Lemma: Exact recovery and a constant complement

A channel $\mathcal{N}:M\to A$ has an exact CPTP left inverse if and only if a complementary channel satisfies

for a fixed density operator $\tau$.

Proof. Suppose a recovery exists. Compose a Stinespring isometry for $\mathcal{N}$ with one for the recovery. This dilates the identity channel. Every Kraus operator in any Kraus family of the identity is proportional to $I_M$, since its Choi operator has rank one. The dilation therefore factors as the unchanged message tensored with a fixed environment vector. Tracing the recovery environment makes the original complement constant.

Conversely, diagonalize $\tau ={\sum }_j{\lambda }_j|j⟩⟨j|$, omitting zero eigenvalues. A Stinespring isometry can be written

The constant-complement condition gives $⟨{\varphi }_{i^{\prime }{j}^{\prime }}|{\varphi }_{ij}⟩={\delta }_{i{i}^{\prime }}{\delta }_{j{j}^{\prime }}$. An isometry on their span sends ${\varphi }_{ij}$ to $|i⟩|j⟩$. Trace the second factor and define an arbitrary trace-preserving action on the orthogonal complement. This is an exact recovery.