ScalingStacks

Proof. Let $\alpha:S\sqcup(-n,-1)\to T\sqcup(1,n)$ be a non-zero element of $F_{n}$. Let $s\in S^{\prime\prime}$. Given $1\leq m\leq n$, we put $i_{m}(s)=\beta^{m-1}\circ\cdots\circ\beta^{1}\circ\alpha(s)$. We put $d_{s}=\min\{m|i_{m+1}(s)\in T_{m+1}\}$.

Let $s,s^{\prime}$ be two distinct elements of $S$ and let $\theta=\beta^{n}_{|\beta^{n-1}\circ\cdots\circ\beta^{1}\circ\alpha(\{s,s^{\prime}\})}\cdots\beta^{1}_{|\alpha(\{s,s^{\prime}\})}\cdot\alpha_{|\{s,s^{\prime}\}}$.

$\bullet\$If $s,s^{\prime}\in S^{\prime}$, then $\theta=\alpha_{|\{s,s^{\prime}\}}\neq 0$.

$\bullet\$Assume $s\in S^{\prime}$ and $s^{\prime}\in S^{\prime\prime}$. We have $\theta=\theta^{1}\cdot\alpha_{|\{s,s^{\prime}\}}$ where

$$\theta^{1}=(\operatorname{id}\nolimits_{\alpha(s)}\boxtimes(\alpha_{-n+i_{d_{s^{\prime}}}(s^{\prime})-1}\cdot[1\to-n+i_{d_{s^{\prime}}}(s^{\prime})-1]\cdot\kappa^{d_{s^{\prime}}-1}\cdot[i_{1}(s^{\prime})\to 1])).$$

We have

$$i(\theta^{1}\circ\alpha_{|\{s,s^{\prime}\}})=i(\alpha_{s},\alpha_{s^{\prime}})+i(\alpha_{s},\alpha_{-n+i_{d_{s^{\prime}}}(s^{\prime})-1})+d_{s^{\prime}}-1=i(\alpha_{|\{s,s^{\prime}\}})+i(\theta^{1}).$$

Ir follows that $\theta\neq 0$.

$\bullet\$Assume finally $s,s^{\prime}\in S^{\prime\prime}$ and $d_{s^{\prime}}\geq d_{s}$. We have $\theta=\theta^{1}\cdot\theta^{2}\cdot\theta^{3}\cdot\alpha_{|\{s,s^{\prime}\}}$ where

$$\theta^{1}=(\alpha_{-n+i_{d_{s^{\prime}}}(s^{\prime})-1}\cdot[1\to-n+i_{d_{s^{\prime}}}(s^{\prime})-1]\cdot\kappa^{d_{s^{\prime}}-d_{s}}\cdot[i_{d_{s}+1}(s^{\prime})\to 1])\boxtimes\operatorname{id}\nolimits_{i_{n}(s)}$$

$$\theta^{2}=[-n+i_{d_{s}}(s^{\prime})+1\to i_{d_{s}+1}(s^{\prime})]\boxtimes(\alpha_{n-i_{d_{s}}(s)+1}\cdot[-n\to n-i_{d_{s}}(s)+1])$$

$$\theta^{3}=(([1\to-n+i_{d_{s}}(s^{\prime})+1]\cdot\kappa^{d_{s}-1}\cdot[i_{1}(s^{\prime})\to 1])\boxtimes([1\to-n]\cdot\kappa^{d_{s}-1}\cdot[i_{1}(s)\to 1])).$$

We have

$$i(\theta^{1}\circ\theta^{2}\circ\theta^{3}\circ\alpha_{|\{s,s^{\prime}\}})=i(\alpha_{i_{d_{s}}(s)},\alpha_{i_{d_{s^{\prime}}}(s^{\prime})})+d_{s^{\prime}}-d_{s}+i(\alpha_{s},\alpha_{s^{\prime}})=i(\theta^{1})+i(\theta^{2})+i(\theta^{3})+i(\alpha_{|\{s,s^{\prime}\}}).$$

It follows that $\theta\neq 0$.

It follows from Remark 7.4.11 that $q(\alpha)\neq 0$.

Define $S^{\prime}$ and $S^{\prime\prime}$ as above. Let $r=|S^{\prime\prime}|$. We have $\alpha(S^{\prime\prime})=(n-r+1,n)$ and $\alpha^{-1}(i)<\alpha^{-1}(i^{\prime})$ for $i<i^{\prime}$ in $(n-r+1,n)$.

Given $i<i^{\prime}$ in $(-n,-1)$ with $\alpha(i),\alpha(i^{\prime})\in(1,n)$, we have $\alpha(i)<\alpha(i^{\prime})$.

Consider now $\tilde{\alpha}:S\sqcup(-n,-1)\to T\sqcup(1,n)$ another non-zero element of $F_{n}$ and assume $q(\alpha)=q(\tilde{\alpha})\neq 0$. We have $S\cap\tilde{\alpha}^{-1}(T)=S^{\prime}$ and $T\cap\tilde{\alpha}(S)=T^{\prime}$. The discussion above shows that $\alpha(s)=\tilde{\alpha}(s)$ for $s\in S^{\prime\prime}$. Note also that $\alpha_{s}=\tilde{\alpha}_{s}$ for $s\in S^{\prime}$. As a consequence, $\alpha=\tilde{\alpha}$ if $\mu(q(\alpha))=0$.

