Lemma 4

Let $n\in \mathbf{N}$. Put

$ni-1(n)i-1$

$q_n=\sum(-1)q\circ (p\circ q)$.

$i=1$

We have $u^n= 1_E-q_n\circ p$ and $v^n= 1_F-p\circ q_n$.

We have $(q\circ p)^i=q\circ (p\circ q)^{i-1}\circ p$ for every integer $i\geqslant 1$. We then calculate

$nnni-1(n)i$ $u= (1_E-q\circ p)= 1_E-\sum(-1)(q\circ p)= 1_E-q_n\circ p$ $nnni-1(n)i$

$v= (1_F-p\circ q)= 1_F-\sum(-1)(p\circ q)= 1_F-p\circ q_n$,

$i=1$

whence the result.