ScalingStacks

Proof. If $\alpha\in B_{n}$, then $\alpha\sim 0$ and we are done. Assume now $\alpha\in A_{n}$. We proceed by induction on $M(\alpha)=\frac{1}{2}|L(\alpha_{|(-n,-1)})|$ and then on $N(\alpha)=n-\max\{i\ |\ [-n+i-1\to-n+i]\in L(\alpha)\}$ if $M(\alpha)\neq 0$ to show that there exists $\sigma\in E_{n}$ with $\alpha\sim\sigma$.

If $M(\alpha)=0$, then $\alpha\in E_{n}$ and we are done. Assume now $M(\alpha)>0$. By Lemma 8.2.4, there are $i\in(1,n-1)$ and $\beta\in G_{n}$ such that $\alpha=\beta T_{i}$, and we choose $i$ maximal with this property, so that $N(\alpha)=n-i$. We have $\alpha\sim T_{i}\beta$. If $T_{i}\beta\in B_{n}$ then we are done. We assume now $T_{i}\beta{\not\in}B_{n}$. We have $L(\beta_{|(-n,-1)})=L(\alpha_{|(-n,-1)})\setminus\{[-n+i-1\to-n+i],[-n+i\to-n+i-1]\}$.

If $\beta^{-1}(\{i,i+1\}){\not\subset}(-n,-1)$, then $L(T_{i}\beta_{|(-n,-1)})=L(\beta_{|(-n,-1)})$, hence $M(T_{i}\beta)<M(\alpha)$. By induction, there is $\sigma\in E_{n}$ with $T_{i}\beta\sim\sigma$, hence $\alpha\sim\sigma$.

Assume now there are $j,k\in(1,n)$ with $\beta(-n+j-1)=i$ and $\beta(-n+k-1)=i+1$. Since $T_{i}\beta\neq 0$, we have $j<k$. Since $\beta\in A_{n}$, we have $j>i$ and $k>i+1$. We have $M(T_{i}\beta)\leq M(\beta)+1=M(\alpha)$. On the other hand, $[j\to k]\in L(T_{i}\beta)$ (cf Lemma 7.4.20), hence $N(T_{i}\beta)<N(\alpha)$. We conclude by induction.

The case of $F_{n}$ follows by applying Remark 8.2.6. ∎