unit modulo $(Y)$. Let $M = q\tau(g)$. Thus, we can write $$\begin{aligned}
 \tau(f) = \tau\left(M\frac{\alpha(g)}{\tau(g)}\right) + M = \left(I + \tau\circ \frac{\alpha(g)}{\tau(g)}\right)M.\end{aligned}$$ We want to show that the map $\left(I + \tau\circ \frac{\alpha(g)}{\tau(g)}\right)^{-1}$ exists.

Suppose $z$ is an element of $F[X; Y, \lambda]_c.$ We first claim that $\|\tau(z)\|_{\lambda, c} \leq \frac{\|z\|_{\lambda, c}}{p^{c_ns}}$. Indeed, there exists $\mu$ such that $$\begin{aligned}
 \|z\|_{\lambda, c} &=& |z_\mu|_\lambda p^{c\cdot \mu} \\
&\geq& |z_\nu|_\lambda p^{c\cdot\nu} \quad \text{ for all } \nu\end{aligned}$$ Thus, $$\begin{aligned}
 \frac{\|z\|_{\lambda, c}}{p^{c_ns}} &=& |z_\mu|_\lambda p^{c\cdot \mu - c_ns} \\
&\geq& |z_\nu|_\lambda p^{c\cdot \nu - c_ns} \quad \text{ for all } \nu.\end{aligned}$$ The maximum over all $\nu$ with $\nu_n \geq s$ is equal to $\|\tau(z)\|_{\lambda, c}$, as asserted. Thus $\|\tau(g)\|_{\lambda, c} \leq \frac{p^{c_ns}}{p^{c_ns}} = 1$, so by the lemma 2.8 $\|\tau(g)^{-1}\| \leq 1$. Let $h = \frac{\alpha(g)}{\tau(g)}$. Then, $$\begin{aligned}
 \|h\|_{\lambda, c} \leq \|\alpha(g)\|_{\lambda, c}\|\tau(g)^{-1}\|_{\lambda, c} < p^{c_ns}.\end{aligned}$$

Next we claim that $\|(\tau\circ h)^m (z)\|_{\lambda, c} < \frac{\|z\|_{\lambda, c}\|h\|_{\lambda, c}^{m}}{p^{mc_ns}}$, for all $m$ in ${\mathbf{N}}$. Indeed, $\|\tau(zh)\|_{\lambda, c} \leq \frac{\|zh\|_{\lambda, c}}{p^{c_ns}} \leq \frac{\|z\|_{\lambda, c}\|h\|_{\lambda, c}}{p^{c_ns}}$ by what we just proved. Now, assume that this is true for $m$, then $$\begin{aligned}
 \|\tau\left((\tau\circ h)^{m}(z) h \right)\|_{\lambda, c} &\leq& \frac{\|(\tau\circ h)^m(z)\|_{\lambda, c}\|h\|_{\lambda, c}}{p^{c_ns}} \\
&\leq& \frac{\|z\|_{\lambda, c}\|h\|_{\lambda, c}^{m}}{p^{mc_ns}}\frac{\|h\|_{\lambda, 