Lemma: Snapshot–channel–adjoint comparison

Let $\Phi$ be a complete channel between finite tensor-product hybrid interfaces. Fix 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 on $I$ give equal output snapshots on $Y$, for all input states.

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

The factorization is stable under tensoring with an untouched reference. To infer equality of joint output–reference states, the two inputs must agree on their joint $I$–reference reductions.

Proof. (i) implies (ii). For (ii) implies (i), choose a fixed complementary state $\sigma$, put $E_{\sigma }({\omega }_I)={\omega }_I\otimes \sigma$, and define $K=R_Y\Phi E_{\sigma }$. This is a complete channel. Condition (ii) identifies $R_Y\Phi (\omega )$ with $K{R}_I(\omega )$ on all states, and therefore on the whole linear input space.

Taking adjoints gives (iii). Conversely, (iii) makes the expectation of every output observable on $Y$ a function of the snapshot on $I$; this is (ii).

Tensor the linear channel identity in (i) with a reference identity. The resulting factorization involves $I$ together with the reference, which proves the final statement.