Lemma: Oriented perfect-tensor isometries

Both $F_L,F_R$ are isometries from two qubits to four qubits. For either map, erasure of any one output qubit has constant complete complement $I_2/2$ on the entire four-dimensional input. Consequently any three of its four output qubits recover both input qubits, including arbitrary references.

Proof. For $F_R$, privacy of the original boundary leg $1$ gives

which proves ${F_R}^{†}{F}_R=I_4$. The same calculation applies to $F_L$.

The normalized Choi vector of $F_R$ is the original six-leg perfect vector with its legs rearranged: its amplitudes are $F_R/2=V/\sqrt{2}$. Its reduction to the two input-reference legs and any one output leg is $I_8/8$ by Proposition prop:perfect. Expanding that identity on input matrix units gives the constant complement $M\mapsto \mathrm{Tr}(M)I_2/2$. Theorem thm:exact, now with $d=4$, constructs a decoder on the remaining three qubits.