Proposition: Classical-channel obstruction inside the pilot

For the witness family, every first link with $e\ge 4$ is entanglement breaking. Nevertheless its signalling strength is positive.

Without an additional quantum side channel or preshared entanglement, no local decoder after such a link approximates the unknown-qubit identity channel with half-diamond error below $1/2$.

Proof. For $e\ge 4$ and $s\le 2$, $p_e=s/(e+s)\le 1/3$. The normalized Choi state of ${\mathcal{D}}_p$ is

At $p=1/3$ it is the separable mixture

For smaller $p$ it is a convex mixture of that state and $I_4/4$. Equivalently, ${\mathcal{D}}_p$ has a measure-and-prepare implementation and is entanglement breaking. The strength remains $p_e>0$.

Local decoding preserves separability from an untouched reference. Every separable two-qubit state has Bell overlap at most $1/2$: for pure product states this follows by Cauchy–Schwarz, and mixtures preserve the bound. Testing the Bell projector against the identity channel output gives half-diamond error at least $1/2$.