then follow the program of [@HuefClarkRaeburn2015] to generalise [@LacaRaeburn1996 Proposition 6.6]. We state this generalisation in Theorem \[p-controlled map implies amenable\] below; its proof will take up the remainder of this section.

\[expanded controlled map definition\] Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups. Let $\mu:G\to K$ be an order-preserving group homomorphism. For each $k\in Q$, let $\Sigma_k$ be the set of $\sigma\in \mu^{-1}(k)\cap P$ which are minimal in the sense that $$x\in \mu^{-1}(k)\cap P \text{\ and\ }x\leq\sigma\Rightarrow \sigma=x.$$ We say $\mu$ is a *controlled map* if it has the following properties:

1.  \[cm-1\] For all $x,y\in G$ such that $x\vee y<\infty$ we have $\mu(x)\vee \mu(y)=\mu(x\vee y)$.

2.  \[cm-2\] For all $k\in Q$, $\Sigma_k$ is complete in the following sense: for every $x\in \mu^{-1}(k)\cap P$ there exists $\sigma\in \Sigma_k$ such that $\sigma\leq x$.

3.  \[cm-3\] For all $k\in Q$ and $\sigma,\tau\in\Sigma_k$ we have $\sigma\vee\tau<\infty\Rightarrow\sigma=\tau$.

\[p-controlled map implies amenable\] Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups. Suppose that $\mu:G\to K$ is a controlled map. If $K$ is an amenable group and $\big(\mu^{-1}(e),\mu^{-1}(e)\cap P\big)$ is an amenable quasi-lattice ordered group, then $(G,P)$ is amenable.

We start by showing that the kernel of a controlled map is a quasi-lattice ordered group.

Let 