Suppose for some $i$ we have $\gamma_{i}>0$ but $\beta_{i}=0$. Let $\{w_{i}=0\}=D_{\ell_{i}}$. Then $\beta_{i}=0$ means that $\pi^{*}\exp(\varphi_{2})\not\equiv 0$ on $D_{\ell_{i}}$, while $\gamma_{i}>0$ means that $\pi^{*}\exp(F)\equiv 0$ on $D_{\ell_{i}}$. We have that $\pi(D_{\ell_{i}})\subset V\cap U$, since $F$ is smooth outside $V$. However, we can also expand

$$\pi^{*}\exp(\varphi_{1})=U_{1}(w)\prod_{j=1}^{n}|w_{j}|^{2\alpha_{j}},$$

and the fact that $\varphi_{1}\not\equiv-\infty$ on $V\cap U$ again implies that $\alpha_{i}=0$. Since $\varphi_{1}|_{V}=\varphi_{2}|_{V}$, and $\pi(D_{\ell_{i}})\subset V$, we see again that

$$\pi^{*}\exp(\varphi_{1})=\pi^{*}\exp(\varphi_{2})\quad\text{ on }D_{\ell_{i}}.$$

Recall that $\mathcal{S}=\{\varphi_{2}>\varphi_{1}-\eta\}\cap U$. This shows that $\pi(D_{\ell_{i}})$ is contained in the interior of $\mathcal{S}$, which is impossible because we assumed $q\in\overline{\pi^{-1}(\mathcal{S}^{c})}$. ∎