A \!$}\left/\raisebox{-.1em}{$R^{ \mathbb A }$}\right.}} $ is irreducible.

$(1)\Rightarrow (3)$. Let $K \subset \mathbb A $ be a proper closed subset. Then there is $x \in M\setminus K$, so that $p(x)\notin p[K]$. Thus $p$ is irreducible.

$(3)\Rightarrow (2)$. Let $n \in {\mathbb N}$ and $a \in A_{n}$. By irreducibility of $p$, $$O=p^{-1}(p^{\#}({\varphi}_{n}^{-1}(a)))= \setnew{x\in \mathbb A }{[x]_{R^{ \mathbb A }}\subseteq {\varphi}_n^{-1}(a)}$$ is an open, nonempty, and $R^{ \mathbb A }$-invariant set contained in $ {\varphi}_{n}^{-1}(a)$. Let $m>n$ and $b \in A_{m}$ be such that $ {\varphi}_{m}^{-1}(b) \subseteq O$, which exist since such sets are a basis for the topology on $ \mathbb A $. If $b' {\mathbin{R}}^{A_m}b$, there are $x \in {\varphi}_{m}^{-1}(b), x' \in {\varphi}_{m}^{-1}(b')$ such that $x {\mathbin{R}}^{ \mathbb A }x'$. But $x \in {\varphi}_{m}^{-1}(b) \subseteq O$, which is $R^{ \mathbb A }$-invariant, so also $x' \in O$. It follows that $ {\varphi}_{n}(x') = a$ and thus ${\varphi}^{m}_{n}(b') = a$, for ${\varphi}_{n}= {\varphi}^{m}_{n}{\varphi}_{m}$.

$(2)\Rightarrow (1)$. Since $\setnew{{\varphi}_{n}^{-1}(a)}{n \in {\mathbb N}, a \in A_{n}}$ is a basis for the topology on $ \mathbb A $ it suffices to fix $n \in {\mathbb N}$ and $a \in A_{n}$ and prove that there is $x \in M$ with ${\varphi}_n(x) = a$. We construct 