Exercise 4.1.4 from Brown and Ozawa The question is to show the following: Let $(\mathcal{A},\alpha)$ and $(\mathcal{B},\beta)$ be $\Gamma$-$C^*$-algebras. Let $\varphi:\mathcal{A}\to \mathcal{B}$ be a c.c.p map which is $\Gamma$-equivariant. Show that the map $\tilde{\varphi}: C_c(\Gamma,\mathcal{A})\to C_c(\Gamma,\mathcal{B})$, defined by
$$\tilde{\varphi}\left(\sum_s a_ss\right)=\sum_s\varphi(a_s)s$$ extends to a continuous map from $\mathcal{A}\rtimes_{\alpha,r}\Gamma$ into $\mathcal{B}\rtimes_{\alpha,r}\Gamma$.
Of course, such a map is going to be $\Gamma$-equivariant since it is $\Gamma$-equivariant at the level of $C_c(\Gamma,\mathcal{A})$. I want to know if $\Gamma$-equivariance is necessary for this map to be continuous or not? It is clear that if I remove the assumption of $\Gamma$-equivariance of $\varphi:\mathcal{A}\to \mathcal{B}$, the map $\tilde{\varphi}$ is no longer going to be $\Gamma$-equivariant. I am concerned with the continuity of this map.
Here is my attempt at the proof:
Without any loss of generality, assume that $\mathcal{A}$ and $\mathcal{B}$ have a faithful covariant representation inside $\mathbb{B}(\mathcal{H})$ and $\mathbb{B}(\mathcal{K})$ respectively. Using Fell’s absorption principle, we can view $\mathcal{A}\rtimes_{\alpha,r}\Gamma$ and $\mathcal{B}\rtimes_{\beta,r}\Gamma$ as sub-algebras of $\mathbb{B}(\mathcal{H})\otimes C_r^*(\Gamma)$ and $\mathbb{B}(\mathcal{K})\otimes C_r^*(\Gamma)$ respectively. Using Arveson’s Extension theorem, we can view $\varphi$ as a c.c.p map from $\mathbb{B}(\mathcal{H})\to\mathbb{B}(\mathcal{K})$ which I denote by $\varphi$ again. Now, let us observe that $\tilde{\varphi}$ is nothing but the map $\varphi\otimes\text{id}|_{\mathcal{A}\rtimes_{\alpha,r}\Gamma}$.
Please let me know if I missed something here.
Thank you for the help.
 A: I think this solution is problematic, regardless of the question in hand about $G$-equivariance. The reason is the use of Arveson's extension theorem; yes, the reduced crossed product $A\rtimes_rG$ sits inside $B(H)\otimes C^*_r(G)$, but how? Let us see:
Fell's absorption principle tells us that, given a covariant representation $(H,\pi,u)$ for a $C^*$-dynamical system $\alpha:G\curvearrowright A$, the inflated representation $(H\otimes\ell^2G, \pi\otimes1, u\otimes\lambda^G)$ is unitarily equivalent to the regular representation $\pi\rtimes\lambda^G:C_c(G,A)\to B(H\otimes\ell^2G)$ given by
$$\pi\rtimes\lambda^G\bigg(\sum_sa_ss\bigg):=\sum_s\pi^{\alpha}(a_s)\cdot(1_H\otimes\lambda_s),$$
where $\lambda^G:G\to B(\ell^2G)$ denotes the left regular representation and  $\pi^\alpha:A\to B(H\otimes\ell^2G)$ is the representation defined by $\pi^\alpha(a)(h\otimes\xi_t)=\pi(\alpha_{t^{-1}}(a))(h)\otimes\xi_t$, where $\{\xi_t\}_{t\in G}$ is the canonical ONB of $\ell^2G$.
In particular, if $\pi$ is faithful to begin with, then, since $\pi\rtimes\lambda^G$ is faithful and unitarily equivalent to $(\pi\otimes1)\times(u\otimes\lambda^G)$, we have that $(\pi\otimes1)\times(u\otimes\lambda^G)$ is faithful. Notice that $(\pi\otimes1)\times(u\otimes\lambda^G)(as)=\pi(a)u_s\otimes\lambda^G_s\in B(H)\otimes C^*_r(G)$ for all $a\in A$ and $s\in G$, and thus we have an embedding
$$\iota:=(\pi\otimes1)\times(u\otimes\lambda^G):A\rtimes_rG\hookrightarrow B(H)\otimes C^*_r(G)$$
satisfying $\iota(\sum_sa_ss)=\sum_s\pi(a_s)u_s\otimes\lambda_s$ on $C_c(G,A)$.
