\rightarrow S$ a morphism to a smooth surface. Let $p 
\in C$ be a node, denote by $C_a$ and $C_b$ the two irreducible components of $C$ meeting at $p$. Let $\nu : \tilde C \rightarrow C$ be the normalization of $C$ at $p$ and let $\tilde f = f \circ \nu $. Suppose that:

1.  the valences of the vertices $C_a$ and $C_b$ in the dual graph of $C$ are at most 3, and

2.  the map $f_* : {\mathcal T}_{C_a, p} + {\mathcal T}_{C_b, p} \longrightarrow {\mathcal T}_{S, f(p)} $ is surjective.

Then $${\rm H}^1 (C, f^* {\mathcal T}_S ) \cong {\rm H}^1 (\tilde C, \tilde f ^* {\mathcal T}_S )$$

[*Proof.*]{} Consider the sequence on $C$ $$0 \longrightarrow f^* {\mathcal T}_S \longrightarrow \tilde f ^* {\mathcal T}_S \stackrel {\varepsilon} 
{\longrightarrow } {\mathcal T}_{S, f(p)} \longrightarrow 0$$

Because ${\mathcal T}_{S, f(p)}$ is supported in dimension 0, ${\rm H}^1 (C, {\mathcal T}_{S, f(p)}) = 0$, and it is enough to prove that the sequence is exact on global sections. Let $\{ p, q_a, r_a \}$ contain all the nodes of $C$ on $C_a$ and let $\{p, q_b, r_b \}$ contain the nodes on $C_b$. Consider now the following diagram: $$\xymatrix 