the maps $\dfrac {u_j}{v_j}$ become submersions whose restrictions to $\tilde C$ are meromorphic maps of degree $k.d- k^2$, which is a multiple of $d^{\prime}$ because of (i).

A neighborhood of $\tilde C$ is not equivalent to a neighborhood of $C^{\prime}$ in the plane. In fact, let us take a linear pencil $\mathcal L$ in the plane with base point outside $C$ and transverse to the branches at each nodal point (this is before blow-up‘s and separation of branches). We have $2d=Tang(\mathcal L,C)+ \chi(C)$; since $tang(\mathcal L,C,P)=2$ for each nodal point $P$, we get $2d = 2s+\chi(C)+ tang(\mathcal L,C)$, where the last term counts the tangencies with the regular part of $C$. These tangencies persist when we blow up and separate branches; therefore, if a neighborhood of $\tilde C$ is equivalent to a neighborhood of $C^{\prime}$, we get in this neighborhood a foliation $\mathcal L^{\prime}$ with $Tang(\mathcal L^{\prime}, C^{\prime})=tang(\mathcal L,C)=2d-2s-\chi(C)$. It follows that $(deg(\mathcal L^{\prime})+2)d^{\prime}=2d-2s-\chi(C)+\chi(C^{\prime})=2d-2s+2g(C)-2g(C^{\prime})$ and since $g(C)-s=g(C^{\prime})$, we conclude that $(deg(\mathcal L^{\prime})+2)d^{\prime}=2d= 2d^{\prime}+2k^2$, a contradiction because of (ii).

We remark that when $d=4, d^{\prime}=3$ or $d=3,d^{\prime}=2$ the construction can be done with $k=1$ because $d>2g(C^{\prime})$ (and obviously (i) is satisfied in both cases). When $d=4, d^{\prime}=3$ we have also that 