components in common. Let $\{ p_1, \ldots , p_r \} = C_1 \cap C_2$ be the nodes of $C$ lying on $C_1$ and $C_2$. If $h^1 (C_1, {\mathcal F}| _{C_1} (-p_1 - \ldots -p_r)) = 0$, then ${\rm H}^1 (C, {\mathcal F}) \cong {\rm H}^1 (C_2, {\mathcal F}| _{C_2})$.

[*Proof.*]{} Simply consider the long exact sequence associated to the “component sequence” $$0 \longrightarrow {\mathcal F}| _{C_1} (-p_1 - \ldots -p_r) \longrightarrow {\mathcal F}\longrightarrow {\mathcal F}| _{C_2} \longrightarrow 0$$ where the first map is extension by zero and the second map is restriction. [ $\Box $ ]{}

\[easyredu\] Let $C$ be a rational tree and let $R \subset C$ be a connected union of irreducible components of $C$. Let ${\mathcal F}$ be a locally free sheaf on $C$ such that the restriction of ${\mathcal F}$ to each irreducible component of $C$ which is not in $R$ is generated by global sections. Then $h^1 (C, {\mathcal F}) = h^1 (R, {\mathcal F}|_R )$.

[*Proof.*]{} Proceed by induction on the number $\ell $ of irreducible components of $C$ not in $R$. If $\ell = 0$, there is nothing to prove. Suppose $\ell \geq 1$. Let $C_1$ be an end of $C$ which is 