Let $1\leq i\leq k$. The rational function $x:=t\varpi^{-a+i}$ has a zero of order one along the component $E_{i-1}$ and the support of its divisor does not contain the component $E_{i}$. On the other hand, the rational function $y:=t^{-1}\varpi^{a-i+1}$ has a zero of order one along the component $E_{i}$ and the support of its divisor does not contain the component $E_{i-1}$. Thus $\{x,y\}$ is a system of parameters in a neighbourhood of $q_{i}$. We denote

The local ring at the point $q_{i}$ is $A_{(x,y)}$. Let $p$ be a point such that $|\varpi|^{a-i+1}<|t(p)|<|\varpi|^{a-i}$. Therefore, for $f\in A$ we have $|f(p)|\leq 1$. Moreover, if $f\in(x,y)$, then $|f(p)|<1$. Since the ideal $(x,y)$ is maximal, we deduce that, for $f\in A$, the condition $|f(p)|<1$ is equivalent to the condition $f\in(x,y)$. This implies that ${\operatorname{red}}(p)=q_{i}$. A similar argument works for $q_{0}$ and $q_{k+1}$.

Assume now that $p\in\operatorname{im}(\theta_{\Sigma})$ and that $|t(p)|=|\varpi|^{a-i}$. If $i\not=0$ we consider again the ring $A$, but in this case $|x(p)|=|t(p)\varpi^{-a+i}|=1$. Let $I=\{f\in A\mid|f(p)|<1\}$. It is clear that $(y,\varpi)\subset I$. For $f=\sum_{m\in\mathbb{Z}}\beta_{m}t^{m}\in A$, since $p\in\operatorname{im}(\theta_{\Sigma})$, we have

This implies that $I\subset(y,\varpi)$. Hence $I$ is the ideal that defines the component $E_{i}$ and this is equivalent to ${\operatorname{red}}(p)=\eta_{i}$. The case $i=0$ is analogous. ∎