Proposition: Exact snapshot–channel–adjoint comparison

Let ${\mathcal{C}}_Y$ be a channel from the full input $X$ to an output factor $Y$, and let $A\subseteq X$. The following are equivalent:

1. Whenever ${\mathrm{Tr}}_{X\setminus A}\rho ={\mathrm{Tr}}_{X\setminus A}\sigma$, one has ${\mathcal{C}}_Y(\rho )={\mathcal{C}}_Y(\sigma )$.

2. There is a unique channel ${\Psi }_A$ such that

   $${\mathcal{C}}_Y={\Psi }_A\circ {\mathrm{Tr}}_{X\setminus A}.$$
3. $${\mathcal{C}}_Y^{\ast }(\mathcal{B}({\mathcal{H}}_Y))\subseteq \mathcal{A}(A).$$

The factorization is stable under tensoring with an untouched finite-dimensional reference.

Proof. Assume (i), and fix a state $\omega$ on $X\setminus A$. Define

It is a channel. By (i), for every state $\rho$,

Density matrices linearly span the full matrix algebra: a Hermitian matrix is a difference of positive matrices, and each nonzero positive matrix is a scalar multiple of a density matrix. Extending first over Hermitian and then over complex matrices proves (ii). Uniqueness follows because partial trace is onto, using $a\mapsto a\otimes \omega$.

Taking adjoints in (ii) gives

which proves (iii). Conversely, (iii) makes all output expectations depend only on the reduction to $A$; states are separated by matrix observables, giving (i).

The equality of linear maps in (ii) can be tensored with ${\mathrm{id}}_Q$ for any reference $Q$. Its input snapshot is then the joint $AQ$ reduction, not merely the separate marginals.