%% starts on pg 6 (and a half) Fix a basepoint \(y_{0} \in Y\) and let \(K = E_{y_{0}}\). For each \(y \in Y\) choose \(p_{yy_{0}} \in E(X)\) such that \[ p_{yy_{0}}:y_{0} \mapsto y \] and note that \(p_{yy_{0} }:E_{y_{0}} \to E_{y}\) is a homeomorphism. Define an action \(\kappa\) of \(K\) on \(X\) by \[ \kappa_{k}(x) = p_{\pi(x)y_{0}}k{{p}^{-1}_{\pi(x)y_{0}}} (x). \] This should define a continuous action of \(K\) on \(\ell^\infty(Y;X)\). Let \(E_{\ell^\infty}\) be the averaging map for \(K\); the induced map \((id \otimes E_{\ell^{\infty}})\rtimes_{\max} G\) is faithful by a twisted version of the usual argument for compact groups. % fixme: write proof? Brown & Ozawa pg 133 On the other hand, \(E_{C}: C(X) \to C(Y)\) is defined as the dual ucp map to the map \begin{align*} Y &\to P(X) \text{(Probability measures on \(X\) )} \\ y &\mapsto \mu_{y} \end{align*} as in \cite[Theorem 7.20]{auslander:minimal}. We first claim that the family \(\left \{ p|_{{\pi(y)}^{-1}} \mid p \in E(X), y \in Y \right \}\) is equicontinuous (so in particular, the family \(\left \{ p_{yy_{0}}|_{{\pi(y)}^{-1}} \mid p \in E(X), y \in Y \right \}\) is). Indeed, let \(\epsilon > 0\). Then by definition of an equicontinuous extension (Auslander, pg 95) there is \(\delta > 0\) such that for all \(x,z \in X\) and \(g \in G\), \[ \pi(x) = \pi(z) \text{ and } d(x,z) < \delta \implies d(gx, gz) < \epsilon. \] This condition is preserved by taking pointwise lists of \(g\)’s (up to replacing \(< \epsilon\) by \(\leq \epsilon\)) so holds for all \(p \in E(X)\) too; this gives the claim. now, let \(f \in \ell^\infty(y;x)\) and let \(k \in k\). we need to show that if \((\kappa_{k}(f)(x) := f({k}^{-1}x)\), then \(\kappa_{k}(f) \in \ell^\infty(y;x)\) too. \(\kappa_{k}(f)\) is clearly bounded. The equicontinuous condition is satisfied as \(\kappa_{k}(f)\) restricted to a given fiber \({\pi}^{-1}(y)\) is the composition \(f \circ p_{yy_0} \circ k \circ {p_{yy_0}}^{-1}\); all the maps in the composition are from equicontinuous families, so \(f\) is too. We now have that \(\kappa_{k}\) is a well-defined map \(\ell^\infty(Y;X) \to \ell^\infty(Y;X)\); straightforward algebraic checks show that it is a group action by \(*\)-homomorphisms as \(k\) ranges over \(K\). We also need that \(\kappa_{k}(f)\) varies continuously in \(k\) for each fixed \(f \in \ell^\infty(Y;X)\). This follows very similarly to the arguments sketched above for equicontinuity of \(\kappa_{k}(f)|_{{\pi(y)}^{-1}}\) as \(y\) varies. We can deduce that the map \[ (id \otimes E_{\ell^\infty}) \rtimes_{\max}G: (\ell^\infty(G) \otimes\ell^\infty(Y;X))\rtimes_{\max}G \to (\ell^\infty(G)\otimes\ell^\infty(Y) \rtimes_{\max}G \] is faithful from a general lemma as follows \begin{lemma} Let \(K\) be a compact group acting by \(\alpha\) (point-norm) continuously on a \(C^*\)-algebra \(A\) and let \(G\) act on \(A\) by \(\gamma\). Assume moreover that the conditional expectation \[ E:A \to A, a \mapsto \int_{k}\alpha_{k}(a)dk \] defined by averaging over \(K\) is \(G\)-equivariant (we do {\bf not } assume that each \(\alpha_{k}\) is \(G\)-equivariant). Then the induced map \[ (id \otimes E)\rtimes_{\max}G: (\ell^\infty(G)\otimes A)\rtimes_{\max}^{\lambda \otimes \gamma} \to (\ell^\infty(G)\otimes A)\rtimes_{\max}^{\lambda \otimes \gamma} \] is faithful. \end{lemma} Sketch proof: Define an action \(\beta\) of \(K\) on \(\ell^\infty(G)\otimes A = \ell^\infty_{\text{rc}}(G,A)\) by \[ (\beta_{k}f)(g) = (\gamma_{g}\circ \alpha_{k}\circ\gamma_{{g}^{-1}}(f(g)) \] This commutes with the action \(\lambda \otimes \gamma\) of \(G\) whence are check it induces an action of \(K\) on \((\ell^\infty(G) \otimes A ) \rtimes_{\max}G\). One checks moreover that \((id \otimes E)\rtimes_{\max}G\) is the same as the averaging map for this latter action of \(K\). Hence it is faithful, as averaging over a compact group is always faithful. The fact that the Diagram~\ref{eqn:commutative-diagram} commutes follows as \(E_{\ell^\infty}\) is defined by averaging over \(P_{y y_{0}}K{P_{y y_{0}}}^{-1} = E_{y}\) in each fiber, which is exactly how \(E_{C}\) is defined.