ScalingStacks

Proof. We have

	$\displaystyle a_{3}\lambda_{13}\circ\eta a_{3}$	$\displaystyle=a_{3}\varepsilon a_{3}a_{1}\circ a_{3}a_{1}\lambda_{33}a_{1}\circ a_{3}a_{1}a_{3}\eta\circ\eta a_{3}$
		$\displaystyle=a_{3}\varepsilon a_{3}a_{1}\circ\eta a_{3}^{2}a_{1}\circ\lambda_{33}a_{1}\circ a_{3}\eta$
		$\displaystyle=\lambda_{33}a_{1}\circ a_{3}\eta$

	$\displaystyle\lambda_{13}a_{1}\circ a_{1}\eta$	$\displaystyle=\varepsilon a_{3}a_{1}^{2}\circ a_{1}\lambda_{33}a_{1}^{2}\circ a_{1}a_{3}\eta a_{1}\circ a_{1}\eta$
		$\displaystyle=\varepsilon a_{3}a_{1}^{2}\circ a_{1}a_{3}^{2}\lambda_{11}\circ a_{1}a_{3}\eta a_{1}\circ a_{1}\eta$
		$\displaystyle=a_{3}\lambda_{11}\circ\varepsilon a_{3}a_{1}^{2}\circ a_{1}a_{3}\eta a_{1}\circ a_{1}\eta$
		$\displaystyle=a_{3}\lambda_{11}\circ\eta a_{1}\circ\varepsilon a_{1}\circ a_{1}\eta$
		$\displaystyle=a_{3}\lambda_{11}\circ\eta a_{1}$

	$\displaystyle\lambda_{23}a_{1}\circ a_{2}\eta$	$\displaystyle=a_{3}a_{2}\varepsilon a_{1}\circ a_{3}\lambda_{12}a_{3}a_{1}\circ\eta a_{2}a_{3}a_{1}\circ a_{2}\eta$
		$\displaystyle=a_{3}a_{2}\varepsilon a_{1}\circ a_{3}a_{2}a_{1}\eta\circ a_{3}\lambda_{12}\circ\eta a_{2}$
		$\displaystyle=a_{3}\lambda_{12}\circ\eta a_{2}$

It follows that the first statement of the lemma holds when $n=1$. Consider now $n\geq 2$. We prove the first statement of the lemma by induction on $n$. We have

	$\displaystyle\lambda_{(n+2\cdots 2)}\circ\eta G_{1}\cdots G_{n}$	$\displaystyle=\lambda_{(n+2\cdots 3)}\circ(\lambda_{(23)}\circ\eta G_{1})G_{2}\cdots G_{n}$
		$\displaystyle=\lambda_{(n+2\cdots 3)}\circ(\lambda_{(12)}\circ G_{1}\eta)G_{2}\cdots G_{n}$
		$\displaystyle=\lambda_{(12)}\circ G_{1}(\lambda_{(n+1\cdots 2)}\circ\eta G_{2}\cdots G_{n})$
		$\displaystyle=\lambda_{(12)}\circ G_{1}(\lambda_{(1\cdots n)}\circ G_{2}\cdots G_{n}\eta)$
		$\displaystyle=\lambda_{(1\cdots n+1)}\circ G_{1}\cdots G_{n}\eta$

The second statement of the lemma follows by applying the duality of ${\mathcal{M}}^{\prime}$. ∎