Proposition: Entanglement-breaking boundary and exact recovery

The noisy accessible channel is entanglement breaking if and only if $\eta \lambda =0$. If $\lambda =0$, every deterministic decoder obeys

Exact deterministic recovery is possible if and only if $(\eta ,\lambda )=(1,1)$.

Proof. At $\lambda =0$, the channel is measure and prepare:

Its referenced output is a sum of positive product operators, hence separable. At $\eta =0$, it is vacuum replacement. Lemma lem:witnesses gives the floor $1/2$.

For the Bell input, the partial-transpose calculation of Proposition prop:negativity now has coherence $\lambda \eta /2$, so

It is strictly positive if $\eta \lambda >0$, proving the converse entanglement-breaking statement.

If $\eta <1$, a left inverse of ${{\mathcal{Z}}^{\mathcal{R}}}_{\lambda }{\mathcal{N}}_{\eta }$ would make $\mathcal{D}{{\mathcal{Z}}^{\mathcal{R}}}_{\lambda }$ a left inverse of ${\mathcal{N}}_{\eta }$, contradicting Corollary cor:exactEndpoint. The full-transfer case is settled by Theorem thm:fullNoise below.