$x_1 ^{k_1} U _1 (x_1)$
   $\tilde v $   $x_1 ^{k_1} V _1 (x_1)$
  ------------- -------------------------

. f\_2 \^\* : {

  ------------- -------------------------
   $\tilde u $   $x_2 ^{k_2} U _2 (x_2)$
   $\tilde v $   $x_2 ^{k_2} V _2 (x_2)$
  ------------- -------------------------

. $$ with $\bigl( U _1 (0), V _1 (0) \bigr) , \bigl( U _2 (0), V _2 (0) \bigr)$ linearly independent. By changing $v$ to $v - \frac{V_1 (0)} {U_1 (0)} u$ and $\tilde v$ to $\tilde v - \frac{V_1 (0)} {U_1 (0)} \tilde u$, we may assume that $V_1 (0) = 0 $, while preserving the equations of $b$. With these assumptions, $(1,0)$ and $(\star , 1)$ are tangent vectors at $\tilde q$ to $C_1$ and $C_2$ respectively. Moreover, since $\tilde f _i$ is not ramified at $p_i$, necessarily $k_i = 1$. We have therefore $$\begin{aligned}
\tilde f_1 ^* : \left\{ \begin{tabular}{c@{ $\longmapsto $ }c}
$\tilde u $&$x_1 U _1 (x_1)$ \vphantom{$\frac{\frac{1}{1}}{\frac{1}{1}}$} \\
$\tilde v $&$x_1 ^2 \overline V _1 (x_1)$
\end{tabular} \right. & &
f_1 ^* = \tilde f_1 ^* \circ b ^*: \left\{ \begin{tabular}{c@{ $\longmapsto $ }c}
$u $&$x_1 U _1 (x_1)$ \vphantom{$\frac{\frac{1}{1}}{\frac{1}{1}}$} \\
$v $&$x_1 ^3 U _1 (x_1) \overline V _1 (x_1)$
\end{tabular} \right. 