done. So suppose that $w_pw^*_qw_rw^*_s\neq 0$. Then $q\vee r<\infty$. Since $\mu$ is a controlled map and $\mu(p)=\mu(q)$, by Definition \[expanded controlled map definition\], $$\mu(pq^{-1}(q\vee r))=\mu(q\vee r)=\mu(q)\vee \mu(r).$$ Similarly, $\mu(sr^{-1}(q\vee r))=\mu(q)\vee \mu(r)$. Since $I$ is closed under $\vee$ we have $\mu(q)\vee \mu(r)\in I$, and hence $w_pw^*_qw_rw^*_s\in {\operatorname{span}}\{w_pw^*_q:\mu(p)=\mu(q)\in I\}$. It follows that $C_I$ is a $C^*$-subalgebra.

For each $k\in I$, we have $B_k\subseteq C_I$, and so ${\operatorname{span}}\{B_k:k\in I\}\subseteq C_I$. To show the reverse inclusion observe that for $w_pw^*_q\in C_I$ we have $w_pw^*_q\in B_{\mu(p)}$. Since the finite span of closed subalgebras is closed, $\overline{\operatorname{span}}\{w_pw^*_q:\mu(p)=\mu(q)\in I\}\subseteq{\operatorname{span}}\{B_k:k\in I\}$. Thus $C_I={\operatorname{span}}\{B_k:k\in I\}$.

\[piT is faithful on core\] Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ is a controlled map. If $\big(\mu^{-1}(e),\mu^{-1}(e)\cap P\big)$ is amenable, then $\pi_T$ is faithful on $\overline{\operatorname{span}}\{w_pw^*_q:\mu(p)=\mu(q)\}$.

By Lemma \[core is limit of CI\], $\overline{\operatorname{span}}\{w_pw^*_q:\mu(p)=\mu(q)\}=\overline{\cup_{I\in{\mathcal{I}}}C_I}$. Thus it suffices to show that $\pi_T$ is isometric on each $C_I$. Fix $I\in {\mathcal{I}}$. Suppose that $\pi_T(R)=0$ for some $R\in C_I$. Then there exist $R_k\in B_k$ such that $R=\sum_{k\in I}R_k$ and then $\sum_{k\in I}\pi_T(R_k)=0$.

We claim that if $k\not\leq j$, then $\pi_T(B_k)|_{H_j}=0$ (it then follows that $\pi_T(R_k)|_{H_j}=0$). To prove the claim, it suffices to show that $\pi_T(w_q^*)\epsilon_{\gamma z}=0$ for all $q\in \mu^{-1}(k)\cap 