Lemma: Regional determination is adjoint inclusion

Let $\Psi$ be a channel on the finite site interface, and let $N,Y\subseteq X$. Its output reduction on $Y$ is determined by the input reduction on $N$, for all input states, if and only if

The implication from inclusion remains valid with an untouched reference, provided equality is imposed on the joint $N$–reference input reductions.

Proof. The inclusion makes every output expectation an input expectation of an operator in $\mathcal{A}(N)$, so equal reductions give equal outputs.

Conversely, suppose a Hermitian output observable has pullback $B\notin \mathcal{A}(N)$. Let $K\ne 0$ be the Hermitian Hilbert–Schmidt orthogonal component of $B$ outside $\mathcal{A}(N)$. Since the identity belongs to $\mathcal{A}(N)$, $\mathrm{Tr}K=0$. For sufficiently small $\epsilon >0$,

are states with equal reductions on $N$, but their tested output expectations differ by $2\epsilon \mathrm{Tr}(KB)=2\epsilon \mathrm{Tr}(K^2)>0$. This contradicts determination. Hermitian observables span the algebra.

Tensoring the inclusion with the reference matrix algebra proves the final assertion.