We apply a similar argument to the one used in the case of the positive fibre (see Proposition 4.11). Clearly, since $\gamma_{1}$ is represented by the orbits of the $S^{1}$ action

$$-\int_{\gamma_{1}(b)}\eta=b_{1},$$

which is continuous. We now prove that, for $j=2,3$

$$\alpha_{j}(b)=-\int_{\gamma_{j}(b)}\eta$$

(77)

extends continuously to points in $\Gamma_{d}$ or in $\Gamma_{e}$. As we did in Proposition 4.11, we can think of $\alpha_{j}(b)$ as

$$\alpha_{j}(b)=\int_{S}\omega,$$

where $S$ is a surface spanned by the cycles $\gamma_{j}(b^{\prime})$ as $b^{\prime}$ moves along a curve joining $\bar{b}$ and $b$. Suppose $b\in\Gamma_{e}$ (or $\Gamma_{d}$), then we need to show that $\alpha_{j}(b)$ is independent of the curve from $\bar{b}$ to $b$, or equivalently that

$$\int_{S_{1}-S_{2}}\omega=0,$$

where $S_{1}$ and $S_{2}$ are the surfaces corresponding to two different paths from $\bar{b}$ to $b$. The boundary $\partial(S_{1}-S_{2})$ is determined by monodromy. It is easy to see that $\partial(S_{1}-S_{2})$ is a multiple of $\gamma_{1}(b)$, therefore for some integer $k$ we have

$$\int_{S_{1}-S_{2}}\omega=-\int_{\partial(S_{1}-S_{2})}\eta=k\int_{\gamma_{1}(b)}\eta=0,$$

where the last equality follows from the fact that $b\in\Gamma_{d}$ or $\Gamma_{e}$. To show that $\alpha$ extends continuously also to points of $\Delta$ we can argue that (77) makes sense also over singular fibres, since both $\eta$ and $\gamma_{j}(b)$ are well defined when $b\in\Delta$. ∎