Corollary: No simultaneous exact bleaching in a pure complete dilation

For $d\ge 2$, the two complementary channels of an isometry cannot both be constant on the entire code. Nor can both outputs separately recover the entire unknown code.

Proof. If one is constant, the other is exactly recoverable. A constant channel cannot recover two distinct inputs. If one output is exactly recoverable, its complement is constant by Theorem thm:exact, and therefore is not itself recoverable.