Let $s\in S^{\prime\prime}$, $t=q(\alpha)(s)$, $\tilde{t}=q(\tilde{\alpha})(s)$, $i=\alpha^{-1}(t)$ and $\tilde{i}=\tilde{\alpha}^{-1}(t)$. Since $q(\alpha)_{s}=q(\tilde{\alpha})_{s}$, it follows that $\tilde{t}=t$, $\mu(q(\alpha)_{s})=\mu(q(\tilde{\alpha})_{s})$, $[\alpha(s)\to 1]\cdot\alpha_{s}=[\tilde{\alpha}(s)\to 1]\cdot\tilde{\alpha}_{s}$, and $\alpha_{i}\cdot[-1\to i]=\tilde{\alpha}_{\tilde{i}}\cdot[-1\to\tilde{i}]$. We deduce that $\tilde{\alpha}_{s}=\alpha_{s}$ for $s\in S^{\prime\prime}$.

We proceed now by induction on $\mu(q(\alpha))$ to show that $q(\alpha)$ determines $\alpha$, for $\alpha\in F$.

Assume there is $s\in S^{\prime\prime}$ such that $\mu(q(\alpha)_{s})=1$. Let $j=\alpha(s)\in(1,n)$ and $i=-n+j-1$. We have $t=\alpha(i)=q(\alpha)(s)\in T$. Define $\alpha^{\prime}:S\setminus\{s\}\sqcup(-n+1,-1)\to T\setminus\{t\}\sqcup(1,n-1)$ an element of $F_{n-1}$ as follows. Given $s^{\prime}\in S\setminus\{s\}$, we put $\alpha^{\prime}_{s^{\prime}}=\alpha_{s^{\prime}}$ if $\alpha(s^{\prime})<j$, $\alpha^{\prime}_{s^{\prime}}=[\alpha(s^{\prime})\to\alpha(s^{\prime})-1]\cdot\alpha_{s^{\prime}}$ if $\alpha(s^{\prime})>j$. Given $i^{\prime}\in(-i+1,-1)$, we put $\alpha^{\prime}_{i^{\prime}}=\alpha_{i^{\prime}}$. Given $i^{\prime}\in(-n+1,-i)$, we put $\alpha^{\prime}_{i^{\prime}}=\alpha_{i^{\prime}-1}$. This defines an element of $F_{n-1}$. Furthermore, $q(\alpha^{\prime})=q(\alpha)_{|S\setminus\{s\}}$.

We define similarly $\tilde{i}$, $\tilde{j}$, $\tilde{t}$ and $\tilde{\alpha}^{\prime}$ starting with $\tilde{\alpha}$ and $s$. We have $\tilde{j}=j$ and $\tilde{t}=t$, hence also $\tilde{i}=i$. We have $q(\alpha^{\prime})=q(\tilde{\alpha}^{\prime})$, hence $\alpha^{\prime}=\tilde{\alpha}^{\prime}$ by induction. Since $\alpha_{s}=\tilde{\alpha}_{s}$ and $\alpha_{i}=\tilde{\alpha}_{i}$, it follows that $\alpha=\tilde{\alpha}$.

Assume $\mu(q(\alpha)_{s})\geq 2$ for all $s\in S^{\prime\prime}$. We have $\alpha^{-1}((1,n-r))=\{i_{1}<\cdots<i_{n-r}\}\subset(-n,-1)$. Note that $\alpha_{-i_{d}}=[-i_{d}\to d]$ for $1\leq d\leq n-r$. Let $\varphi:(-r,-1)\to(-n,-1)\setminus\alpha^{-1}((1,n-r))$ be the unique increasing bijection. We define $\alpha^{\prime}:S\sqcup(-r,-1)\to T\sqcup(1,r)$ and an element of $F_{r}$ as follows. We put $\alpha^{\prime}_{s}=\alpha_{s}$ for $s\in S^{\prime}$, $\alpha^{\prime}_{s}=[\alpha(s)\to\alpha(s)-n+r]\cdot\alpha_{s}$ for $s\in S^{\prime\prime}$ and $\alpha_{i}=\alpha_{\varphi(i)}\cdot[i\to\varphi(i)]$ for $i\in(-r,-1)$.

Let $s\in S^{\prime\prime}$, $t=q(\alpha)(s)$ and $i=\alpha^{-1}(t)$. We have

$$q(\alpha^{\prime})_{s}=\alpha_{i}\cdot[1\to i]\cdot[\alpha(s)\to 1]\cdot\alpha_{s}.$$

Define $\tilde{\alpha}^{\prime}$ similarly, starting with $\tilde{\alpha}$ instead of $\alpha$. We have $q(\alpha^{\prime})=q(\tilde{\alpha}^{\prime})$. By induction, we deduce that $\alpha^{\prime}=\tilde{\alpha}^{\prime}$, hence $\alpha=\tilde{\alpha}$.

This completes the proof that the restriction of $q$ to $F$ is injective.

We deduce that the restriction of $q$ to $E$ is injective using Remark 8.2.6 ∎