and surjective. To prove (\[Prp\_magic\_a.1\]), let $\sum_{g\in G} a_g u_g$ be an arbitrary element in ${C^* (G, A, {\alpha})}$. Then $$\begin{aligned}
\Bigg\| \psi \Bigg(\sum_{g\in G} a_g u_g \Bigg)\Bigg\|
= 
\Bigg\| \sum_{g \in G }  a_g z_g \Bigg\| 
\leq 
\sum_{g\in G} \|  a_g \| 
\leq 
{{\operatorname{card}}}(G) \Bigg\| \sum_{g\in G} a_g u_g \Bigg\|.\end{aligned}$$ Part (\[Prp\_magic\_a.2\]) is immediate from the fact that $z_1=1_A$. To prove (\[Prp\_magic\_a.3\]), let $$x=\sum_{g \in G} a_g u_g, 
\quad
y= \sum_{h\in G} b_h u_h, 
\quad
\mbox{and} 
\quad
z= \sum_{t\in G} c_t u_t$$ be arbitrary elements in $F$. So $z^* =\sum_{t\in G} \alpha_{t^{-1}} (c^*_t) u_{t^{-1}}$. A short computation then shows that $$\begin{aligned}
\label{Eq1100.190708}
\psi( xyz^* )  
=
 \sum_{g,h, t  \in G } a_g \alpha_{g} (b_h) \alpha_{ght^{-1}} (c^*_t) z_{gh t^{-1}}\end{aligned}$$ and $$\label{Eq11100.190708}
\psi(x) \psi(y) \psi(z)^*
=
  \sum_{g,h, t \in G } a_g z_g b_h z_h z^{*}_{t} c^{*}_{t}.$$ Therefore, using (\[Eq1100.190708\]) and (\[Eq11100.190708\]) at the first step and using (\[Ap\_inner.a\]) at the second step, $$\begin{aligned}
&\| \psi (xy z^*) - \psi (x) \psi(y) \psi(z)^*\| 
\\
&\hspace*{3em} {\mbox{}}\leq
\sum_{g,h, t  \in G} 
\|  a_g \| \cdot \| \alpha_{g} (b_h) -  z_g b_h z^*_g \| \cdot \|\alpha_{ght^{-1}} (c^{*}_t) z_{ght^{-1}} \|
\\&
\hspace*{6em} {\mbox{}}+
\sum_{g,h, t  \in G}
 \|  a_g z_g b_h z^*_g \| \cdot 