The composite is $f=e\times m$. Exchanging two composites exchanges the two $e$ constituents and the two $m$ constituents and produces one mutual winding between unlike constituents. The identical-particle phases are both $+1$, while the mutual contribution is $-1$, hence
$$
e^{i\theta_{ff}}=
e^{i\theta_{ee}}e^{i\theta_{mm}}e^{2i\theta_{em}}=-1.
$$
The <topological spin> obeys $e^{2\pi i s_f}=e^{i\theta_{ff}}$, so
$$
s_f=\frac12\pmod 1.
$$
This is precisely the two-dimensional spin-statistics relation for a fermion, so the result is consistent.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 / 1 / b / Solution

{title2=$\theta_a$}
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The topological spin $\theta_a$ is the phase produced by a $2\pi$ rotation of anyon $a$, equivalently its self-exchange phase in an Abelian anyon theory. The two-dimensional spin-statistics relation is $\theta_a=e^{2\pi i s_a}$.


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 Incoming links: Topological spin