ScalingStacks

Proof. The non-zero coefficients of $\pi^{\prime\prime}$ are

	$\displaystyle\pi^{\prime\prime}_{11}$	$\displaystyle=\sigma E_{2}\circ E_{2}\sigma\circ E_{2}^{2}\pi\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}$
	$\displaystyle\pi^{\prime\prime}_{22}$	$\displaystyle=\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}$
	$\displaystyle\pi^{\prime\prime}_{33}$	$\displaystyle=\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}$
	$\displaystyle\pi^{\prime\prime}_{44}$	$\displaystyle=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}$

$$\pi^{\prime\prime}_{12}=\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1},\ \pi^{\prime\prime}_{13}=\sigma E_{2},\ \pi^{\prime\prime}_{24}=\sigma E_{1},\ \pi^{\prime\prime}_{34}=\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}.$$

Let $a=E_{1}\tau\circ\pi^{\prime\prime}$ and $b=\pi^{\prime\prime}\circ E_{2}\tau$. We have

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

$$a_{12}=E_{1}\tau_{2}\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}=0=b_{12}$$

$$a_{13}=E_{1}\tau_{2}\circ\sigma E_{2}=b_{13}$$

	$\displaystyle a_{23}$	$\displaystyle=E_{1}\sigma^{-1}\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ E_{1}\tau_{2}\circ\sigma E_{2}$
		$\displaystyle=E_{1}\sigma^{-1}\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ E_{1}E_{2}\pi\circ\sigma E_{2}\circ E_{2}\sigma\circ\tau_{2}E_{1}\circ E_{2}\sigma^{-1}$
		$\displaystyle=E_{1}\sigma^{-1}\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{2}\circ E_{2}E_{1}\pi\circ\circ E_{2}\sigma\circ\tau_{2}E_{1}\circ E_{2}\sigma^{-1}$
		$\displaystyle=\sigma E_{1}\circ E_{2}\tau_{1}\circ E_{2}E_{1}\pi\circ E_{2}\sigma\circ\tau_{2}E_{1}\circ E_{2}\sigma^{-1}=b_{23}$

$$a_{24}=E_{1}\sigma^{-1}\circ\tau_{1}E_{2}\circ E_{1}\sigma\circ\sigma E_{1}=\sigma E_{1}\circ E_{2}\tau_{1}=b_{24}$$

	$\displaystyle a_{44}$	$\displaystyle=E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}$
		$\displaystyle=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ E_{1}^{2}\pi\circ E_{1}\sigma\circ\sigma E_{1}\circ E_{2}\tau_{1}=b_{44}$

All the other coefficients of $a$ and $b$ vanish. We deduce that $a=b$, hence $\tau$ is an endomorphism of $E^{2}(m,\pi)$. It follows easily that $\tau$ defines an endomorphism of $E^{2}$.

We have $\tau^{2}=0$ and

	$\displaystyle d(\tau)$	$\displaystyle=\left(\begin{matrix}d(\tau_{2})&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&d(\tau_{1})\end{matrix}\right)+\tau\circ\partial+\partial\circ\tau$
		$\displaystyle=\left(\begin{matrix}\operatorname{id}\nolimits&&&\\ 2E_{2}\pi\circ\tau_{2}&\operatorname{id}\nolimits&&\\ \sigma\circ E_{2}\pi\circ\tau_{2}^{2}&&\operatorname{id}\nolimits\\ &\tau_{1}^{2}\circ E_{1}\pi&2\tau_{1}\circ E_{1}\pi\circ\sigma&\operatorname{id}\nolimits\end{matrix}\right)$
		$\displaystyle=\operatorname{id}\nolimits.$

We have $E^{3}(m,\pi)=([m^{\prime\prime\prime},\delta^{\prime}],\pi^{\prime\prime\prime})$, where

$$m^{\prime\prime\prime}=E_{2}^{3}(m)\oplus E_{2}^{2}E_{1}(m)\oplus E_{2}E_{1}E_{2}(m)\oplus E_{2}E_{1}^{2}(m)\oplus E_{1}E_{2}^{2}(m)\oplus E_{1}E_{2}E_{1}(m)\oplus E_{1}^{2}E_{2}(m)\oplus E_{1}^{3}(m).$$

We have

$$\tau E=\left(\begin{matrix}\tau_{2}E_{2}&0&0&0&0&0&0&0\\ 0&\tau_{2}E_{1}&0&0&0&0&0&0\\ 0&0&0&0&\sigma^{-1}E_{2}&0&0&0\\ 0&0&0&0&0&\sigma^{-1}E_{1}&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&\tau_{1}E_{2}&0\\ 0&0&0&0&0&0&0&\tau_{1}E_{1}\end{matrix}\right)$$

and

$$E\tau=\left(\begin{matrix}E_{2}\tau_{2}&0&0&0&0&0&0&0\\ 0&0&E_{2}\sigma^{-1}&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&E_{2}\tau_{1}&0&0&0&0\\ 0&0&0&0&E_{1}\tau_{2}&0&0&0\\ 0&0&0&0&0&0&E_{1}\sigma^{-1}&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&E_{1}\tau_{1}\end{matrix}\right)$$

Let $a=(E\tau)\circ(\tau E)\circ(E\tau)$ and $b=(\tau E)\circ(E\tau)\circ(\tau E)$. We have

$$a_{11}=E_{2}\tau_{2}\circ\tau_{2}E_{2}\circ E_{2}\tau_{2}=\tau_{2}E_{2}\circ E_{2}\tau_{2}\circ\tau_{2}E_{2}=b_{11}$$

$$a_{44}=E_{1}\tau_{1}\circ\tau_{1}E_{1}\circ E_{1}\tau_{1}=\tau_{1}E_{1}\circ E_{1}\tau_{1}\circ\tau_{1}E_{1}=b_{44}$$

$$a_{25}=E_{2}\sigma^{-1}\circ\sigma^{-1}E_{2}\circ E_{1}\tau_{2}=\tau_{2}E_{1}\circ E_{2}\sigma^{-1}\circ\sigma^{-1}E_{2}=b_{25}$$

$$a_{47}=E_{2}\tau_{1}\circ\sigma^{-1}E_{1}\circ E_{1}\sigma^{-1}=\sigma^{-1}E_{1}\circ E_{1}\sigma^{-1}\circ\tau_{1}E_{2}=b_{47}$$

and all the other coefficients of $a$ and $b$ vanish. It follows that $a=b$. This completes the proof of the theorem. ∎