Lemma: Snapshot–channel–adjoint equivalence

Let $\Phi$ be a complete channel between finite tensor-product hybrid interfaces. Fix an input region $I$ and output region $Y$. The following are equivalent:

1. There is a complete channel $K$ such that

   $$R_Y\Phi =K{R}_I.$$

2. Equal input snapshots in $I$ give equal output snapshots in $Y$, for all input states.

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

When they hold, factorization is stable under every untouched reference: equal input reductions on $I$ together with that reference give equal output reductions on $Y$ together with it.

Proof. (i) implies (ii). For (ii) implies (i), choose a fixed complementary state $\sigma$ and define

The map $K$ is complete and trace preserving. By (ii), $R_Y\Phi (\omega )=K{R}_I(\omega )$ on every state. States linearly span the hybrid input space, so the equality holds as a linear-map identity.

Taking adjoints in (i) gives

which is (iii). Conversely, (iii) implies that expectations of every output observable depend only on the input snapshot in $I$, giving (ii).

Finally, tensor the linear-map identity in (i) with a reference identity. This proves the reference statement. Equality on $I$ alone is not enough to conclude equality jointly with a reference unless the corresponding joint input reductions also agree.