\\  &  \Longrightarrow & \\
\tilde f_2 ^* : \left\{ \begin{tabular}{c@{ $\longmapsto $ }c}
$\tilde u $&$x_2 U _2 (x_2)$ \vphantom{$\frac{\frac{1}{1}}{\frac{1}{1}}$} \\
$\tilde v $&$x_2 V _2 (x_2)$
\end{tabular} \right. & &
f_2 ^* = \tilde f_2 ^* \circ b ^* : \left\{ \begin{tabular}{c@{ $\longmapsto $ }c}
$u $&$x_2 U _2 (x_2)$ \vphantom{$\frac{\frac{1}{1}}{\frac{1}{1}}$} \\
$v $&$x_2 ^2 U _2 (x_2) V _2 (x_2)$
\end{tabular} \right.\end{aligned}$$ where $U_1 (0), V_2 (0) \neq 0$.

In order to conclude we still need to show that $U_2 (x_2)$ is not identically zero, but this is clear, since otherwise the morphism $f _2$ would be constant (i.e. the morphism $\tilde 
f_2$ would map $C_2$ to the exceptional divisor $E$). [ $\Box $ ]{}

\[semtang\] In the situation described by the previous lemma, the two curves $C_1$ and $C_2$ are [simply tangent at $q$]{}.

We will see later (Lemma \[conota\]) that being simply tangent is closely related to the local structure of the conormal sheaf.

\[conotra\] Suppose that $X$ is a smooth surface and let $f: C \rightarrow X$ be a morphism from a curve $C$ consisting of two irreducible components $C_1$ and $C_2$, meeting in a node $p$. Denote by $f_i$ the restriction of $f$ 