, there is a decomposition $\theta^{u}=r^{\prime}(\theta^{u})\cdot r(\theta^{u})$ as in the lemma. Now $r(\theta)=r(\theta^{u})\cdot u$ and $r^{\prime}(\theta)=r^{\prime}(\theta^{u})$ satisfy the requirements of the lemma.

Assume now that given any set $I^{\prime}$ satisfying the assumptions of Lemma 7.4.26, we have $\mu(u)\geq 1$.

Let $s\in I$ with $\mu(\theta_{s})\geq 1$ such that given $s^{\prime}\in I$ with $\mu(\theta_{s^{\prime}})\geq 1$, we have $\mathrm{supp}(\theta_{s}^{r})\subset\mathrm{supp}(\theta_{s^{\prime}}^{r})$. Given $s^{\prime}\in I\setminus\{s\}$, we have $\mu(\alpha^{s^{\prime}})=0$ (notations of §7.4.5).

Let $I^{\prime}$ be the set of $s^{\prime}\in I$ such that there is a sequence $s_{0}=s,s_{1},\ldots,s_{r}=s^{\prime}$ of elements of $I$ such that $s_{i}\to s_{i+1}$ is an arrow of $\Gamma(\theta)$ for $0\leq i<r$. Assume there exist $s_{1},\ldots,s_{d}$ in $I^{\prime}\setminus\{s_{0}\}$ such that $s_{d}=s_{1}$ and $s_{i}\to s_{i+1}$ is an arrow of $\Gamma(\theta)$ for $1\leq i<d$. Then $I^{\prime\prime}=\{s_{1},\ldots,s_{d}\}$ satisfies the assumptions of Lemma 7.4.26. On the other hand, we have $\mu(\alpha^{s^{\prime}})=0$ for $s^{\prime}\in I^{\prime\prime}$, hence we get a contradiction. It follows that $I^{\prime}$ is a cycle or a line and it satisfies the assumptions of Lemma 7.4.26. The braids $r^{\prime}(\theta)=\theta^{u}$ and $r(\theta)=u$ of Lemma 7.4.26 satisfy the requirements of the lemma. ∎