ScalingStacks

Proof. Note first that, given $i^{\prime}_{1}$ and $i^{\prime}_{2}$ two distinct elements of $\{1,\ldots,n\}$, then $\lambda$ induces a bijection

$$\bigl((i^{\prime}_{1}+n{\mathbf{Z}})\times(i^{\prime}_{2}+n{\mathbf{Z}})\bigr)/n{\mathbf{Z}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{\Pi(S^{1})}(a_{i^{\prime}_{1}},a_{i^{\prime}_{2}}).$$

Consider $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)$ and $i_{1},i_{2}\in\tilde{I}$ with $i_{2}{\not\in}i_{1}+n{\mathbf{Z}}$. Note that $\lambda(i_{1},i_{2})=\lambda(i_{2},i_{1})^{-1}$ for any $i_{1},i_{2}$.

Let $\zeta_{r}=F(\sigma)_{a_{i_{r}}}$ and $\zeta=\lambda(i_{1},i_{2})$. We have $\bar{\zeta}=\lambda(\sigma(i_{1}),\sigma(i_{2}))$. So, $\zeta\in L(F(\sigma))$ if and only if $i_{1}-i_{2}$ and $\sigma(i_{1})-\sigma(i_{2})$ have opposite signs. On the other hand, $(i_{1},i_{2})\in L(\sigma)$ if and only if $i_{1}<i_{2}$ and $\sigma(i_{2})<\sigma(i_{1})$. This shows that $\lambda(L(\sigma))\subset L(F(\sigma))$ and $\lambda$ induces a bijection $L(\sigma)/n{\mathbf{Z}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}L(F(\sigma))/\mathrm{inv}$.

Consider $(i_{1},i_{2})\in L(\sigma)$. Let $r=\lfloor\frac{i_{2}-i_{1}}{n}\rfloor$ and $s=\lfloor\frac{\sigma(i_{1})-\sigma(i_{2})}{n}\rfloor$. We have $r>0$ if and only if $\operatorname{supp}\nolimits(\lambda(i_{1},i_{2}-rn))\subsetneq\operatorname{supp}\nolimits(\lambda(i_{1},i_{2}))$ and $s>0$ if and only if $\operatorname{supp}\nolimits(\lambda(i_{1},i_{2}+sn))\subsetneq\operatorname{supp}\nolimits(\lambda(i_{1},i_{2}))$. There is $i$ such that $(i_{1},i)$ and $(i,i_{2})$ are in $L(\sigma)$ if and only if there are $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ with the same orientations in $L(F(\sigma))$ such that $\lambda(i_{1},i_{2})=\zeta^{\prime\prime}\circ\zeta^{\prime}$. We have $i_{2}-i_{1}>n$ if and only if there is $\zeta^{\prime}$ such that $\zeta$, $\zeta^{\prime}$ and $\zeta\circ\zeta^{\prime-1}$ have the same orientation. We have $\sigma(i_{1})-\sigma(i_{2})>n$ if and only if there is $\zeta^{\prime\prime}$ such that $\bar{\zeta}$, $\zeta^{\prime\prime}$ and $\bar{\zeta}\circ\zeta^{\prime\prime-1}$ have the same orientation.

We deduce that $(i_{1},i_{2})\in D(\sigma)$ if and only if $\lambda(i_{1},i_{2})\in D(F(\sigma))$.

Assume now $(i_{1},i_{2})\in L(\sigma)$. Let $i^{\prime}_{r}\in[1,n]\cap(i_{r}+n{\mathbf{Z}})$ for $r\in\{1,2\}$. We have

$$(F(\sigma)^{\lambda(i_{1},i_{2})})_{a_{i^{\prime}_{1}}}=F_{i^{\prime}_{2}}(\sigma(i_{2})-i_{2})\circ F_{i^{\prime}_{1}}(i_{2}-i_{1})=F_{i^{\prime}_{1}}(\sigma(i_{2})-i_{1})=(F(\sigma^{i_{1},i_{2}}))_{a_{i_{1}}}.$$

Similarly, $(F(\sigma)^{\lambda(i_{1},i_{2})})_{a_{i^{\prime}_{2}}}=(F(\sigma^{i_{1},i_{2}}))_{a_{i_{2}}}$. It follows that $F(\sigma)^{\lambda(i_{1},i_{2})}=F(\sigma^{i_{1},i_{2}})$. This completes the proof of the lemma. ∎