The standard root-of-unity sl2 tilting ribbon quotient is modular

By [F2], the full actual tilting category has its semisimple quotient with exactly the simple labels $0,\dots ,\mathrm{\ell }-2$ and the complete truncated fusion multiplicities in [F1]. The all-endomorphism trace calculation kills every excluded indecomposable, not just its identity trace. Thus the quotient here is the actual category in the Statement, rather than a fusion ring substituted for it.

By [F3], its full braiding and ribbon structure descend from the integral finite R-series, and the pivotal conversion gives $d_a=[a+1{]}_q=\sin ((a+1)\pi \mathrm{/}\mathrm{\ell })\mathrm{/}\sin (\pi \mathrm{/}\mathrm{\ell })$ and ${\theta }_a=s^{a(a+2)}=q^{a(a+2)\mathrm{/}2}$ with $s=e^{\pi i\mathrm{/}(2\mathrm{\ell })}$. The self-dual cup/cap loop is $-\delta$ and the positive pivotal trace uses the proved parity conversion; no equivalence with unmodified positive-loop pivotal diagrams is assumed. The actual category therefore satisfies every hypothesis of [F1], for all $\mathrm{\ell }\ge 3$, including even $\mathrm{\ell }$.

Apply [F1] to this actual spherical ribbon fusion category. It gives $s_{ab}=\sin ((a+1)(b+1)\pi \mathrm{/}\mathrm{\ell })\mathrm{/}\sin (\pi \mathrm{/}\mathrm{\ell })$ and $D^2=\mathrm{\ell }\mathrm{/}(2{\sin }^2(\pi \mathrm{/}\mathrm{\ell }))$. Choose its positive square root $D=\sqrt{\mathrm{\ell }\mathrm{/}2}\mathrm{/}\sin (\pi \mathrm{/}\mathrm{\ell })$; division gives exactly the displayed normalized $S$. Finite sine orthogonality gives $S^2=I$, so $S$ and $s$ are invertible. This establishes modularity of the actual quotient and all its stated data, without a source-only category premise or a TQFT assumption.