Fusion multiplicity spaces are finite-dimensional inner-product spaces

The category is finite and semisimple, so $V_{ab}^c$ is a finite-dimensional complex vector space by [F1]. By [F2], every $f\circ g^{†}\in \mathrm{End}(c)$ is a unique scalar multiple of the nonzero identity; hence the scalar pairing in [F3] exists and is unique.

For scalars $\lambda ,\mu \in \mathbb{C}$ and $f,h,g\in V_{ab}^c$, linearity of composition gives $(\lambda f+\mu h)g^{†}=\lambda f{g}^{†}+\mu h{g}^{†}$, while conjugate-linearity of dagger gives $f(\lambda g+\mu h{)}^{†}=\bar{\lambda }f{g}^{†}+\bar{\mu }f{h}^{†}$. Taking the unique scalar coefficients from [F2] proves linearity in the first argument and conjugate-linearity in the second. Taking daggers of $f{g}^{†}=⟨f,g⟩{\mathrm{id}}_c$ gives $g{f}^{†}=\overline{⟨f,g⟩}{\mathrm{id}}_c$, so $⟨g,f⟩=\overline{⟨f,g⟩}$.

For each $f\in V_{ab}^c$, $f{f}^{†}=(f^{†}{)}^{†}\circ f^{†}$ is positive in $\mathrm{End}(c)$ by [F1], so its scalar coefficient $⟨f,f⟩$ is real and nonnegative. If $⟨f,f⟩=0$, then $(f^{†}{)}^{†}\circ f^{†}=0$; the nondegeneracy clause of [F1], applied to the morphism $f^{†}$, gives $f^{†}=0$ and then $f=0$. Thus $⟨f,f⟩>0$ whenever $f\mathrm{\ne }0$.

The pairing is finite-dimensional by step 1.1, sesquilinear and Hermitian by step 1.2, and positive definite by step 1.3. If $V_{ab}^c=0$, scalar uniqueness in step 1.1 still holds and positive definiteness is vacuous because there are no nonzero vectors. These are exactly the inner-product axioms, with the stated linear-first convention.