Corollary: The precise no-hiding conclusion

The complete residual $B$ exactly recovers $C$ if and only if $\mathcal{N}$ is a replacement channel. The accessible factor $R$ exactly recovers $C$ if and only if $\mathcal{E}$ is a replacement channel. For $d\ge 2$, the two factors cannot both independently recover the entire code, and cannot both have constant channels.

Proof. Apply Theorem thm:exact with the output factors interchanged. A replacement channel cannot recover two distinct code states. If one factor recovers, the other is constant; this rules out both independent recovery and simultaneous constancy.