 Solution

Use an anyon fusion tree in which the two crossing charges combine first. With external charges in the order $d,j,h,e$, define the normalized treesThe desired value is the matrix element of anyon monodromy between $|T_{bc}⟩$ and $|T_{fg}⟩$. Their closure is understood in the physics normalization of anyon diagrams.

First apply the anyon F-matrix to the three charges $d,j,c$, and then an inverse anyon F-matrix to $j,h,e$. With the source convention $|((uv{)}_{r}w{)}_t⟩={\sum }_s[F_t^{uvw}{]}_{rs}|(u(vw{)}_s{)}_t⟩$, this givesOnly allowed anyon fusion channels enter the sums: in the second tree these require $N_{jh}^x{N}_{xe}^y{N}_{dy}^a=1$. The complex conjugate in $U$ comes from the inverse change of parentheses, not from reversing a crossing.

Reading the actual overcrossings from bottom to top gives two positive exchanges in the source's anyon R-matrix convention. Their product isThe order of $h,j$ is restored after the second exchange. In the $X$ basis this anyon monodromy is diagonal, soEquivalently, entirely in terms of anyon F-matrices and anyon R-matrices,This expression is also meaningful for $b\ne f$ or $c\ne g$: it then carries the unmatched anyon vertex gauge transformation factors identified in a.

To check the transparent limit, put $m_x=1$ in every allowed channel. The unitary matrix identity for $F_y^{jhe}$ first givesOnce $c=g$, the unitary matrix identity for $F_a^{djc}$ givesas in c. The same cancellation proves that the recoupling expression is an identity operator whenever the full anyon monodromy is trivial, including when the individual exchanges have fermionic signs.