Recall the construction of the reduced crossed product:



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> Let $\Gamma$ be a discrete group and $A$ be a $C^\*$-algebra with an action $\alpha: \Gamma\to \operatorname{Aut}(A)$. Consider the $\*$-algebra $C\_c(\Gamma,A)$ of finitely supported functions $\Gamma \to A$ with the $\alpha$-twisted multiplication and involution. Then we can build a canonical faithful representation of $C\_c(\Gamma,A)$ as follows: start with a faithful representation $A \subseteq B(H)$. This induces a new faithful representation $\pi: A \to B(H \otimes \ell^2(\Gamma))$ by $\pi(a)(\xi \otimes \delta\_g) = \alpha\_{g}^{-1}(a)\xi \otimes \delta\_g$. Considering the left regular representation $\lambda: \Gamma \to U(\ell^2(H)): g \mapsto (\delta\_h \mapsto \delta\_{gh})$, we obtain an induced faithful representation
> $$C\_c(\Gamma,A) \to B(H \otimes \ell^2(\Gamma)): \sum\_{s \in \Gamma} a\_s s \mapsto \sum\_{s \in \Gamma} \pi(a\_s)(1 \otimes \lambda\_s)$$
> which induces a $C^\*$-norm on $C\_c(\Gamma,A)$. The reduced crossed product $A \rtimes\_r \Gamma$ is the $C^\*$-completion of $C\_c(\Gamma,A)$ with respect to this norm, and does not depend on the choice of faithful representation $A\subseteq B(H)$.
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Let $\Gamma$ be a discrete group and let $\varphi: A \to B$ be a $\Gamma$-equivariant completely positive contraction between the $\Gamma$-$C^\*$-algebras $A$ and $B$. I want to show the following:



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> The induced map $C\_c(\Gamma,A) \to C\_c(\Gamma,B): \sum\_{s \in \Gamma} a\_s s \mapsto \sum\_{s \in \Gamma}\varphi(a\_s)s$ is bounded, hence extends uniquely to a map $\varphi \rtimes\_r \Gamma: A \rtimes\_r \Gamma \to B \rtimes\_r \Gamma.$
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**Attempt:** Let $\pi\_A: A \to B(H\_A \otimes \ell^2(\Gamma))$ and $\pi\_B: B \to B(H\_B \otimes \ell^2(\Gamma))$ be faithful representations as above. Then by the $C^\*$-identity.
$$\|\sum\_s \varphi(a\_s)s\|^2 = \|\sum\_s \pi\_B(\varphi(a\_s))(1 \otimes \lambda\_s)\|^2$$
$$=\|\sum\_{s,t} (1 \otimes \lambda\_{s^{-1}}) \pi\_B(\varphi(a\_s^\*)\varphi(a\_t)) (1\otimes \lambda\_t)\|.$$


This looks like something we could apply Cauchy-Schwarz for completely positive maps on, but the surrounding factors $1 \otimes \lambda\_{s^{-1}}$ and $1 \otimes \lambda\_t$ complicate this. Maybe I need to apply Cauchy-Schwarz on some carefully crafted matrix. Does anybody see how I can continue?


Of course, other approaches are also welcome! I am also interested in the following: if $\varphi$ is injective, then is the extension $A \rtimes\_r \Gamma \to B \rtimes\_r \Gamma$ also injective?



A completely positive equivariant map $\varphi: A \to B$ induces a map $A \rtimes_r \Gamma \to B \rtimes_r \Gamma.$
This is Exercise 4.1.4 in the book by Brown+Ozawa (is that where this question ultimately comes from?) The "Hint" is "The reduced case is easy." Hmm.


Well, it might perhaps help to look at this section of the book. Indeed, Proposition 4.1.5, *and its proof*, shows that if $F\subseteq\Gamma$ is a finite-set, and $P:\ell^2(\Gamma)\rightarrow \ell^2(F)$ the projection, then
$$ (1\otimes P)x(1\otimes P) = \sum\_{s\in\Gamma} \sum\_{p\in F\cap sF} \alpha\_{p^{-1}}(a\_s)\otimes e\_{p, s^{-1}p} \in A\otimes M\_F(\mathbb C)
\cong M\_F(A). $$
Here $x = \sum a\_s\lambda\_s \in C\_c(\Gamma, A) \subseteq A \rtimes\_r \Gamma \subseteq \mathcal B(H\otimes\ell^2(\Gamma))$ and $e\_{p,q}$ are the matrix units of $M\_F(\mathbb C)$. The point of this is to show that the norm of $x$, acting on $H\otimes\ell^2(\Gamma)$, does not depend upon the particular representation $A\subseteq\mathcal B(H)$ (because $M\_F(A)$ has a unique norm).


However, this formula also gives us a convenient way to compute the norm of $\varphi\rtimes 1$. For, if $y=(\varphi\rtimes 1)x$, then $y=\sum\_s \varphi(a\_s)\lambda\_s$, and so
$$ (1\otimes P)y(1\otimes P) = \sum\_{s\in\Gamma} \sum\_{p\in F\cap sF} \beta\_{p^{-1}}(\varphi(a\_s))\otimes e\_{p, s^{-1}p}
= \sum\_{s\in\Gamma} \sum\_{p\in F\cap sF} \varphi(\alpha\_{p^{-1}}(a\_s))\otimes e\_{p, s^{-1}p}, $$
the first equality by applying to above to $B\rtimes\_r\Gamma$, and the second equality using that $\varphi$ is equivariant. However, this then equals
$$ (\varphi\otimes 1\_F)\big( (1\otimes P)x(1\otimes P) \big), $$
where $(\varphi\otimes 1\_F)$ is the dilated map $M\_F(A)\rightarrow M\_F(B)$. As $\varphi$ is CCP by assumption, $\varphi\otimes 1\_F$ is contractive, and so
$$ \| (1\otimes P)y (1\otimes P) \| \leq \|(1\otimes P)x(1\otimes P)\|
\leq \|x\|. $$
Taking the SOT limit as $F$ increases, so $P\rightarrow 1$, shows that $\|y\| \leq \|x\|$ as required.


