ScalingStacks

Proof. Put $\sigma^{\prime}=\sigma T_{i}$ and assume $\sigma^{\prime}\in C_{n}\setminus\{0\}$. There are $\gamma\in C_{n-i-1}$, $\beta\in C_{2}$ and $\alpha\in C_{i-1}$ such that $\sigma^{\prime}=\alpha\ast\beta\ast\gamma$.

Lemma 8.2.4 shows that $[-n+i-1\to-n+i]\in D(\sigma^{\prime})$. We have $\sigma^{\prime}_{|\{-n+i-1,-n+i\}}=(\alpha\ast\beta)_{|(-i-1,-i)}\circ([-n+i-1\to-i-1]\boxtimes[-n+i\to-i])$. It follows from Lemma 8.2.4 that $[-i-1\to-i]\in D(\alpha\ast\beta)$. Since $[-i-1\to-i]\in L((\alpha\ast\beta)_{|(-i-1,-i)})$, it follows that $\beta(-1)\neq 1$, hence $\beta\in D_{2}$.

$\bullet\$Assume $[-1\to-2]\in D(\beta)$. We have $\beta=\beta^{\prime}T_{1}$ for some $\beta^{\prime}\in G_{2}$ by Lemma 8.2.4. Since $\beta\in D_{2}$, we have $\beta^{\prime}\in D_{2}\subset A_{2}$. We deduce that $\beta^{\prime}\in E_{2}$, hence $T_{1}\beta^{\prime}\in E_{2}\subset C_{2}$ (Corollary 8.2.13). So, $\sigma T_{i}=\alpha\ast(\beta^{\prime}T_{1})\ast\gamma\sim^{\prime}\alpha\ast(T_{1}\beta^{\prime})\ast\gamma=T_{i}\sigma$.

$\bullet\$Assume now $[-1\to-2]{\not\in}D(\beta)$, i.e., $\beta\in E_{2}$. We have $T_{1}\beta,\beta T_{1}\in D_{2}\subset A_{2}$ and $T_{1}\beta\subset E_{2}\subset C_{2}$ (Corollary 8.2.13).

$\ \ \diamond\$Assume $T_{1}\beta=0$. There is $\beta^{\prime\prime}\in G_{2}$ such that $\beta=T_{1}\beta^{\prime\prime}$ (Lemma 8.2.4). Since $\beta\in E_{2}\cap D_{2}$, we have $\beta^{\prime\prime}\in E_{2}\cap D_{2}\subset C_{2}$, hence also $\beta^{\prime\prime}\in F_{2}$. As a consequence, $\beta^{\prime\prime}T_{1}\in F_{2}\subset C_{2}$. We deduce that $\alpha\ast\beta\sim^{\prime}\alpha\ast(\beta^{\prime\prime}T_{1})$. We have $L(\alpha_{|\beta((-2,-1))})\neq\emptyset$ and $L((\beta^{\prime\prime}T_{1})_{|(-2,-1)})\neq\emptyset$, hence $(\alpha\ast(\beta^{\prime\prime}T_{1}))_{|(-2,-1)}=0$ and $\alpha\ast(\beta^{\prime\prime}T_{1})=0$. We have $\sigma T_{i}=\alpha\ast(T_{1}\beta^{\prime\prime})\ast\gamma\sim^{\prime}\alpha\ast(\beta^{\prime\prime}T_{1})\ast\gamma=0$. Since $T_{i}\sigma T_{i}=0$ and $\sigma T_{i}\neq 0$, it follows that $L((\sigma T_{i})_{|(\sigma T_{i})^{-1}(\{i,i+1\})})\neq\emptyset$, by applying Lemma 8.2.4 to $Z^{{\operatorname{opp}\nolimits}}$. Since $\sigma T_{i}\in A_{n}$, we deduce that $L(\sigma_{|\sigma^{-1}(\{i,i+1\})})\neq\emptyset$, hence $T_{i}\sigma=0\sim^{\prime}\sigma T_{i}$ (using Lemma 8.2.4 for $Z^{{\operatorname{opp}\nolimits}}$ again).

$\ \ \diamond\$Assume now $T_{1}\beta\neq 0$. It follows that $\beta\in F_{2}$, hence $\beta T_{1}\in F_{2}\subset C_{2}$.

There are $\alpha^{1},\ldots,\alpha^{i-1}\in C_{1}$ with $\alpha=\alpha^{i-1}\ast\cdots\ast\alpha^{1}$. Let $s_{i}=\beta(-i)$ for $i\in\{1,2\}$. Consider $j\geq 1$ minimal such that $L((\alpha^{j}\ast\cdots\ast\alpha^{1})_{|\{s_{1},s_{2}\}})\neq\emptyset$.

Define $u^{\prime}=\alpha^{j}\boxtimes([l\to l+1])_{1\leq l\leq j+1}$ and $u^{\prime\prime}=(\alpha^{j-1}\ast\cdots\alpha^{1}\ast\beta)\boxtimes[-j-2\to-1]$.

Let $\zeta=u^{\prime\prime}_{-2}\circ[-1\to-2]\circ(u^{\prime\prime}_{-1})^{-1}$. Define $I$ and $J$ to be the domain and codomain of $u^{\prime\prime}$, intersected with $M$. Note that $\zeta(0),\zeta(1)\in M$. Let $v^{\prime}=(u^{\prime})^{\zeta}=(\alpha^{j})^{\zeta}\boxtimes([l\to l+1])_{1\leq l\leq j+1}$ and define $v^{\prime\prime}:I\sqcup(-j-2,-1)\to J\sqcup(1,2)\sqcup\{-1\}\sqcup(1,j+1)$ by

$$v^{\prime\prime}_{s}=\begin{cases}u^{\prime\prime}_{-2}\circ[-1\to-2]&\text{ if }s=-1\\ u^{\prime\prime}_{-1}\circ[-2\to-1]&\text{ if }s=-2\\ u^{\prime\prime}_{s}&\text{ otherwise.}\end{cases}$$

Lemma 7.4.35 shows that $v^{\prime}$ and $v^{\prime\prime}$ are braids and $\alpha^{j}\ast\cdots\ast\alpha^{1}\ast\beta=u^{\prime}\cdot u^{\prime\prime}=v^{\prime}\cdot v^{\prime\prime}$. We have $v^{\prime\prime}=(\alpha^{j-1}\ast\cdots\alpha^{1}\ast(\beta T_{1}))\boxtimes[-j-2\mapsto-1]$ and we deduce that $\alpha\ast\beta=\alpha^{\prime}\ast(\beta T_{1})$, where $\alpha^{\prime}=\alpha^{i-1}\ast\cdots\ast\alpha^{j+1}\ast(\alpha^{j})^{\zeta}\ast\alpha^{j-1}\cdots\ast\alpha^{1}\in C_{i-1}$. We have $\sigma T_{i}=\alpha^{\prime}\ast(\beta T_{1})\ast\gamma\sim^{\prime}\alpha^{\prime}\ast(T_{1}\beta)\ast\gamma=T_{i}\sigma$. This completes the proof of the first statement of the lemma.

The second statement of the lemma follows from the first one applied to $Z^{\operatorname{opp}\nolimits}$ thanks to Remark 8.2.6. ∎