Theorem: Exact recovery criterion

A finite-dimensional channel $\mathcal{N}$ has a CPTP left inverse if and only if a complementary channel has the form

for a fixed density operator $\tau$.

Proof. Let $V:{\mathcal{H}}_{\mathrm{in}}\to {\mathcal{H}}_{\mathrm{out}}\otimes E$ be a Stinespring isometry.

Suppose a recovery exists and let $W:{\mathcal{H}}_{\mathrm{out}}\to {\mathcal{H}}_{\mathrm{in}}\otimes F$ dilate it. Then $(W\otimes I_E)V$ dilates the identity channel. Every Kraus operator of the identity is proportional to the identity by the rank-one Choi argument. Consequently this composite isometry has the form

for one fixed unit vector. Tracing over the recovered input and $F$ shows that the original complementary output is constant.

Conversely, diagonalize $\tau ={\sum }_j{\lambda }_j|j⟩⟨j|$, omitting zero eigenvalues. Write $V={\sum }_j{V}_j^{\mathrm{raw}}\otimes |j⟩$. The complementary identity implies

Hence ${V_j}^{\mathrm{raw}}=\sqrt{{\lambda }_j}{V}_j$, where the $V_j$ are isometries with mutually orthogonal ranges. Let $P={\sum }_j{V}_j{V}_j^{†}$, and choose a density operator ${\omega }_{\ast }$ on the input. Define

This is completely positive and trace preserving. Since $\mathcal{N}(X)={\sum }_j{\lambda }_j{V}_{j}X{V}_j^{†}$, orthogonality gives $\mathcal{RN}(X)=X$.