, and the obstruction space of $\zeta'$ equals the obstruction space of $\zeta$.  \[lem-def0b\] Suppose that $h:B'\rightarrow B$ is type (IIIb). Then we have exact sequences: $$\begin{aligned} 0 \rightarrow {\mathbb{E}\text{xt}_{{\mathcal O}_{B'}}}^0(L_{\zeta'},{\mathcal O}_{B'}) \rightarrow {\mathbb{E}\text{xt}_{{\mathcal O}_{B}}}^0(L_\zeta,{\mathcal O}_B) \rightarrow 0 \\ 0 \rightarrow 