Why linear map from here from $V \to V^*$ is surjective.

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Given a 2-covector $\omega$ over finite dimension vector space $V$,

We assume the linear map $i(\omega):V\to V^* $,which is defined as $i(\omega)(v) = \omega(v,\star)\in V^*$,with $\omega$ is nondegenerate,that is any $v\ne 0$ there always exist a $u\in V$ such that $\omega(v,u) \ne 0$

We need to show $i(\omega)$ is linear isomorphism.

To show the linear map $i(\omega):V \to V^*$ is injective is easy,we need to show now it's onto,that is given any linear functional $\ell\in V^*$ there exist one $v$ such that $\omega(v,\star) = \ell$.

My attempt,it's sufficient to construct some $v$ such that $\omega(v,e_i) = \ell(e_i)$ for all basis of $V$ but then the $v$ is not obvious to find.is it Riesz representation here?but the $(,)$ is not inner product?