Theorem: Full-code recovery and a constant complement

Let $\mathcal{N}:\mathcal{B}(H)\to \mathcal{B}(Q)$ be CPTP, with Kraus operators $E_i:H\to Q$. The following are equivalent:

1. A CPTP map $D:\mathcal{B}(Q)\to \mathcal{B}(H)$ satisfies $D\mathcal{N}={\mathrm{id}}_H$.

2. Every product ${E_j}^{†}{E}_i$ is a scalar multiple of $I_H$.

3. A complementary channel of $\mathcal{N}$ has the form

   $$\hat{\mathcal{N}}(\rho )=\mathrm{Tr}(\rho )\omega$$

   for a fixed state $\omega$.

Proof. The standard Kraus complementary channel has entries

Hence (ii) and (iii) are equivalent, with the scalar matrix positive and of trace one.

For (i)$\Rightarrow$(ii), let $K_a$ be Kraus operators of $D$. The Kraus operators $K_a{E}_i$ represent the identity channel. Its Choi matrix has rank one, so every $K_a{E}_i=c_{ai}{I}_H$. Trace preservation of $D$ gives

For (ii)$\Rightarrow$(i), change the Kraus basis to diagonalize the positive scalar matrix. Remove zero Kraus operators and write

The ranges of the isometries $W_i$ are orthogonal. With $\Pi ={\sum }_i{W}_i{W}_i^{†}$, define

for an arbitrary fixed state $\tau$ on $H$. This map is CPTP and $D\mathcal{N}={\mathrm{id}}_H$.