Modular Hom state spaces have exact sewing inverses and the dimension formula

Choose the right-duality maps on $i^{\ast }$ induced by the pivotal structure. Concretely, from the left maps $b_i,e_i$ and pivotal $j_i$, take $b_i^R=(1_{i^{\ast }}\otimes j_i^{-1})b_{i^{\ast }}$ and $r_i=e_{i^{\ast }}(j_i\otimes 1_{i^{\ast }})$. The left snakes for $i^{\ast }$, with naturality and tensor compatibility of $j$, give $(r_i\otimes 1_i)(1_i\otimes b_i^R)=1_i$ and $(1_{i^{\ast }}\otimes r_i)(b_i^R\otimes 1_{i^{\ast }})=1_{i^{\ast }}$. These also follow directly from the right-dual data in [F2]. Fix all parentheses through [F4]. Finite semisimplicity gives $p_{i\alpha }{g}_{j\beta }={\delta }_{ij}{\delta }_{\alpha \beta }{1}_{i^{\ast }}$ and $\sum g_{i\alpha }{p}_{i\alpha }=1_X$. The maps $G,F$ are therefore typed finite sums between the specified Hom spaces.

Substitute the displayed $F$ into $G$. Interchange separates the middle composite into the right snake $(1_{i^{\ast }}\otimes r_i)(b_i^R\otimes 1_{i^{\ast }})=1_{i^{\ast }}$. Thus $GF(z)={\sum }_{i,\alpha }(g_{i\alpha }{p}_{i\alpha }\otimes 1_Y)z=z$. No dimension scalar appears because the projections are composition-dual, not trace-dual.

Conversely, for $f:\mathbf{1}\to X\otimes i$ put $g_f=(1_X\otimes r_i)(f\otimes 1_{i^{\ast }}):i^{\ast }\to X$. The other right snake gives $f=(g_f\otimes 1_i)b_i^R$. In $F(G_i(f\otimes k))$, the $j\beta$ term contains $p_{j\beta }{g}_f:i^{\ast }\to j^{\ast }$, which is zero for $j\mathrm{\ne }i$ and a scalar ${\gamma }_{\beta }$ for $j=i$. The same snake gives its second factor as ${\gamma }_{\beta }k$. The sum of the first factors multiplied by those scalars is $({\sum }_{\beta }{g}_{i\beta }{p}_{i\beta }{g}_f\otimes 1_i)b_i^R=f$. Hence $F{G}_i(f\otimes k)=f\otimes k$. Decomposable tensors span each finite tensor product, so $FG=1$ on the entire direct sum, including zero Hom blocks. A basis change $g^{\mathrm{\prime }}=gA$, $p^{\mathrm{\prime }}=A^{-1}p$ cancels in the finite sum for $F$, proving independence of all auxiliary summand bases.

For a nonseparating cut, [F1] and direct-sum additivity identify the target with the stated Hom space into $X\otimes H$. Its inclusions and projections satisfy ${\sum }_i{u}_i{v}_i=1_H$ and $v_i{u}_j={\delta }_{ij}{1}_{i\otimes i^{\ast }}$. Thus $GF=1$ and the $(i,j)$ block of $FG$ is ${\delta }_{ij}1$, exactly as in the Statement. All multiplicity coordinates remain inside their Hom spaces; none is removed by these simple-label sums.

Naturality on every uncut factor follows by composing the written evaluation, inclusion and projection words and using bifunctoriality; the inverse is natural because it is the inverse of that same map. For two disjoint cut pairs, expand their two finite resolutions before any contraction. Evaluations or inclusions on distinct leg pairs commute by interchange. For separating cuts, insert the actual identities $\sum gp=1$ at their two seams; both orders give the same product of coupon inclusions and projections, regrouped as finite sums. If one summand is produced by the first cut, applying its resolution of identity gives the same all-cuts coupon word. Associators comparing the two fixed tensor words agree by [F4]. Adjacent swaps therefore identify every ordering of finitely many disjoint cuts. Composing the reverse cutting maps cancels each adjacent $F\mathrm{/}G$ pair by steps 2.1–3.1. This proves iterated algebraic sewing and its full inverse, with empty sums and empty cut lists included.

The three-marked sphere has dimension $N_{i_1{i}_2}^{i_3^{\ast }}$; by [F5] this equals ${\sum }_x{S}_{i_{1}x}{S}_{i_{2}x}{S}_{i_{3}x}\mathrm{/}{S}_{0x}$. The zero-, one- and two-marked spheres follow from the corresponding entries of $S^2=C$: their dimensions are $1$, ${\delta }_{i0}$ and ${\delta }_{i,j^{\ast }}$. Sewing pants and handles using the proved inverse maps gives all finite products with $H^{\otimes g}$. At a separating seam the simple-label sum is ${\sum }_a{S}_{ax}{S}_{a^{\ast }y}={\delta }_{xy}$, by symmetry, dual-label symmetry and $S^2=C$. At a nonseparating seam its same-index value is $1$. Combining the powers of $S_{0x}$ gives exactly $2-2g-n$. Induction over the finite seam list proves the displayed dimension formula for every $g,n\ge 0$. In particular the unmarked torus dimension is $\mathrm{\mid }I\mathrm{\mid }$. This proves an algebraic identity from actual Hom decompositions and Verlinde; it makes no unproved assertion about surgery independence.