Rigid structure of $\operatorname{Vec}(G)$

Question

Apparently if $G$ is a finite group $\operatorname{Vec}(G)$ is a Rigid monoidal category (whereby I mean the category of graded vector spaces over a group $G$). I struggle to see why it is rigid. It is easy to see that that some objects are rigid by taking the dual of for example $V_g$ (Where I mean the vector space $V$ localized at $g$) to be $V^*_{g^{-1}}$ and the evaluation and coevaluation to be the same as that in Vec. However when you take direct sums of these elements I don't see how you can extend this rigidity as you won't have something localized at $1$. How could you extend this to a rigid structure on all of the category?

Answer 1

Just use $0$ on the components that don't have grading $1$ when defining the (co)evaluation. For instance, if $V=V_g\oplus V_h$ is nontrivial in two degrees $g$ and $h$, define the dual $V^*$ to be $V^*_{g^{-1}}\oplus V^*_{h^{-1}}$ where $V^*_{g^{-1}}=(V_g)^*$ and $V^*_{h^{-1}}=(V_h)^*$. The evaluation is then the map $V\otimes V^*\to k$ which is the evaluation map for $V_g$ on $V_g\otimes V^*_{g^{-1}}$, the evaluation map for $V_h$ on $V_h\otimes V^*_{h^{-1}}$, and simply $0$ on the other summands $V_g\otimes V^*_{h^{-1}}$ and $V_h\otimes V^*_{g^{-1}}$. Similarly, the coevaluation is the map $k\to V\otimes V^*$ which is given by the coevaluations for $V_g$ and $V_h$ on two of summands of $V\otimes V^*$ and is $0$ on the others.