Corollary: Two limiting cases

Under the hypotheses of thm:luders:

1. if $q$ is injective, exact record compatibility already implies full Lüders compatibility;

2. if there is one coarse outcome, full compatibility is equivalent to $T=T\circ {\mathcal{D}}_P$.

Proof. For injective $q$, the two dephasings in eq:dephasings coincide. For one coarse outcome, ${\Pi }_{\ast }=I$, so ${\mathcal{D}}_{\Pi }=\mathrm{id}$. Apply thm:luders.