$$||{\overline{\b}}^s|| \leq (\mu + O(\delta))||\b^s||+O(\rho).$$ Since $\mu<1$ and $||\b^s||\leq \delta/2$, for appropriately chosen $\rho=O(\delta)$ the above can be made $\leq \delta/2$. This proves . As for , note that $$\label{eqn:estlip} \mLip {\overline{\b}^s} \le
\mLip (\pi_u \circ \G_b^{-1}\circ \gamma(\b^s)) \cdot \mLip
G_{\overline\b, b}.$$ We need to show that the right hand side of the above is $\leq D\delta/2$. It is enough to do this for $\b^s$ and $\overline\beta^s$ of class $C^1$, since these maps are dense in $\B^s$. In this $C^1$ case, we have:

$$d(\pi_u \circ \G_b^{-1} \circ \gamma(\b^s)) =  d(\pi_u \circ \G_b^{-1})\circ \gamma(\b^s) \cdot d\gamma(\b^s) \leq$$

$$\leq \left[d(\pi_u \circ \F_b^{-1})\circ \gamma(\b^s) +O(\rho)\right]\cdot d\gamma(\b^s) \leq$$

$$\leq \left[\left( \begin{array}{ccc} \frac {\partial \pi_u \circ
h^{-1}_b}{\partial x_s} & \frac {\pi_u \circ \partial
h^{-1}_b}{\partial x_u} & 0 \\ \end{array} \right) \circ
\gamma(\b^s) +O(\rho) \right] \cdot \left( \begin{array}{cc} 1 & 0
\\ \frac {\partial \beta^s}{\partial x_s} & \frac {\partial
\beta^s}{\partial m} \\ 0 & 1 \\ \end{array} \right) \leq$$

$$\leq \left[\left( \begin{array}{ccc} 0 & \mu & 0 \\ \end{array}
\right)+O(\delta)+O(\rho) \right] \cdot \left( \begin{array}{cc} 1 &
0 \\ \frac {\partial \beta^s}{\partial x_s} & \frac {\partial
\beta^s}{\partial m} \\ 0 & 1 \\ \end{array} \right) \leq$$

$$\label{eqn:ppp} \leq \left( \begin{array}{cc}
\mu\cdot \frac {\partial \beta^s}{\partial x_s} & \mu\cdot 