Theorem: Exact accessible recovery and the complete complement

The following are equivalent:

1. a CPTP map $\mathcal{D}:R\to C$ satisfies $\mathcal{DN}={\mathrm{id}}_C$;

2. $\mathcal{E}(X)=\mathrm{Tr}(X){\sigma }_B$ for one fixed state ${\sigma }_B$;

3. $\mathcal{E}(E_{ij})={\delta }_{ij}{\sigma }_B$ on all logical matrix units.

If $\mathrm{rank}{\sigma }_B=r$, then $\dim R\ge dr$. All identities preserve arbitrary untouched references.

Proof. The equivalence of (ii) and (iii) is linearity.

Assume (iii), and diagonalize

Write

The matrix-unit identity implies

Thus $dr$ orthonormal vectors occur in $R$. Let $K\simeq {\mathbb{C}}^r$, put

Then

An explicit deterministic decoder is

It is CPTP by the same trace argument as in Lemma lem:global, and it exactly recovers $X$.

Conversely, dilate an exact decoder as an isometry $L:R\to C\otimes G$. The composite $Z=(L\otimes I_B)V$ dilates the identity. Since its decoded marginal on input $|i⟩$ is pure,

Recovery of $E_{ij}$ gives $⟨{\xi }_j,{\xi }_i⟩=1$; hence all normalized ${\xi }_i$ are the same vector. Tracing $C,G$ gives (iii). Every assertion is an operator identity, so reference extension preserves it.