Lemma: Snapshot, channel, and adjoint comparison

For a complete finite channel $\Phi$, input region $I$, and output region $Y$, the following are equivalent:

1. $R_Y\Phi =K{R}_I$ for a complete channel $K$;

2. equal input snapshots on $I$ give equal output snapshots on $Y$, for every pair of input states;

3. ${\Phi }^{\ast }(\mathcal{A}(Y))\subseteq \mathcal{A}(I)$.

The factorization is stable under tensoring with an untouched reference identity.

Proof. The first assertion implies the second. Conversely choose a fixed complementary state $\sigma$, put $E_{\sigma }({\rho }_I)={\rho }_I\otimes \sigma$, and define $K=R_Y\Phi E_{\sigma }$. The second assertion gives $R_Y\Phi (\rho )=K{R}_I(\rho )$ for every state. States linearly span the finite hybrid operator space, so this is a linear-map identity. Taking adjoints proves the third assertion. Conversely, the third assertion says that every output expectation depends only on $R_I\rho$, which proves the second.

Tensoring the linear-map identity with ${\mathrm{id}}_R$ proves the reference claim. Equality on $I$ alone does not imply equality jointly with a reference; the appropriate joint reductions on $IR$ must agree.