Left and right dual objects are isomorphic in a semisimple multitensor category

By semisimplicity, decompose $X$ into simple summands. For a simple summand $V$ and a simple object $Z$, dual transposition from [F2] gives $\mathrm{Hom}(\mathbf{1},V\otimes Z)\cong \mathrm{Hom}(V^{\vee },Z),\mathrm{Hom}(V\otimes Z,\mathbf{1})\cong \mathrm{Hom}(Z,{}^{\vee }V)\mathrm{.}$ Rigidity makes left and right dualization quasi-inverse contravariant equivalences, and exactness in [F3] therefore preserves simple objects. Hence the first space is nonzero exactly when $Z\cong V^{\vee }$, and the second exactly when $Z\cong {}^{\vee }V$.

This comparison does not require $k$ to be algebraically closed. Write the semisimple objects as $\mathbf{1}\cong {⨁}_i{S}_i^{\oplus a_i}$ and $V\otimes Z\cong {⨁}_i{S}_i^{\oplus b_i}$ with pairwise nonisomorphic simples $S_i$. Then both ${\dim }_k\mathrm{Hom}(\mathbf{1},V\otimes Z)$ and ${\dim }_k\mathrm{Hom}(V\otimes Z,\mathbf{1})$ equal ${\sum }_i{a}_i{b}_i{\dim }_k\mathrm{End}(S_i)$. Hence the two unique simples detected in step 1.1 agree: $V^{\vee }\cong {}^{\vee }V$. Taking finite direct sums over the simple summands gives the result for $X$.