The proof is the same as in [1] Proposition 3.10. Let $s=b_{1}+\sqrt{-1}b_{2}$ and $r_{3}=b_{3}$. Roughly speaking, one considers the maps given by $\sigma_{1}(s,r)=\left(\bar{s}/\penalty\epsilon,r,\theta_{0}\right)$ and $\sigma_{2}(s,r)=(\epsilon,s/\penalty\epsilon,r,\theta_{0})$ for small $\epsilon>0$ and $\theta_{0}\in S^{1}$ fixed; these define sections of $f|_{U}=F$ disjoint from $\Crit(F)$, where $F$ is as in (14). The Hamiltonian vector fields $\eta_{i}$ of $F_{i}$ extend to $X\setminus U$. One can define a basis $\gamma$ of $H_{1}(F_{\bar{b}},\mathbb{Z})$ in terms of suitable composition of the integral curves of $\eta_{i}$. The period $\lambda_{1}$ is obtained by integrating along the path $\gamma_{1}$ starting at $\sigma_{1}(s,r)$, passing through $\sigma_{2}(s,r)$ and going back to $\sigma_{1}(s,r)$. The contribution of $\gamma_{1}\cap U$ to the period $\lambda_{1}$ is $\lambda_{0}$, whereas the contribution of $\gamma_{1}\cap X\setminus U$ is $dH$. The remaining periods can be computed integrating along classes in $H_{1}(F_{\bar{b}},\mathbb{Z})$ represented by integral curves of $\eta_{2}$ and $\eta_{3}$, respectively. ∎