to $$P_{(i,i+1)}={\mathrm{id}}_{A^{\otimes (i-1)}} \otimes \tau_{A,A} \otimes {\mathrm{id}}_{A^{\otimes (n-i-1)}}.$$

Duality {#sect-duality}
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Assume that ${\mathcal{S}}$ is *left rigid*, i.e. to each object $X \in {\mathrm{Ob}}({\mathcal{S}})$ is associated an object $X^* \in {\mathrm{Ob}}({\mathcal{S}})$, called the *left dual* of $X$, together with morphisms $${\mathrm{ev}}_X{\colon}X^* \otimes X \to {\mathbb{1}}\quad \text{and} \quad {\mathrm{coev}}_X{\colon}{\mathbb{1}}\to X \otimes X^*$$ respectively called *left evaluation* and *left coevaluation*, such that $$({\mathrm{ev}}_X \otimes {\mathrm{id}}_{X^*})({\mathrm{id}}_{X^*} \otimes {\mathrm{coev}}_X)={\mathrm{id}}_{X^*} \quad \text{and} \quad ({\mathrm{id}}_X \otimes {\mathrm{ev}}_X)({\mathrm{coev}}_X \otimes {\mathrm{id}}_X)={\mathrm{id}}_X.$$ The symmetry of ${\mathcal{S}}$ endows ${\mathcal{S}}$ with a canonical structure of a pivotal category with *right evaluation* and *right coevaluation* associated to $X \in {\mathrm{Ob}}({\mathcal{S}})$ defined respectively by $${\widetilde{\mathrm{ev}}}_X={\mathrm{ev}}_X \tau_{X,X^*}{\colon}X \otimes X^* \to {\mathbb{1}}\quad \text{and} \quad {\widetilde{\mathrm{coev}}}_X=\tau_{X,X^*}{\mathrm{coev}}_X{\colon}{\mathbb{1}}\to X^* \otimes X.$$ Then ${\mathcal{S}}$ is spherical (meaning that the left and right traces induced by the (co)evaluation morphisms coincide) and even ribbon (with trivial twist). The *trace* of an endomorphism $f \in {\mathrm{End}}_{\mathcal{S}}(X)$ is $${\mathrm{tr}}(f)={\mathrm{ev}}_X({\mathrm{id}}_{X^*} \otimes f){\widetilde{\mathrm{coev}}}_X={\widetilde{\mathrm{ev}}}_X(f \otimes {\mathrm{id}}_{X^*}){\mathrm{coev}}_X \in {\mathrm{End}}_{\mathcal{S}}({\mathbb{1}}).$$ The *dimension* of an object $X$ is $$\dim(X)={\mathrm{tr}}({\mathrm{id}}_X)={\mathrm{ev}}_X{\widetilde{\mathrm{coev}}}_X={\widetilde{\mathrm{ev}}}_X{\mathrm{coev}}_X ={\mathrm{ev}}_X\tau_{X,X^*}{\mathrm{coev}}_X \in {\mathrm{End}}_{\mathcal{S}}({\mathbb{1}}).$$ The trace is symmetric: for any morphisms $p{\colon}X \to Y$ and $q{\colon}Y
\to X$ in ${\mathcal{S}}$, $${\mathrm{tr}}(pq)={\mathrm{tr}}(qp).$$ The trace and dimension are $\otimes$-multiplicative: $${\mathrm{tr}}(f\otimes g) ={\mathrm{tr}}(f) \, {\mathrm{tr}}(g)   \quad {\text{and}}\quad    \dim(X\otimes Y) = \dim(X) \dim(Y)$$ for any 