Proof that the coordinate map is a linear isomorphism

Proof

We check linearity from the uniqueness of coordinates, and then exhibit a two-sided inverse.

1. Let $\varphi \colon V \to F^n$ be the map $\varphi(v) = [v]_B$. It is well defined, because each $v \in V$ equals $a_1 b_1 + \cdots + a_n b_n$ for exactly one $(a_1, \dots, a_n) \in F^n$, since $B$ is a basis.
2. $\varphi$ is additive. Let $[v]_B = (a_1, \dots, a_n)$ and $[w]_B = (c_1, \dots, c_n)$. Then $v + w = \sum_i (a_i + c_i) b_i$.
3. By uniqueness of coordinates, $[v + w]_B = (a_1 + c_1, \dots, a_n + c_n) = [v]_B + [w]_B$.
4. $\varphi$ is homogeneous. For $c \in F$, $cv = \sum_i (c a_i) b_i$, so by uniqueness $[cv]_B = c\,[v]_B$. Hence $\varphi$ is a linear map.
5. Let $\psi \colon F^n \to V$ be the map $\psi(a_1, \dots, a_n) = a_1 b_1 + \cdots + a_n b_n$ on the coordinate space.
6. For $v \in V$ with $[v]_B = (a_1, \dots, a_n)$, the definition of coordinates gives $\psi(\varphi(v)) = \sum_i a_i b_i = v$. So $\psi \circ \varphi = \mathrm{id}_V$.
7. For $(a_1, \dots, a_n) \in F^n$, let $v = \psi(a_1, \dots, a_n) = \sum_i a_i b_i$. By uniqueness of coordinates, $[v]_B = (a_1, \dots, a_n)$. So $\varphi \circ \psi = \mathrm{id}_{F^n}$.
8. By the two-sided inverse criterion, $\varphi$ is bijective, and its inverse is $\psi$.
9. So $\varphi$ is a bijective linear map, that is, a linear isomorphism. Its inverse sends $(a_1, \dots, a_n)$ to $a_1 b_1 + \cdots + a_n b_n$.