Proof. [04NS]

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Proof.

Write $W\coloneqq\sum_{j\in J}W^{\sigma}_{j}+\sum_{i\in L_{\sigma}}W^{\sigma}_{i}\in\Div_{0}(\mathscr{X}).$ We have

	$\displaystyle\textrm{for }j\in J\quad\ord_{D_{j}}(W)$	$\displaystyle=\ord_{D_{j}}(W^{\sigma}_{j})=-1$
	$\displaystyle\textrm{for }i\in L_{\sigma}\quad\ord_{D_{i}}(W)$	$\displaystyle=\ord_{D_{i}}(W^{\sigma}_{i})=-1$
	$\displaystyle\textrm{for }l\in L\setminus L_{\sigma}\quad\ord_{D_{l}}(W)$	$\displaystyle=\sum_{j\in J}d_{j,l}+\sum_{i\in L_{\sigma}}c_{i,l}=-\sum_{j\in J}\lambda_{j,l}+\sum_{i\in L_{\sigma}}c_{i,l}(1-\sum_{j\in J}\lambda_{j,i})=-1$

by Lemma 2.2.1, Eq. 2.2.2 and Eq. 2.1.5. ∎