An ordered basis for a finite-dimensional $F$-vector space $V$ establishes a bijection between $V$ and $F^n$. 
To summarize, each ordered basis of $V$ determines a one-one correspondence $\alpha \to (x_1,…,x_n)$ between the set of all vectors in $V$ and the set of all $n$-tuples in $F^n$.

Question: How to prove bijective map between $V$ and $F^n$ explicitly?
My attempt: $(V,F,+,\cdot)$ is finite dimensional vector space. Let $B=\{\alpha_1,…,\alpha_n\}$ be an ordered basis of $V$. $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$$\iff$$\forall \alpha \in V$, $\exists !x_\alpha =(x_{\alpha ,1},…,x_{\alpha ,n})\in F^n$ such that $\sum_{i\in J_n} x_{\alpha ,i}\cdot \alpha_i=\alpha$. So $f:V\to F^{n}$ defined by $f(\alpha)=x_\alpha = (x_{\alpha ,1},…,x_{\alpha ,n})$ is a well-defined map, i.e. $\forall \alpha \in V$, $\exists !x_\alpha \in F^n$ such that $f(\alpha)=x_\alpha$. Claim: $f$ is invertible, i.e. $\exists g:F^n \to V$ such that $g\circ f=\text{id}_V$ and $f\circ g=\text{id}_{F^n}$. Proof: We define $g:F^n \to V$ such that $g(y)=\sum_{i\in J_n}y_i \cdot \alpha_i$. Let $\alpha \in V$. Then $g\circ f(\alpha)=g(f(\alpha))=g(x_\alpha)$$= \sum_{i\in J_n} x_{\alpha ,i}\cdot \alpha_i=\alpha$. Thus $g\circ f=\text{id}_V$. Let $y=(y_1,…,y_n)\in F^{n}$. Then $f\circ g(y) =f(g(y))=f(\sum_{i\in J_n}y_i \cdot \alpha_i)$. Let $\sum_{i\in J_n}y_i \cdot \alpha_i =\beta \in V$. Since $\exists !x_\beta =(x_{\beta ,1},…,x_{\beta ,n})$ such that $\sum_{i\in J_n} x_{\beta ,i} \cdot \alpha_i =\beta$, we have $y=(y_1,…,y_n)=(x_{\beta ,1},…,x_{\beta ,n})=x_\beta$. So $f\circ g(y)=f(\sum_{i\in J_n}y_i \cdot \alpha_i)=f(\beta)=x_\beta=y$. Thus $f\circ g=\text{id}_{F^n}$. Hence $f$ is invertible, or, equivalently, bijective. Is my proof correct?
 A: Your proof is overall correct. However, you can enhance it by

*

*From the form standpoint, separate topics in lines with headers. As currently written, it is hard to read.

*From the substance, you're missing the point that $f$ is in fact a linear map as $\mathbb F^n$ is a linear space. From this, you can study bijectivity by looking at the kernel as we're dealing with finite-dimensional linear spaces.

A: Alternative : You can put your whole argument in the form :
$\dim(V) =n$ and let $\mathcal{B}=\{v_1,v_2,\ldots ,v_n\}$ basis for $V$.
$v\in V$ implies $v=\sum_{i=1}^{n} x_{v,i}v_i$ where $x_{v,i}\in F$
Now define a map $f:V\to F^n$ by
$f(v) =(x_{v,1},x_{v,2}, \ldots ,x_{v,n})$
• Then $f$ is linear as for all $ v, u\in V$ and $\lambda\in F$ $\begin{align}f(v+\lambda u) &=(x_{v,1}+\lambda x_{u,1},x_{v,2}+\lambda x_{u,2}, \ldots +x_{v,n}+\lambda x_{u,n)}\\&=f(v)+\lambda f(u) \end{align}$
• $f$ is injective.
$\ker(f) =\{v\in V : f(v) =(0,0,\ldots, 0)\}$
Let $v\in \ker(f)$. Then $f(v) = (x_{v,1},x_{v,2}, \ldots ,x_{v,n}) =(0,0,\ldots ,0)$
Which implies $x_{v,i}=0 , \forall 1\le i\le n$. Thus $v=\sum_{i=1}^n x_{v,i}v_i=0$.
Hence $\ker(f) =\{\textbf{0}\}$
•$f$ is onto.
Since $f$ is a linear map between two same dimensional vector spaces, $f$ is injective iff $f$ is onto iff $f$ is bijective (consequence of Rank-nullity theorem)
Hence $f$ is an invertible linear map i.e an isomorphism.

