function $$a \in A\mapsto {\llbracket a \rrbracket}_{ {\varphi}}\in {\llbracket A \rrbracket}_{ {\varphi}}$$ is a bijection.

\[clmboundary\] Suppose that the quotient map $p: \mathbb A \to {{\raisebox{.1em}{$ \mathbb A \!$}\left/\raisebox{-.1em}{$R^{ \mathbb A }$}\right.}}$ is irreducible. For every $n \in {\mathbb N}$, $a\in A_{n}$, $$\begin{gathered}
\partial ( {\llbracket a \rrbracket}_{ {\varphi}_n}) = \setnew{x \in {\llbracket a \rrbracket}_{{\varphi}_n}}{ \exists a'\ne a, a' {\mathbin{R}}^{A_n}a, x \in {\llbracket a' \rrbracket}_{{\varphi}_n}} = \\
=\setnew{x\in {\llbracket a \rrbracket}_{{\varphi}_n}}{\exists a'\ne a, x\in {\llbracket a' \rrbracket}_{{\varphi}_n}} .\end{gathered}$$ Moreover, regardless of the irreducibility of $p$, if $p$ is at most $2$-to-$1$ then for each $x$ there are at most two $a\in A_n$ such that $x\in {\llbracket a \rrbracket}_{ {\varphi}_n}$.

Let $x\in\partial ( {\llbracket a \rrbracket}_{ {\varphi}_n})$, so that $x = p(u)$ for some $u \in {\varphi}_{n}^{-1}(a)$. As each $ {\llbracket a' \rrbracket}_{ {\varphi}_n}$ is closed, this implies that there exists $a'\in A_{n}, a'\ne a$ such that $x \in {\llbracket a' \rrbracket}_{ {\varphi}_n}$, so that there is $v\in {\varphi}_{n}^{-1}(a')$ with $u {\mathbin{R}}^{ \mathbb A }v$; in turns, this entails that $a {\mathbin{R}}^{A_n}a'$.

Let now $x\in {\llbracket a \rrbracket}_{ {\varphi}_n}$, and assume that there exists $a'\in A_{n}$, with $a'\ne a, x\in {\llbracket a' \rrbracket}_{ {\varphi}_n}$. Since $ {\llbracket a \rrbracket}_{ {\varphi}_n}\cap {\llbracket 