A proposition in C*-algebra

Question

Problem: Let $A$ and $B$ be C*-algebra and $\varphi:A \rightarrow B$ be a contractive completely positive map. $A_{\varphi}=\{a\in A: \varphi(a^{\ast}a)=\varphi(a)^{\ast}\varphi(a)$ and $\varphi(aa^{\ast})=\varphi(a)\varphi(a)^{\ast}\}$, how to proof $A_{\varphi}$ is self-adjoint (i.e., if $a\in A_{\varphi}$, then $a^{\ast} \in A_{\varphi}$).

There is only word in the proof of this proposition: Following from the Bimodule property, that is, Given $a\in A$, if $\varphi(a^{\ast}a)=\varphi(a)^{\ast}\varphi(a)$ and $\varphi(aa^{\ast})=\varphi(a)\varphi(a)^{\ast}$, then $\varphi(ba)=\varphi(b)\varphi(a)$ and $\varphi(ab)=\varphi(a)\varphi(b)$.

I do not know how to use the Bimodule property. Could someone give me some hints?

Answer 1

It follows just from the definition and the fact that $\varphi$ is $*$-preserving: if $a\in A_\varphi$, then $$ \varphi((a^*)^*a^*)=\varphi(aa^*)=\varphi(a)\varphi(a^*)=\varphi(a^*)^*\varphi(a^*). $$