Canonical F-move paths give the same channel operator

For a canonical associator map $A:T\to T^{\mathrm{\prime }}$, [F1] defines the split Hom map $A_{\ast }:h\mapsto A\circ h$. If $B:T^{\mathrm{\prime }}\to T^{\mathrm{\prime }\mathrm{\prime }}$, then $B_{\ast }(A_{\ast }(h))=(B\circ A)\circ h$. Thus each finite F-move sequence is the matrix of postcomposition by its canonical composite, including tensor products with identities and inverse associators. This argument applies in any monoidal category with the fixed coordinate spaces and bases posited in the Given data.

The two sequences in the Statement have the same source and target parenthesized words, so their canonical morphisms are equal by [F2]. Coordinates of equal morphisms in the fixed fusion-tree bases are equal; hence the two matrix composites agree.

Conjugating the same channel projection by equal coordinate maps gives equal operators, since matrix multiplication preserves equality and the maps are invertible. This is exactly the definition in [F3].