ScalingStacks

Proof. Denote by $\tilde{\epsilon}$ the map defined by the right hand side of the equality of the lemma.

Let $N$ (resp. $N^{\prime}$) be the cardinality of the set of $a\in{\mathbf{Z}}/n$ such that $v_{a}+v_{a+1}$ (resp. $v^{\prime}_{a}+v^{\prime}_{a+1}$) is odd, where $v^{\prime}_{a}=v_{a}+\delta_{ab}$. The integers $N$ and $N^{\prime}$ are even. We have

	$\displaystyle\tilde{\epsilon}(r,\sum_{a}v_{a}\alpha_{a})+\tilde{\epsilon}((r,\sum_{a}v_{a}\alpha_{a})(s,\alpha_{b}))$	$\displaystyle=\tilde{\epsilon}(r,\sum_{a}v_{a}\alpha_{a})+\tilde{\epsilon}(r+s+\frac{1}{2}(v_{b-1}-v_{b+1}),\alpha_{b}+\sum_{a}v_{a}\alpha_{a})$
		$\displaystyle=2s+v_{b+1}+v_{b-1}+\frac{1}{2}(N+N^{\prime}).$

We have

$$N^{\prime}=\begin{cases}N+2&\text{ if }v_{b-1},\ v_{b}\text{ and }v_{b+1}\text{ have the same parity}\\ N-2&\text{ if }v_{b-1},\ v_{b}+1\text{ and }v_{b+1}\text{ have the same parity}\\ N&\text{otherwise}.\end{cases}$$

It follows that

$$\tilde{\epsilon}(r,\sum_{a}v_{a}\alpha_{a})+\tilde{\epsilon}((r,\sum_{a}v_{a}\alpha_{a})(s,\alpha_{b}))=2s+1.$$

We deduce by induction on $\sum_{a\in{\mathbf{Z}}/n}|v_{a}|$ that $\tilde{\epsilon}(r,\sum_{a}v_{a}\alpha_{a})=\epsilon(r,\sum_{a}v_{a}\alpha_{a})$.

Given $a,b\in{\mathbf{Z}}/n$ and $i\in I$, we have

$$\alpha_{a,b}\cdot\varepsilon_{i}=\delta_{\begin{subarray}{c}i\in\{a,b\}\\ a\neq b\end{subarray}}\ \varepsilon_{i}\mod 2L_{n}.$$

It follows that $m(\sigma)\equiv\sum_{i\in I\setminus(I\cap J)}\varepsilon_{i}\mod{2L_{n}}$. Write $\llbracket\sigma\rrbracket=\sum_{a}v_{a}\alpha_{a}$. Given $a\in{\mathbf{Z}}/n$, the integer $v_{a}+v_{a+1}$ is odd if and only if $a\in I\Delta J$, hence $\epsilon(0,\llbracket\sigma\rrbracket)=\frac{1}{2}|I\Delta J|=|I\setminus(I\cap J)|$. It follows that $\epsilon(\deg_{[1,n]^{+}}(\sigma))=0$. ∎