= 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 trees
$$
|T_{bc}\rangle=|((d j)_b(h e)_c)_a\rangle,\qquad
|X_{xy}\rangle=|(d((j h)_x e)_y)_a\rangle.
$$
The desired value is the matrix element of <anyon monodromy> between $|T_{bc}\rangle$ and $|T_{fg}\rangle$. 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 $|((uv)_r w)_t\rangle=\sum_s[F^{uvw}_t]_{rs}|(u(vw)_s)_t\rangle$, this gives
$$
|T_{bc}\rangle=\sum_{x,y}U_{bc;xy}|X_{xy}\rangle,\qquad
U_{bc;xy}=[F^{djc}_a]_{by}\,[F^{jhe}_y]^*_{xc}.
$$
Only allowed <anyon fusion> channels enter the sums: in the second tree these require $N^x_{jh}N^y_{xe}N^a_{dy}=1$. The complex conjugate in $U$ comes from the inverse change of parentheses, not from reversing a crossing.
\Image[/past-exam-of-the-anyons-and-topological-quantum-field-theories-course-of-the-university-of-oxford/2025-fusion-recoupling.png]
{title=Recoupling exposes the inner pair on which the full braid acts diagonally}
Reading the actual overcrossings from bottom to top gives two positive exchanges in the source's <anyon R-matrix> convention. Their product is
$$
m_x=R^{jh}_xR^{hj}_x.
$$
The order of $h,j$ is restored after the second exchange. In the $X$ basis this <anyon monodromy> is diagonal, so
$$
\boxed{\mathcal A_{bc;fg}=\sum_{x,y}U_{bc;xy}\,R^{jh}_xR^{hj}_x\,U^*_{fg;xy}.}
$$
Equivalently, entirely in terms of <anyon F-matrices> and <anyon R-matrices>,
$$
\boxed{\mathcal A_{bc;fg}=\sum_{x,y}[F^{djc}_a]_{by}[F^{jhe}_y]^*_{xc}\,R^{jh}_xR^{hj}_x\,[F^{djg}_a]^*_{fy}[F^{jhe}_y]_{xg}.}
$$
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 <1/a>.
To check the transparent limit, put $m_x=1$ in every allowed channel. The <unitary matrix> identity for $F^{jhe}_y$ first gives
$$
\sum_x[F^{jhe}_y]^*_{xc}[F^{jhe}_y]_{xg}=\delta_{cg}.
$$
Once $c=g$, the <unitary matrix> identity for $F^{djc}_a$ gives
$$
\mathcal A_{bc;fg}=\delta_{cg}\sum_y[F^{djc}_a]_{by}[F^{djc}_a]^*_{fy}=\boxed{\delta_{cg}\delta_{bf}},
$$
as in <1/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.
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