It is often stated that a choice of a basis for a vector space over $R$ gives an isomorphism between $V$ and $R^n$. However, I am having a hard time showing this and confused over some issues. For simplicity, let's stick to a two-dimensional space $V$. Let's choose a basis $\{e_1, e_2\}$ for the vector space. Then we have the linear map $\phi: R^2 \to V; \phi(v_1,v_2)=v_1 e_1+v_2 e_2$. To prove the isomorphism between $R^2$ and $V$, I need to construct a map $\phi^{-1}: V\to R^2: \phi^{-1} \phi=Identity$. I don't see any obvious way to construct such a map. However, if there is an inner product (denoted here by ".") on $V$, we can construct such a map in the following way. Suppose $e_1,e_2$ is an orthonormal basis. We then set $\phi^{-1}: V\to R^2; \phi^{-1}(v)=(e_1.v, e_2.v)$. It is obvious that $\phi^{-1}\phi(v_1,v_2)=(v_1,v_2)$. So my question is, is it true that we need an inner product and the existence of an orthonormal basis to prove the isomorphism between $V$ and $R^2$? Or I have misunderstood something here?
Edit: I also would like to mention that this question originated from noticing the fact that although a given vector can be written as $v=v_1e_1+v_2e_2$, there is no natural way to find the coefficients $v_1,v_2$. The inner product seems to determine the values of $v_1,v_2$.