ScalingStacks

Proof. Note that $d(\pi^{\prime})=0$.

Let $a=\tau_{1}\circ E_{1}\pi^{\prime}\circ\sigma(m^{\prime})\circ E_{2}\pi^{\prime}$ and $b=E_{1}\pi^{\prime}\circ\sigma(m^{\prime})\circ E_{2}\pi^{\prime}\circ\tau_{2}$. We have

	$\displaystyle a_{11}$	$\displaystyle=\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}$
		$\displaystyle=\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}$
		$\displaystyle=\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ\tau_{2}E_{2}\circ E_{2}\tau_{2}$
		$\displaystyle=E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ\tau_{2}E_{2}\circ E_{2}\tau_{2}$
		$\displaystyle=E_{1}\sigma\circ\sigma E_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}\circ E_{2}\tau_{2}$
		$\displaystyle=E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ\tau_{2}E_{2}\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}$
		$\displaystyle=E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}$
		$\displaystyle=E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}=b_{11},$

	$\displaystyle a_{12}$	$\displaystyle=\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}\circ E_{2}\sigma+\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma$
		$\displaystyle=\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}+\tau_{1}^{2}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma$
		$\displaystyle=\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}$
		$\displaystyle=E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}^{2}E_{1}+E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}$
		$\displaystyle=E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}+E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}=b_{12},$

$$a_{21}=0=b_{21}\text{ and }$$

	$\displaystyle a_{22}$	$\displaystyle=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma$
		$\displaystyle=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma$
		$\displaystyle=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma$
		$\displaystyle=E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma$
		$\displaystyle=E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}\circ E_{2}\sigma$
		$\displaystyle=E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}$
		$\displaystyle=E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}$
		$\displaystyle=E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}=b_{22}.$

The lemma follows. ∎