Proposition: Affine long-range channels do not universally clone

For any finite extension made of complete instruments and wire relocations:

1. the input-to-readout-law map is affine;

2. equal complete operational input states give equal laws for a fixed readout;

3. no deterministic program universally clones unknown pure states.

Proof. (a) follows from linearity of complete channels and classical marginalization. For (b), a fixed readout is a channel on its accessed quantum and classical factors.

For (c), let $\mathcal{C}$ be a hypothetical CPTP cloner. Use root fidelity $F(\rho ,\sigma )=\Vert \sqrt{\rho }\sqrt{\sigma }{\Vert }_1$. CPTP maps do not decrease fidelity Watrous. For two pure states with $0<f=|⟨\psi ,\varphi ⟩|<1$, cloning would give output fidelity $f^2<f$, contradicting monotonicity.