Consider a completely bounded unital map $\Phi: \mathbf M\_h(\mathbb C) \to \mathbf M\_k(\mathbb C)$. Suppose that $\Phi$ has right-inverse $\Psi$ which is completely positive. Is the operator norm of $\Phi$ stable under tensor products with other spaces — *i.e.*, is $$\lVert \Phi \rVert \stackrel?= \lVert \Phi \rVert\_{\mathrm{cb}} = \bigl\lVert \Phi \otimes \mathrm{id}\_{\mathbf A} \bigr\rVert$$ for $\mathbf A = \mathbf M\_h(\mathbb C)$?



Completely bounded norm for unital maps with completely positive sections
Unfortunately, the answer is no.


Suppose $\Phi : M\_4 \rightarrow M\_2$ and $\Psi : M\_2 \rightarrow M\_4$ are given by 
$$\Phi\left(\left[\begin{array}{cc} A & B\\ C& D\end{array}\right]\right) = A + 10B^T \ \ \textrm{and} \ \ \Psi(A) = \left[\begin{array}{cc} A&0\\ 0&A\end{array}\right] $$
where $A,B,C,D\in M\_2$. Then $\Phi$ is unital, completely bounded, $\Psi$ is unital, completely positive and $\Phi\circ\Psi = I\_{M\_2}$ but 
$$\|\Phi\| \leq 11 \ \ \textrm{and} \ \ \|\Phi\|\_{\rm cb} \geq 20$$
since the transpose map is cb-norm 2.


  

There is also a negative answer if you look for $\Phi$ being left-invertible by a ucp map:


In this case a counterexample is given by $\Phi : M\_2 \rightarrow M\_4$ and $\Psi : M\_4 \rightarrow M\_2$ defined by
$$
\Phi(A) = \left[\begin{array}{cc}A & 0 \\ 0 & A^T\end{array}\right]\ \ \textrm{and} \ \ \Psi\left(\left[\begin{array}{cc} A & B \\ C & D\end{array}\right]\right) = A
$$
where $A,B,C,D \in M\_2$. Then $\Phi$ is ucb, $\Psi$ is ucp and $\Psi\circ\Phi = I\_{M\_2}$ but
$$
\|\Phi\| = 1 < 2 = \|\Phi\|\_{\rm cb}.
$$


