of the fiber of $\bar \Psi$ at that point. The second intersection number involves in addition the multiplication by the degree of the covering $\Gamma \stackrel{\bar\Psi}{\to} 
\Delta$, for each irreducible component of $\Gamma$ and of $\Delta$. This shows that $\alpha > 0$, which is our first claim. Let us emphasize that $\lambda_{p,t_0}(0)$ was identified with the intersection number between $\Gamma_{II}$ and the hyperplane $\{ \tau = t_0\}$.

To prove the second part of the statement, let us first observe that the right-hand side of the claimed inequality is equal to the following intersection numbers, taken with appropriate multiplicities: $${{\rm{int}}}_{(p,t_0)}(\Gamma_P, \{ \sigma = 0\} ) = {{\rm{int}}}_{(0,t_0)}(\Delta_P, \{ \sigma = 0\} ) 
.$$ By Proposition \[l:tang\], $\Delta_P$ is tangent to $\{ \sigma = 0\}$. This tangency implies that: $${{\rm{int}}}_{(0,t_0)}(\Delta_P, \{ \sigma = 0\} ) > {{\rm{int}}}_{(0,t_0)}(\Delta_P, \{ \tau = t_0\} 
),$$ which gives the claimed inequality.

\[c:a1\] In the conditions of Theorem \[t:lambdaexchange\], if $\deg f_0 > 2$ and the singularities of some fiber ${{\mathbb Y}}_{0,t}$ at $H^{{\infty}}$ are of type $A_1$ only, then $\Gamma_{II} = \emptyset$.

Since $\deg f > 2$, the type $A_1$ of the singularities at $H^{{\infty}}$ of ${{\mathbb Y}}_{0,t}$ is independent of $t$. Consequently, there is no 