Let $V$ be an object in the category (Vect$_{\mathbb{C}},\otimes,\mathbb{C}$), and let $\{ e_i\}$ be a basis for $V$. I need to show that if $V$ admits a dual $V^*$ then $V$ must be finite dimensional. I have been given the hint that I must show that the coevaluation is necessarily of this form: $$1 \mapsto \sum_{i,j}c_{i,j}e_i \otimes e^j,$$ where $c_{i,j} \in \mathbb{C}$ and $\{ e^{j}\}$ is a basis for $V^*$, but I don't know how to show this or why this will be sufficient. Help with the proof or some kind of explanation would be great!
Thanks in advance.
First, the coevaluation takes this form because $e_i\otimes e^j$ is a basis for $V\otimes V^*$, so it's just the expression of the coevaluation in terms of this basis.
As for why this answers the question, observe that the sum is necessarily finite.
I.e., only finitely many $c_{ij}\ne 0$.
Then the fact that $$V\to k\otimes V\to V\otimes V^*\otimes V\to V\otimes k \to V$$ is supposed to be the identity will be a problem.
Since finitely many $c_{ij}$ are nonzero, there are only finitely many indices $i$ such that $c_{ij}$ is nonzero. Suppose $k$ is such that $c_{kj}=0$ for all $j$.
Then $$e_k \mapsto 1\otimes e_k \mapsto \sum_{i,j} c_{ij}e_i\otimes e^j\otimes e_k \mapsto \sum_{i,j} c_{ij}d_{jk}e_i,$$ where $d_{jk}= \epsilon(e^j\otimes e_k)$ and $\epsilon$ is the evaluation. Then the coefficient of $e_k$ in this last term is $\sum_j c_{kj} d_{jk} = 0\ne 1$.