Consider a non-singular, completely positive, unital map $\Psi: \mathbf M\_k(\mathbb C) \to \mathbf M\_h(\mathbb C)$. This map will have one or more retractions. Does $\Psi$ admit a retraction $\Phi: \mathbf M\_h(\mathbb C) \to \mathbf M\_k(\mathbb C)$, such that $\lVert \Phi \rVert = \lVert \Phi \rVert\_{\mathrm{cb}}$?


(This question is a follow-up to a [previous question](https://mathoverflow.net/questions/283787/completely-bounded-norm-for-unital-maps-with-completely-positive-sections), in which it was established that $\Psi$ may have retractions which do *not* have this property.)



Retractions for completely positive unital maps, with particularly nice norms
The answer is no again. When $h=k$, non-singular is the same as bijective so any $\Psi$ has only one map $\Phi$ such that $\Phi\circ\Psi = I\_{M\_k}$, namely the inverse. In the example below we find a bijective, unital, completely positive $\Phi$ such that its inverse $\Phi = \Psi^{-1}$ is unital, completely bounded but $\|\Phi\| < \|\Phi\|\_{cb}$.


Let $\Psi : M\_2 \rightarrow M\_2$ such that
$$
\Psi\left(\left[\begin{array}{cc} a&b \\ c&d \end{array}\right]\right)
= \frac{2}{5}\left[\begin{array}{cc}\frac{3}{2}a+d &c \\ b & \frac{3}{2}d+a \end{array}\right].
$$
Consider that $\Psi$ is unital and invertible and
$$
\left[\begin{array}{cc}\Psi(E\_{11}) & \Psi(E\_{12}) \\ \Psi(E\_{21}) & \Psi(E\_{22})\end{array}\right]
= \frac{2}{5}\left[\begin{array}{cccc}\frac{3}{2}\\ &1&1\\&1&1\\&&&\frac{3}{2}\end{array}\right] \geq 0
$$
gives by Choi's theorem that $\Psi$ is completely positive.
Because it is invertible there is only one unique choice for a retract: $\Phi = \Psi^{-1}$. However for $\|\left[ \begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right] \| \leq 1$
$$
\begin{align\*}
\left\|\Phi\left(\left[\begin{array}{cc} a&b \\ c&d \end{array}\right]\right)\right\|
&= \left\|\frac{5}{2}\left[\begin{array}{cc} (6a-4d)/5 &c \\ b & (6d-4a)/5 \end{array}\right]\right\|
\\[1ex]&= \frac{5}{2}\left\|\left[\begin{array}{cc} a & c \\ b & d \end{array}\right] + \left[\begin{array}{cc} (a-4d)/5 \\ & (d-4a)/5 \end{array}\right]\right\| \\[1ex]&\leq \frac{5}{2}(1 + 1) = 5
\end{align\*}
$$
but
$$
\left\lVert\, \Phi^{(2)}\left( \left[ \begin{array}{cccc} 1 \\ &&1 \\ &1\\ &&&1\end{array} \right] \right)^{\phantom \vert} \right\rVert
= \frac{5}{2}\left\lVert\; \left[\begin{array}{cccc} 6/5& &&1 \\ &-4/5 \\ &&-4/5 \\ 1& &&6/5\end{array}\right]^{\phantom\vert}\right\rVert > 5
$$
which implies that $\|\Phi\|\_{cb} > 5$. Therefore, $\|\Phi\| < \|\Phi\|\_{cb}$.


For the case when $h > k$ I still feel that one could find a non-singular ucp map with all retractions not realizing their cb-norms at the first level. 


