Let $V$ be a finite dimensional vector space, and $\psi : V \rightarrow V^{**}$ be a map defined by
$$ \psi(v)(f) = f(v)$$ for all $v \in V$ and $f \in V^{**}$.
We can easily show that $\psi$ is a linear map. And in fact, it is an isomorphism.
Indeed, if we choose a basis $\mathcal{B} = \{v_1, \ldots, v_n\}$ of $V$, then $\mathcal{B}^* = \{v_1^*, \ldots, v_n^*\}$ is a basis of $V^*$, and $\mathcal{B}^{**} = \{v_1^{**}, \ldots, v_n^{**}\}$ is a basis of $V^{**}$. We see that $\psi(v_i)(v_j^*) = v_j^*(v_i)=\delta_{ij}$. Hence $\psi(v_i) = v_i^{**}$. Because $\psi$ sends a basis of $V$ to a basis of $V^{**}$, we conclude that $\psi$ is an isomorphism.
First, I tried to find a linear map $\phi : V^{**} \to V$ such that $\phi \circ \psi$ and $\psi \circ \phi$ are identity maps of $V$ and $V^{**}$, respectively. However, I can not find it. Can we construct this $\phi$?