The canonical algebra of a fusion category is Frobenius

The direct sum $K$ represents $(X,Y)\mapsto {\mathrm{Hom}}_{\mathcal{C}}(X\otimes Y,\mathbf{1})$ by semisimplicity and duality. Evaluation $K\cdot \mathbf{1}\to \mathbf{1}$ therefore induces a unit and multiplication by composition. To specify multiplication without an implicit diagram, choose split matrix units $q_a^r:r\to i\otimes j$ and $p_a^r:i\otimes j\to r$ with $p_a^r{q}_b^s={\delta }_{rs}{\delta }_{ab}{\mathrm{i}\mathrm{d}}_r$ and ${\sum }_{r,a}{q}_a^r{p}_a^r={\mathrm{i}\mathrm{d}}_{i\otimes j}$. The $(i,j)$-to-$r$ component of $\mu$ is ${\sum }_a{p}_a^r⊠(q_a^r{)}^{\ast }$. This expression is independent of matrix units. In either iterated triple multiplication the inserted sums of matrix units are identities, so both products are composition of the same three evaluations. This proves associativity; the unit is inclusion of the unit summand.

The pairing $\beta =\lambda \mu$ has no component between the $i$ and $j$ summands unless $j$ is dual to $i$. For a dual pair, the displayed multiplication shows its component is the tensor product of a nonzero projection $i\otimes j\to \mathbf{1}$ with the dual of its splitting $\mathbf{1}\to i\otimes j$. These are perfect evaluation pairings between dual simple objects, by rigidity. Hence every simple block of the induced map from $K$ to its dual is invertible, proving nondegeneracy. Let $c:\mathbf{1}\to K\otimes K$ be its inverse pairing, so $(\beta \otimes \mathrm{i}\mathrm{d})(\mathrm{i}\mathrm{d}\otimes c)={\mathrm{i}\mathrm{d}}_K=(\mathrm{i}\mathrm{d}\otimes \beta )(c\otimes \mathrm{i}\mathrm{d})$. Associativity gives $\beta (\mu \otimes \mathrm{i}\mathrm{d})=\beta (\mathrm{i}\mathrm{d}\otimes \mu )$, and taking its mates gives $\mathrm{\Delta }=(\mu \otimes \mathrm{i}\mathrm{d})(\mathrm{i}\mathrm{d}\otimes c)=(\mathrm{i}\mathrm{d}\otimes \mu )(c\otimes \mathrm{i}\mathrm{d})$. Taking mates once more gives $(\mu \otimes \mathrm{i}\mathrm{d})(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })=\mathrm{\Delta }\mu =(\mathrm{i}\mathrm{d}\otimes \mu )(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})$ and coassociativity; the inverse-pairing equations and $\beta =\lambda \mu$ give $(\lambda \otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }=\mathrm{i}\mathrm{d}=(\mathrm{i}\mathrm{d}\otimes \lambda )\mathrm{\Delta }$. These mate calculations are justified by composing with the perfect pairing; each resulting equality is respectively the associative triple product or one of the two inverse-pairing identities.