# unital and completely isometric imply completely positive?

Let $B, C$ be unital $C^*$-algebras, $S\subset C$ an operator system, $f:S\to C$ a unital linear map.

Then, $f$ is completely isometric if and only if $f$ is isometric and both $f$ and $f^{-1}:f(S)\to S$ are completely positive.

The $\Leftarrow-$direction is not difficult. Since $f$ is completely positive and unital, it is $\|f\|_{c.b.}=\|f\|=\|f(1)\|=1$. Therefore, the amplifications $f^{(n)}:M_n(S)\to M_n(B)$ are contractive for each $n\in \mathbb{N}_0$. The same argument shows that $f^{-1}$ is as well contractive. From there you get that $f$ is completely isometric.

However, for $\Rightarrow$, I don't know how to see that $f$ and $f^{-1}$ are completely positive.

I appreciate any hints.

$$\tag{*}\|f\|_{\rm cb}=\|f(1)\|\implies f\ \text{ is cp}$$
Assume first that $(*)$ holds for functionals, that is when $B=\mathbb C$. Consider, for general $B$, $f^{(n)}:M_n(S)\to M_n(B)$. Let $\varphi$ be a state on $M_n(C)$. Then $\varphi\circ f^{(n)}:M_n(S)\to \mathbb C$ is unital and $$\|\varphi\circ f^{(n)}\|\leq\|f^{(n)}\|\leq \|f(1)\|=\|1\|=1.$$ So the scalar version of $(*)$ applies to $\varphi\circ f^{(n)}$, and we get that $\varphi\circ f^{(n)}$ is positive for all states $\varphi$. Thus $f^{(n)}$ is positive.
To prove $(*)$ in the scalar, case, assume now that $B=\mathbb C$. Extend $f$, by Hahn-Banach, to all of $C$. We still have that $\|f\|=\|f(1)\|=1$ on $B$. Now let $a\in B^{\rm sa}$, with $\|a\|=1$. We have, for $n\in\mathbb Z$, \begin{align} |f(a)+in|&=|f(a+in 1)|\leq \|a+in 1\|=\|(a-in 1)(a+in 1)\|^{1/2}\\ \ \\ &=\|a^2+n^2 1\|^{1/2}=(1+n^2)^{1/2} \end{align} (the last equality, due to $a^2\geq0$). Looking at the imaginary part of $f(a)+in$, we get from above that $$n-(1+n^2)^{1/2}\leq\operatorname{Im} f(a)\leq -n+(1+n^2)^{1/2}.$$ As $n$ is arbitrary, it follows that $\operatorname{Im} f(a)=0$. Now we have $|f(a)|\leq\|f\|=f(1)=1$, so $f(a)\in[-1,1]$.
If we now take $a\geq0$ with $a\leq 1$, we apply the above to $2a-1$ to get $-1\leq 2f(a)-1\leq 1$, or $0\leq f(a)\leq 1$. Thus $f$ is positive.