Lemma 6.1.18 (Continuity of inverse).label The map $x\mto x\inv$ is continuous on $G(A)$
Proof. Using notations as in Proposition 6.1.17 we have
\begin{align*}\norm{x\inv-x_0\inv}{}&=\norm{\sums{\infty}{n=1}\braks{x_0\inv(x_0-x)}^nx_0\inv}{}\\&\leq \norm{x_0\inv}{}\sums{\infty}{n=1}\norm{x_0\inv}{}^{n}\norm{x_0-x}{}^{n}\\&=\norm{x_0\inv}{}^{2}\norm{x_0-x}{}\parens{1-\norm{x_0\inv}{}\norm{x-x_0}{}}\inv\end{align*}
$\square$