Finite-dimensional vector spaces are rigid

By [L3], choose a basis $(v_1,\dots ,v_n)$ of $V$. Let $(v_1^{\ast },\dots ,v_n^{\ast })$ be the dual basis from [L2], and define $\mathrm{ev}:V^{\ast }\otimes V\to k,f\otimes v\mapsto f(v),$ $\mathrm{coev}:k\to V\otimes V^{\ast },1\mapsto {\sum }_{i=1}^n{v}_i\otimes v_i^{\ast }\mathrm{.}$

For each basis vector $v_j$, the first zig-zag sends $v_j$ to ${\sum }_{i=1}^n{v}_i^{\ast }(v_j)v_i={\sum }_{i=1}^n{\delta }_{ij}{v}_i=v_j\mathrm{.}$ By linearity it is the identity on $V$.

For each dual basis vector $v_j^{\ast }$, the second zig-zag sends $v_j^{\ast }$ to ${\sum }_{i=1}^n{v}_j^{\ast }(v_i)v_i^{\ast }={\sum }_{i=1}^n{\delta }_{ji}{v}_i^{\ast }=v_j^{\ast }\mathrm{.}$ By linearity it is the identity on $V^{\ast }$.

Thus $V^{\ast }$ is a left dual of $V$. The same formulas, read in the mirrored order, make $V^{\ast }$ a right dual of $V$ as well, so $V$ is rigid. Since $V$ was arbitrary and [L1] supplies the monoidal structure, every finite-dimensional object of ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k$ is rigid.