Let $G$ be a discrete group. Let $(A,\alpha)$ and $(B,\beta)$ be $G$-$C^\*$-algebras and $\varphi: A \to B$ be $G$-equivariant and completely positive. All crossed products in this post are full (= universal).


I want to prove that there is a completely positive map
$$\varphi \rtimes G: A \rtimes\_\alpha G \to B \rtimes\_\beta G$$
such that $(\varphi\rtimes G)(\sum\_s a\_s s) = \sum\_s \varphi(a\_s)s$.


I managed to prove the following using the $G$-equivariant Stinespring theorem:



> 
> Assume that $u: G\to B(H)$ is a unitary representation and $\sigma: A \to B(H)$ a completely positive map satisfying $\sigma(\alpha\_g(a)) = u\_g \sigma(a)u\_g^\*$ (i.e. $\sigma$ is $G$-equivariant where $B(H)$ has the $G$-action induced by $u$). Then there is a unique completely positive map
> $$\sigma \rtimes G: A \rtimes G \to B(H)$$
> satisfying $\sigma \rtimes G(\sum\_s a\_s s) = \sum\_s \sigma(a\_s)u\_s.$
> 
> 
> 


I think I might be able to use this result to prove the result I want: Maybe the following works:


Let $(u,\pi)$ be a covariant representation of $B \rtimes G$ on $H$ where $\pi$ is chosen faithful. Then consider the composition $\sigma: A \to B(H)$ defined by
$$A \stackrel{\varphi}\to B \stackrel{i}\to B \rtimes G \stackrel{\pi}\to B(H)$$
which is completely positive and satisfies $\sigma(\alpha\_g(a)) = u\_g \sigma(a)u\_g^\*$. Hence, by the above result, we obtain an induced map
$$\sigma \rtimes G: A \rtimes G \to B(H).$$


If we can check that $(\sigma \rtimes G)(A \rtimes G) \subseteq \operatorname{Im}(\pi)$ then we can define
$$\varphi \rtimes G(x) = \pi^{-1}(\sigma \rtimes G)(x) \in B \rtimes G$$
which would yield the desired extension. However, I don't think the above inclusion holds.



A completely positive equivariant map $\varphi: A \to B$ induces a map on the full crossed products
(Too long for a comment). Your notation is confusing you, I think.


A *covariant* representation is not really of $B\rtimes G$, but is of the pair $(B,\beta)$. In particular, $\pi:B\rightarrow B(H)$ (notice the domain!) and not a representation of $B\rtimes G$. The covariant condition, that $\pi(\beta\_t(b)) = u\_g \pi(b) u\_g^\*$, ensures that $\pi$ *extends* to a $\*$-representation, say $\tilde\pi: B\rtimes G\rightarrow B(H)$. (Of course, the point is that *any* $\*$-representation of $B\rtimes G$ comes from some pair $(\pi,u)$, and so we can confuse them!)


Thus, in your argument, we *do not expect* that $(\sigma \rtimes G)(A\rtimes G) \subseteq \operatorname{Im}(\pi)$ but rather in the image of $\tilde\pi$.


Your argument now seems to be to pick $(\pi,u)$ "universal", so that $\tilde\pi$ will be an isometry, and then show that for this $\pi$ we get that $\sigma\rtimes G$ maps into the image of $\tilde\pi$. I think this is essentially obvious. First notice that
$$ \sigma = \tilde\pi \circ i \circ \varphi \implies
\sigma(a) = \pi(\varphi(a)) \quad (a\in A), $$
because by definition $\tilde\pi \circ i = \pi$. By density, we can consider a finite sum $\sum a\_s s \in A\rtimes G$, which has
$$ (\sigma \rtimes G) \Big(\sum a\_s s\Big) = \sum \sigma(a\_s) s
= \sum \pi(\varphi(a\_s)) s, $$
which is clearly a member of the image of $\tilde\pi$.


