any $f \in {\mathrm{End}}_{\mathcal{S}}(I)$, there is a unique ${\left\langlef\right\rangle}_I \in {\mathrm{End}}_{\mathcal{S}}({\mathbb{1}})$ such that $$\label{eq-bij-invert-ob}
f={\left\langlef\right\rangle}_I  {\mathrm{id}}_I.$$

For any invertible object $I$, since $\tau_{I,I} \in  {\mathrm{End}}_{\mathcal{S}}(I \otimes I)$ with $I \otimes I$ invertible, we can consider the scalar $$\sigma_I ={\left\langle\tau_{I,I}\right\rangle}_{I \otimes I}  \in {\mathrm{End}}_{\mathcal{S}}({\mathbb{1}}).$$ By definition, $$\label{eq-sigma-def}
\tau_{I,I}=\sigma_I \, {\mathrm{id}}_{I \otimes I}.$$ The involutivity of a symmetry implies that $$\label{eq-sigma-invert-ob}
\sigma_I^2=1.$$

A left (or a right) dual of an invertible object is invertible. Assume that ${\mathcal{S}}$ is left rigid and endow ${\mathcal{S}}$ with the spherical structure canonically induced by the left duality and symmetry (see Section \[sect-duality\]). The uniqueness of a left dual (up to isomorphism) implies that an object $I$ of ${\mathcal{S}}$ is invertible if and only if the left evaluation ${\mathrm{ev}}_I {\colon}I^* \otimes I \to {\mathbb{1}}$ is an isomorphism. Let $I$ be an invertible object of ${\mathcal{S}}$. Then its dimension $\dim(I) \in {\mathrm{End}}_{\mathcal{S}}({\mathbb{1}})$ is invertible and $$\label{eq-dimen-invert-ob}
\dim(I)^2=1.$$ Indeed, the facts that the dimension of objects is invariant under isomorphisms and is $\otimes$-multiplicative imply that $$1=\dim({\mathbb{1}})=\dim(I^* \otimes I)=\dim(I^*)\dim(I).$$ Then follows from the fact that $\dim(I^*)=\dim(I)$ by the sphericity of ${\mathcal{S}}$. Note that Formulas and imply that for any $f \in {\mathrm{End}}_{\mathcal{S}}(I)$, $${\left\langlef\right\rangle}_I=\dim(I) \, {\mathrm{tr}}(f).$$ Furthermore the scalar $\sigma_I \in 