\noindent (SUFFICIENT TO PROVE) By \cite[Theorem 4.9]{buss:injectivity} it suffices to prove that if \[ \phi:\ell^\infty(G)\otimes C(X) \to B(H) \] is a unital covariant representation, then there is an equivariant ucp map \(\tilde{\phi}\) as below \begin{center} \begin{tikzcd} (\ell^\infty(G)\otimes \ell^\infty(Y;X)) \arrow[dr, dashed, "\tilde{\phi}"]\\ (\ell^\infty(G) \otimes C(X)) \arrow[u]\arrow[r,"\phi" ] & B(H). \end{tikzcd} \end{center} Injectivity of Equation~\ref{eqn:assumed} implies that \(\ell^\infty(G)\otimes C(Y)\) has the \(G\)-WEP~\cite[Theorem 5.4]{buss:injectivity} \noindent (SETTING UP \(\psi\))\hrulefill \begin{center} % todo ? explain the maps \begin{tikzcd} (\ell^\infty(G)\otimes \ell^\infty(Y)) \arrow[d,hookleftarrow]\arrow[dr, "\text{Exists by G-WEP}", dashed]\\ (\ell^\infty(G) \otimes C(Y)) \arrow[d, "\phi"]\arrow[r,hookrightarrow] & {\ell^\infty(G)\otimes C(Y)}^{* *} \arrow[dl, dashed, "\theta"]\\ B(H) \end{tikzcd} \end{center} Let \(\psi:\ell^\infty(G) \otimes \ell^\infty(Y) \to B(H)\) be the composition of the two dashed maps. Then \(\psi\) takes image in \(\phi(\ell^\infty(G) \otimes C(Y))''\) as the map \(\theta\) does. Moreover, \(\phi(C(X)) \subseteq \phi(\ell^\infty(G) \otimes C(Y))'\) whence \[ \phi(\ell^\infty(G)\otimes C(Y))'' \subseteq \phi(C(X))'. \] Hence \(\psi\) takes image in \(\phi(C(X))'\) and so we get an equivariant ucp map % fixme why is this the case? \[ \psi \otimes \phi\vert_{C(X)} : \ell^\infty(G) \otimes \ell^\infty(Y) \otimes C(X) \to B(H). \] \noindent (SETTING UP \(\mu\))\hrulefill On the other hand, there is a map \begin{equation}\label{eqn:mu} \mu: \ell^\infty(Y) \otimes C(X) \to \ell^\infty(Y;X) \end{equation} defined by considering \(\ell^\infty(Y)\) and \(C(X)\) as \(C^*\)-subalgebras of \(\ell^\infty(Y;X)\) and defining \(\mu(g\otimes h) = gh\). It remains to show that \(\psi\otimes id\) factors through \(id\otimes \mu\) {\emph i.e.} we have \begin{center} \begin{tikzcd} \ell^\infty(G)\otimes \ell^\infty(Y) \otimes C(X) \arrow[d,two heads,"id\otimes \mu"]\arrow[r, "\psi\otimes \restrict{\phi}{C(X)}"] & B(H) \\ \ell^\infty(G) \otimes \ell^\infty(Y;X) \arrow[ur] \end{tikzcd} \end{center} \noindent END OF FIRST SECTION OF NOTES \hrulefill \begin{claim} \(\mu\) is surjective. \begin{proof} Note that since \(\mu\) is a \(*\)-homomorphism between \(C^*\)-algebras, it suffices to show that \(\mu\) has dense image. Let \(\epsilon > 0\) and let \(f \in \ell^\infty(Y;X)\). As the family \(\left\{ f\restrict{{\pi}^{-1} (y)} \mid y \in Y \right\}\) is equicontinuous there is a \(\delta > 0\) such that for any \(x,z \in X\) \[ \pi(x) = \pi(z) \text{ and } d(x,z) < \delta \implies |f(x) - f(z)| < \frac{\epsilon}{3}. \] %% Fact from topology: Given any compact Hausdorff space \(X\) with open %% cover \(U_{1},\ldots,U_{n}\) there exist continuous functions %% \(h_{1},\ldots,h_{n}:X\to[0,1]\) so that \(\text{supp}(h_{j}) \subseteq %% U_{j}\) and \(\sum_{j} = 1\). This is a partition of unity. Where %% supp is support. Let \(U_{1},\ldots,U_{n}\) be an open cover of \(X\) by sets of diameter at most \(\delta\) with \(\varphi_{1},\ldots, \varphi_{n}\) a subordinate partition of unity. Let \(M\) be such that \(|f(x)| \leq M\) for all \(x \in X\). Let \(L \in \mathbb{N}\) be such that \(\frac{1}{L} < \frac{\epsilon}{3} \) and let \[ \bb{Q}_{L,M} := \left\{ \frac{r}{L} + \frac{t}{L}i \in \bb{Q}[i] \mid r,t \in \mathbb{Z}; \abs{\frac{r}{L}}, \abs{\frac{t}{L}} \leq M \right\} \] and \[ \mathcal{F} := \left\{ \sum_{i=1}^{N} \lambda_{i}\varphi_{i} \in C(X) \mid \lambda_{i} \in \bb{Q}_{L,M}[i] \right\} \] which is a finite subset of \(C(X)\). %% finite because r, t %% integers and so they are %% only the integers which %% are at most M*L \begin{claim} For each \(y \in Y\) there is a \(g_{y} \in \mathcal{F}\) with \(|f(x) - g_{y}(x)| < \frac{2\epsilon}{3}\) for all \(x \in {\pi}^{-1} (y)\). \begin{proof} Let \[ S = \left\{ i \in \left\{ 1,\ldots, N \right\} \mid U_{i}\cap {\pi(y)}^{-1} \neq \emptyset \right\} \] and choose \(x_{i} \in U_{i} \cap {\pi(y)}^{-1}\) for all \(i \in S\). Note that \(\left\{ \varphi_{i} \restrict{{\pi(y)}^{-1} } \right\}_{i \in S}\) is a partition of unity on \({\pi(y)}^{-1}\) % check with all functions having support diameter at most \(\delta\). For each \(i \in S\) choose \(\lambda_{i} \in \bb{Q}_{L,M}[i]\) such that \(|\lambda_{i} - f(x_{i})|< \frac{\epsilon}{3}\). Then define \(g_{y}:= \sum_{i \in S} \lambda_{i} \varphi_{i} \in \mathcal{F}\) and for any \(x \in {\pi(y)}^{-1}\), \begin{align*} |g_{y}(x) - f(x)| &= \left| \sum_{i \in S}^{}\lambda_{i}\varphi_{i}(x) - \sum_{i \in S}^{}f(x)\varphi_{i}(x) \right| \\ &\leq \sum_{i \in S}^{}\varphi_{i}(x)|\lambda_{i} - f(x)| \\ &\leq \sum_{i \in S}^{}\varphi_{i}(x)\left(|\lambda_{i} - f(x_{i})|+|f(x_{i}) - f(x)| \right) \\ &< \frac{2\epsilon}{3}, \end{align*} completing the proof of the claim. \end{proof} \end{claim} Now, choose \(g_{y}\) as in the claim for each \(y \in Y\). For each \(g \in \mathcal{F}\), let \(E_{g} := \left\{ y \in Y \mid g_{y} = g \right\}\) (possibly empty). Note that \(Y = \sqcup_{g \in \mathcal{F}} E_{g}\). Note finding that \(\sum_{g \in \mathcal{F}}\chi_{E_{g}}\cdot g\) is in the image of the map in Equation~\ref{eqn:mu}, and that for any \(x \in X\) we have that if \(y = \pi(x)\) then \[ \abs{f(x) - \sum_{g \in \mathcal{F}} \chi_{E_{g}}(\pi(x)) g(x)} = \abs{f(x) - g_{y}(x)} < \frac{2\epsilon}{3}. \] Hence \[ \nrm{f - \sum_{g \in \mathcal{F}}^{}\chi_{E_{g}} g} \leq \frac{2\epsilon}{3}< \epsilon \] and we are done. \end{proof} \end{claim} % Ends halfway through pg 12 \noindent (FINISH \(\mu\) AFTER SURJECTIVITY)\hrulefill It remains to show that we have a commutative diagram \begin{equation}\label{eqn:mu-commutative-diagram} \begin{tikzcd} \ell^\infty(G)\otimes \ell^\infty(Y) \otimes C(X) \arrow[d,two heads,"id\otimes \mu"]\arrow[r, "\psi\otimes \phi\restrict{C(X)}"] & B(H) \\ \ell^\infty(G) \otimes \ell^\infty(Y;X) \arrow[ur,dashed] \end{tikzcd} \end{equation} where the dashed arrow is a ucp map. We first compute