ScalingStacks

Proof. Assume first $E=\emptyset$. Given $s\in I$ with $\theta_{s}=\operatorname{id}\nolimits_{s}$, we have a bijection $C(s)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}C(f(s))$. It follows that

	$\displaystyle f(m(\theta))$	$\displaystyle=\sum_{\begin{subarray}{c}s\in I\\ \theta_{s}\neq\operatorname{id}\nolimits_{s}\end{subarray}}\sum_{c\in\theta_{s}(0+)\cup\iota(\theta_{s}(0+))}m_{c}(\llbracket\theta\rrbracket)f(e_{c})+\sum_{\begin{subarray}{c}s\in I\\ \theta_{s}=\operatorname{id}\nolimits_{s}\end{subarray}}\sum_{c\in C(s)}m_{c}(\llbracket\theta\rrbracket)f(e_{c})$
		$\displaystyle=\sum_{\begin{subarray}{c}s^{\prime}\in f(I)\\ f(\theta)_{s^{\prime}}\neq\operatorname{id}\nolimits_{s^{\prime}}\end{subarray}}\sum_{c^{\prime}\in f(\theta)_{s^{\prime}}(0+)\cup\iota(f(\theta)_{s^{\prime}}(0+))}m_{c^{\prime}}(\llbracket f(\theta)\rrbracket)e_{c^{\prime}}+\sum_{\begin{subarray}{c}s^{\prime}\in f(I)\\ f(\theta)_{s^{\prime}}=\operatorname{id}\nolimits_{s^{\prime}}\end{subarray}}\sum_{c^{\prime}\in C(s^{\prime})}m_{c^{\prime}}(\llbracket f(\theta)\rrbracket)e_{c^{\prime}}$
		$\displaystyle=m(f(\theta))$

by Lemma 7.1.24.

Given $s^{\prime}\in f(I)$ such that $f(\theta)_{s^{\prime}}=\operatorname{id}\nolimits_{s^{\prime}}$, we have $s^{\prime}{\not\in}Z^{\prime}_{f}$. We deduce that $i(\theta_{s},\theta_{t})=i(f(\theta)_{f(s)},f(\theta)_{f(t)})$ for all $s\neq t\in I$ by Lemma 7.3.22. So $f(i(\theta))=i(f(\theta))$. We deduce that the lemma holds for $\theta$.

Consider now the case where $E\neq\emptyset$. Let $\bar{\theta}=(\theta_{s})_{s\in I-E}$. We have $\deg_{E^{+}}(\theta)=\deg_{E^{+}}(\bar{\theta})$ by Lemma 7.4.7; taking quotients, we obtain $\deg_{f^{-1}(f(E))^{+}}(\theta)=\deg_{f^{-1}(f(E))^{+}}(\bar{\theta})$. Since $f(\bar{\theta})=(f(\theta)_{t})_{t\in f(I)-f(E)}$, it follows again from Lemma 7.4.7 that $\deg_{f(E)^{+}}(f(\theta))=\deg_{f(E)^{+}}(f(\bar{\theta}))$. Since the lemma holds for $\bar{\theta}$, we deduce that the lemma holds for $\theta$. ∎