Note: If you are not familiar with $\ker$ of a linear map,proof injectivity and surjectivity as the same way we do for any map between two sets.
•Injective
$f(v) =f(u) $ implies $(x_{v,1},x_{v,2}, \ldots ,x_{v,n})=(x_{u,1},x_{u,2}, \ldots ,x_{u,n})$
Then $x_{v,i}=x_{u,i}$ for all $1\le i\le n$.Hence $v=u$
•Surjective:
Given any $(x_1, x_2, \ldots, x_n) $ , consider $v=\sum_{i=1}^{n} x_i v_i$ . Then $f(v) =(x_1, x_2, \ldots, x_n) $ .
A: We can use this result to prove $V$ is isomorphic to $F^n$. Which is precisely theorem 10 section 3.3. Proof: By theorem 7 section 3.2, it is suffice to show, either $f$ or $g$ is a linear map. I will show both $f$ and $g$ are linear map. Let $c\in F$ and $\alpha, \beta\in V$. Then $\exists !x_\alpha =(x_{\alpha ,1},…,x_{\alpha ,n})\in F^n$ such that $\sum_{i\in J_n} x_{\alpha ,i}\cdot \alpha_i=\alpha$ and $\exists !x_\beta =(x_{\beta ,1},…,x_{\beta ,n})\in F^n$ such that $\sum_{i\in J_n} x_{\beta ,i}\cdot \alpha_i=\beta$. So $c\cdot \alpha +\beta$ $=c\cdot (\sum_{i\in J_n} x_{\alpha ,i} \cdot \alpha_i)$ $+ \sum_{i\in J_n} x_{\beta ,i}\cdot \alpha_i$. By distributive axiom of vector space, $c\cdot (\sum_{i\in J_n} x_{\alpha ,i} \cdot \alpha_i)$ $+ \sum_{i\in J_n} x_{\beta ,i}\cdot \alpha_i$ $= \sum_{i\in J_n} (c\cdot x_{\alpha ,i}+x_{\beta ,i})\cdot \alpha_i$. By uniqueness, $c \cdot x_{\alpha ,i}+x_{\beta ,i}=x_{c\cdot \alpha+\beta, i}$, $\forall i\in J_n$. Thus $f(c\cdot \alpha +\beta)=x_{c\cdot \alpha +\beta}$ $=(c\cdot x_{\alpha ,1}+x_{\beta ,1},…, c\cdot x_{\alpha ,n}+x_{\beta ,n})$ $=c\cdot f(\alpha)+f(\beta)$. Hence $f$ is a linear map. Let $c\in F$ and $x,y\in F^n$. Then $x=(x_1,…,x_n)$ and $y=(y_1,…,y_n)$. So $c\cdot x+y$ $=(c\cdot x_1+y_1,…,c\cdot x_n+y_n)$. So $g(c\cdot x+y)$ $= \sum_{i\in J_n} (c\cdot x_{i} +y_{i})\cdot \alpha_i$. By distributive axiom of vector space, $\sum_{i\in J_n} (c\cdot x_{i} +y_{i})\cdot \alpha_i$ $=c\cdot (\sum_{i\in J_n} x_{i}\cdot \alpha_i)$ $+ \sum_{i\in J_n} y_{i}\cdot \alpha_i$ $= c\cdot g(x)+g(y)$. Thus $g(c\cdot x+y)= c\cdot g(x)+g(y)$. Hence $g$ is a linear map.
