Let $T$ be an injective operator system and $U$ be an arbitrary operator system. Let $\varphi: T \to U$ be a unital completely positive map and $\psi: U \to T$ be a unital completely positive map with $\psi \varphi = \iota_T$. Is it true that $\varphi$ is a complete isometry?
Attempt: I think yes. Here is an argument:
It is clear that $\varphi: T \to \varphi(T)\subseteq U$ is a unital complete order isomorphism, as its inverse $\psi: \varphi(T) \to T$ is also a completely positive map. However, it then follows that $\varphi(T)$ is an injective operator system. By a result of Choi and Effros, we can equip $T$ and $\varphi(T)$ with multiplications such that these operator systems become $C^*$-algebras such that the identity maps become unital completely isometric maps. But then the map $\varphi: T \to \varphi(T)$ becomes a $*$-isomorphism (because a unital complete order isomorphism between $C^*$-algebras automatically preserves the $C^*$-algebra structures, this is a result by Choi). In particular, $\varphi$ is completely isometric.
Is the above argument correct? And can we prove it using more "elementary" results?
Context of the question: I try to understand the fact that an injective rigid extension is automatically also an essential extension (see for instance Paulsen's book, theorem 15.8). I tracked down the original proof in the paper "Injective envelopes of operator systems" by Hamana, lemma 3.7, where they end the proof by asserting that a certain composition is the identity map, which motivates my question above. In other papers by Hamana, one can encounter similar statements.
Best Answer
Injectivity of $T$ plays no role. The point is that unital completely positive maps are contractive, so if a strict inequality $\|\phi(a)\|< \|a\|$ holds for some $a\in T$, then $\|a\|=\|\psi(\phi(a))\|\leqslant \|\phi(a)\| < \|a\|$, which is a contradiction. This shows that $\phi$ is an isometry and the same argument applied to matrix amplifications proves that it is in fact a complete isometry.