given by $$0 \longrightarrow k \longrightarrow k + k \longrightarrow {\mathcal Q}\longrightarrow 0$$ and we deduce that the length of ${\mathcal Q}$ is 1. Thus it follows in general that ${\mathcal Q}= 
\oplus {\mathcal O}_{p_i}$, the direct sum of the skyscraper sheaves of the nodes $p_1, \ldots , 
p_r$.

Let us now go back to the sequence (\[defqu\]). Since the sheaf ${\mathcal F}$ is locally free, we may tensor the sequence by ${\mathcal F}$, preserving exactness. To identify the tensor product in the middle we use the projection formula: $$\nu _* {\mathcal O}_{\tilde C} \otimes {\mathcal F}\cong \nu _* ({\mathcal O}_{\tilde C} \otimes \nu ^* {\mathcal F}) 
\cong \nu _* ( \nu ^* {\mathcal F})$$ and we may therefore write the tensored sequence as $$0 \longrightarrow {\mathcal F}\longrightarrow \nu _* \nu ^* {\mathcal F}\longrightarrow 
\mathop {\oplus } \limits _i {\mathcal F}_{p_i} \longrightarrow 0$$ thus proving the lemma. [ $\Box $ ]{}

Given a rational tree $C$ and a node $p \in C$, construct a new curve $C'$ as follows: consider the normalization of $\nu : \tilde C \rightarrow C$ of $C$ at the point $p$, and let $\{p_1, p_2 \} = \nu ^{-1} (p)$. Attach to $\tilde C$ a smooth 