Let $S$ be an operator system, $B$ a $c^*$-algebra and $\phi:S\to B$ a positive linear map. Then $\phi$ is involutive, i.e. $\phi(x^*)=\phi(x)^*$ for all $x\in S$. I want to prove this claim but I'm stuck. My try: Let $a\in S$, there are self-adjoint elements $x,y\in S$ such that $a=x+iy$. Therefore it is $\phi(a^*)=\phi(x-iy)=\phi(x)-i\phi(y)$ and $\phi(a)^*=\phi(x)^*-i\phi(y)^*$. But why is $\phi(x)=\phi(x)^*$ respectively $\phi(y)=\phi(y)^*$?
Could you help me to complete the proof? Regards
Edit: Ok, I have an idea: Every self-ajoint element $p\in S$ can be written as $p=u-v$, witch positive elements $u,v\in S$. So I can write $x=x_{+}-x_{-}$ and $y=y_{+}-y_{-}$ with positive elements $x_{+},x_{-},y_{+},y_{-}\in S$. Therefore it is $\phi(x)=\phi(x_{+})-\phi(x_{-})$ and $\phi(y)=\phi(y_{+})-\phi(y_{-})$. $\phi$ is a positive map, therefore $\phi(x_{+}),\phi(x_{-}),\phi(y_{+}),\phi(y_{-})$ are positive too, then $\phi(x),\phi(y)$ must be sel-adjoint as a difference of two positive elements, i.e. it is $\phi(x)=\phi(x)^*$ respectively $\phi(y)=\phi(y)^*$.
I hope, there is nowhere a mistake.
Yes, your argument is the standard way of proving it.