A proof of a proposition of tensor product

Question

Proposition 3.6.5.(The Trick) Let $A\subset B$ and $C$ be C*-algebras, $||.||_{\alpha}$ be a C*-norm on $B\odot C$ and $||.||_{\beta}$ be the C*-norm on $A\odot C$ obtained by restricting $||.||_{\alpha}$ to $A\odot C \subset B\odot C$. If $\pi_{A}: A\rightarrow B(H)$, $\pi_{C}: C\rightarrow B(H)$ are representations with commuting ranges and if the product *-homomorphism $$\pi_{A}\times \pi_{C}:A \odot C\rightarrow B(H)$$ is continuous with respect to $||.||_{\beta}$, then there exists a c.c.p (contractive completely postive) map $\phi: B \rightarrow \pi_{C}(C)'$ which extends $\pi_{A}$.

Proof. Assume first that $A, B$ and $C$ are all unital and, moreover, that $1_{A}=1_{B}$. Let $$\pi_{A}\times \pi_{B}: A\otimes_{\beta} C\rightarrow B(H)$$ be the extension of the product map to $A\otimes_{\beta} C$. Since $A\otimes_{\beta} C\subset B\otimes_{\alpha} C$, we apply Arveson's Extension Theorem to get u.c.p. (unital completely positive) extension $\Phi: B\otimes_{\alpha} C\rightarrow B(H)$. The desired map is just $\phi(b)=\Phi(b\otimes 1_{C})$.

To see that $\phi$ takes values in $\pi_{C}(C)'$ is a simple multiplicative domain argument. Indeed, $\mathbb{C}1_{B}\otimes C$ lives in the multiplicative domain of $\Phi$ since $\Phi|_{\mathbb{C}1_{B}\otimes C}=\pi_{C}$ is a *-homomorphism. Since $B\otimes \mathbb{C}1_{C}$ commutes with $\mathbb{C}1_{B}\otimes C$ and u.c.p. maps are bimodule maps over their multiplicative domains, a simple calculation completes the proof.

The nonunital case is ......

I can not comprehend the second paragraph of the proof. What does the author mean by "multiplicative domain argument"? The multiplicative domain of c.c.p. $\phi$ is $A_{\phi}=\{b\in B\otimes_{\alpha} C: \phi(b^{*}b)=\phi(b)^{*}\phi(b)$ and $\phi(bb^{*})=\phi(b)\phi(b)^{*}\}$.

Answer 1

As the index in Brown-Ozawa shows, the multiplicative domain is covered on page 12. If you look at Proposition 1.5.7, you see that if $b\in A_\phi$, then $\phi(ba)=\phi(b)\phi(a)$, $\phi(ab)-\phi(a)\phi(b)$ for all $a$.

Since $$\Phi((1_B\otimes c)^*(1_B\otimes c))=\Phi(1_A\otimes c^*c)=\pi_A\times\pi_B(1_A\otimes c^*c)=\pi_A(1_A)\pi_C(c^*c)=\pi_A(1_A)\pi_C(c)^*\pi_C(c)=\pi_A(1_A)\pi_C(c)^*\pi_A(1_A)\pi_C(c)=\Phi(1_B\otimes c)^*\Phi(1_B\otimes c), $$ and the same computation with the star on the right, we get that $1_B\otimes C$ is in the multiplicative domain of $\Phi$ for all $c\in C$. Then $$ \varphi(b)\pi_C(c)=\Phi((b\otimes 1_C)(1_A\otimes 1_C))\pi_C(c)=\Phi(b\otimes 1_C)\pi_A(1_A)\pi_C(1_C)\pi_C(c)=\Phi(b\otimes 1_C)\pi_A(1_A)\pi_C(c)=\Phi(b\otimes 1_C)\Phi(1_A\otimes c)=\Phi(b\otimes 1_C)\Phi(1_B\otimes c)\\ =\Phi((b\otimes 1_C)(1_B\otimes c)) =\Phi((1_B\otimes c)(b\otimes 1_C))\\ =\pi_C(c)\varphi(b) $$ (where in the last line you undo all the operations by reversing the first two lines).