Now let us follow the suggested solution in the post: start by a faithful covariant rep for $\alpha:G\curvearrowright A$, say $(H,\text{id}_A,u)$, (we consider $A\subset B(H)$ without loss of generality). Let us again denote by $\iota$ the embedding $$\iota:A\rtimes_rG\to B(H)\otimes C^*_r(G)$$
satisfying $\iota(\sum_sa_ss)=\sum_sa_su_s\otimes\lambda_s$.
Faithfully represent $B\subset B(K)$. The cpc map $\phi:A\to B\subset B(K)$ extends by Arveson to a cpc $\Phi:B(H)\to B(K)$. We now take the tensor product with $\text{id}_{C^*_r(G)}$ to get a cpc
$$\Phi\otimes\text{id}_{C^*_r(G)}:B(H)\otimes C^*_r(G)\to B(K)\otimes C^*_r(G)$$
and the claim is that $(\Phi\otimes\text{id})\circ\iota\vert_{C_c(G,A)}=\phi$. But,
$$(\Phi\otimes\text{id})\circ\iota(\sum_sa_ss)=\sum_s\Phi(a_su_s)\otimes\lambda_s$$
and there is no way of knowing what $\Phi(a_su_s)$ is, since $\Phi$ is an arbitrary extension. For your solution to work, we would need to fix a priori a faithful covariant rep $(K,\text{id}_B,v)$ for $\beta:G\curvearrowright B$ and we would need an extension $\Phi:B(H)\to B(K)$ satisfying $\Phi(au_s)=\phi(a)v_s$. Maybe, one can get such an extension using equivariance, but I am not sure about this part.
To sum up: the problem of this approach is that $A\rtimes_rG$ sits inside $B(H)\otimes C^*_r(G)$ in general, and not necessarily inside $A\otimes C^*_r(G)$.
A: Suppose that $X$ is a compact Hausdorff $\Gamma$-space, i.e., $\Gamma\curvearrowright X$ by homeomorphisms. Let $x\in X$ be a free point, i.e., $sx\ne x$ for all $s \in \Gamma\setminus\{e\}$. Let us consider the map $\varphi: C(X)\to\mathbb{C}$ by $\varphi(f)=f(x)$, i.e., the evaluation at the point $x$.
Since $x$ is a free point, this map is not $\Gamma$-equivariant. I claim that every extension of $\varphi$ to $C(X)\rtimes_r\Gamma$ factors through the canonical conditional expectation $\mathbb{E}$, i.e, if $\Psi: C(X)\rtimes_r\Gamma\to\mathbb{C}$ is such that $\Psi|_{C(X)}=\varphi$, then $\Psi=\Psi\circ\mathbb{E}$.
Towards this end, it is enough to show that $\Psi(f\lambda(s))=0$ for all $f\in C(X)$ and $s \ne e$. Let $f \in C(X)$ and $s\ne e$.  Let $g\in C(X)$ be such that $g(sx)=0$ and $g(x)=1$ (this follows by Urysohn's Lemma using the fact that $sx\ne x$). Let us now observe that
$$\Psi(f\lambda(s))=g(x)\Psi(f\lambda(s))=\Psi(gf\lambda(s))$$
The second equality follows from the fact that $Psi|_{C(X)}=\delta_x$, hence $C(X)$ falls in the multiplicative domain of $\Psi$.
Now, $$\Psi(gf\lambda(s))=\Psi(fg\lambda(s))=\Psi\left(f\lambda(s)(s^{-1}.g)\right)=\Psi(f\lambda(s))(s^{-1}.g)(x)=\Psi(f\lambda(s))g(sx)=0$$
Therefore, the claim follows.
If my solution would have been correct, then we would have gotten that $\tilde{\varphi}(f\lambda(s))=f(x)\lambda(s)$. $\Gamma$-equivariance is essential it seems ( I am not sure how though!!).