the effect of \(\mu\) on spectra: we claim that the image of the dual map (which is injective by the surjectivity of \(\mu\)) \[ \mu^{*}:\widehat{\ell^\infty(Y;X)}\hookrightarrow \beta Y \times X \] is exactly \(\beta Y \times_{Y} X:= \left\{ (\omega,x) \in \beta Y\times X \mid p(\omega) = \pi(x) \right\}\) where \(p:\beta Y \to Y\) and \(\pi:X \to Y\) are the quotient maps. To see this, first note that as \(\mu(f\otimes 1) = \mu(1 \otimes f)\) for all \(f \in C(Y)\), we have that \(\mu^{*}\) takes image in \(\beta Y \times_{Y} X\). To see that it is surjective, note that \(\widehat{\ell^\infty(Y;X)}\) contains \(Y \times_{Y} X\) as a subspace, and sends this to the corresponding subspace of \(\beta Y \times_{Y} X\) as that subspace is dense and the image of \(\mu^{*}\) is closed (since compact) we are done. Let \(Z\) be such that \(\phi(\ell^\infty(G)\otimes C(X)) = C(Z)\) and consider the duals of the maps in Figure~\ref{eqn:mu-commutative-diagram}: \begin{center} \begin{tikzcd} \text{Prob}(\beta G\times \beta Y \times X) & \arrow[l, "f"'] Z \arrow[dl, dashed]\\ \text{Prob} (\beta G \times \beta Y \times_{Y} X) \arrow[u, hook, "(id \otimes \mu)^{*}"] \end{tikzcd} \end{center} We want the dashed arrow to exist, for which it suffices to show that \(f\) takes image in the image of the vertical map. Now, as \((\psi \otimes\phi\restrict{C(X)})\restrict{\ell^\infty(G)\otimes 1 \otimes 1}\) and \((\psi \otimes\phi\restrict{C(X)})\restrict{1 \otimes 1 \otimes C(X)}\) are \(*\)-homomorphisms, \(f\) actually takes image in \(\beta G \times \text{Prob} (\beta Y) \times X\). Writing \(f=(f_{1}, f_{2}, f_{3})\) it su\(f\)ices to show that the support of \(f_{2}(z) \in \text{Prob}(\beta Y)\) is contained in \({p}^{-1}(\pi(f_{3}(z)))\). Suppose not, for contradiction, so there is \(\omega \in \beta Y\) with \(p(\omega)\neq \pi(f_{3}(z))\) such that for any open \(U \ni \omega\) in \(\beta Y\) there is a \(g \in C_{0}(U)_{+}\) with \(g(\omega) = 1\) and \(f_{2}(z)(g) > 0\). As \(p(\omega) \neq \pi(f_{3}(z))\) there is \(V \subseteq Y\) open with \(V \ni p(\omega)\) and \(\pi(f_{3}(z)) \not\in \overline{V}\). Let \(h \in C(Y)\) satisfy \(f{\restrict{\overline{v}}} = 1\) and \(f(\pi(f_{3}(z))) = 0\). Let \(g \in C(\beta Y)\) have the properties above for \(U = {P}^{-1}(V)\). Now, as \(C(Y)\) is in the multiplicative domain of \(\psi\), we have \[ \left(\psi\otimes\phi\restrict{C(X)}\right)(1\otimes gh \otimes 1) = \psi ( 1 \otimes g) \phi(h) = \left(\psi \otimes \phi\restrict{C(X)})\right(1 \otimes g\otimes h). \] Hence \begin{equation}\label{eqn:rename} \overline{f(z)(1 \otimes gh \otimes 1)} = \overline{f(z)(1 \otimes g \otimes h)} \end{equation} as \(f(z)\) evaluated at a function depends only on its image under \(\psi \otimes \phi\restrict{C(X)}\). By Equation~\ref{eqn:rename}, \[ f_{2}(z)(gh) = f_{2}(z)(g)\cdot h(\pi(f_{3}(z)) \] As \(gh = g\), the left hand side is \(f_{2}(z)(g) > 0\). As \(h(\phi(f_{3}(z))) = 0\) the right hand side is two; this contradiction completes the proof.