rational curve $E$ so that $\tilde C \cap E = \{p_1, p_2 \}$ and $C' := \tilde C \cup E$ is a nodal curve. Clearly we have a morphism $\pi : C' \rightarrow C$, which is an isomorphism away from $E$ and contracts $E$ to the node $p$. We call the morphism $\pi$ the contraction of $E$. The curve $C'$ so obtained is called the “total transform” of $C$ at the node $p$ and $E$ the “exceptional component.”

\[blodo\] Let $C$ be a rational tree, and let ${\mathcal F}$ be a locally free coherent sheaf on $C$. Let $\pi : C' \rightarrow C$ denote the total transform of $C$ at a node $p \in C$; then ${\rm H}^1 (C, {\mathcal F}) \cong {\rm H}^1 (C', \pi ^* {\mathcal F})$.

[*Proof.*]{} The result follows immediately from the Leray’s spectral sequence associated to the map $\pi $ and the fact that $R ^i \pi _* ({\mathcal O}_{C'}) = 0$, for $i > 0$. [ $\Box $ ]{}

\[pieces\] Let $C$ be a rational tree and let ${\mathcal F}$ be a locally free sheaf on $C$. Suppose that $C = C_1 \cup C_2$, where $C_1$ and $C_2$ are unions of components having no 