prove that $P_3\cong \mathbb{R}^4$ prove that $P_3\cong \mathbb{R}^4$
My attempt:
here $P_3=\{f\in \mathbb{R}[x]:\deg(f)\leq3\}$ is vector space over $\mathbb{R}$
So we define $f(x)=a_3x^3+a_2x^2+a_1x+a_0\to (a_3,a_2,a_1,a_0)$  
since clearly $f$ is one-one
how to prove renaming options
 A: Let $\phi : P_3 \to \Bbb R^4 $ be the map as defined by you i.e. 
$\phi(a_3 x^3 + a_2 x^2 + a_1 x + a_0) = (a_3,a_2,a_1,a_0)$
then it is clear that $\phi$ is well-defined.
$\phi((a_3 x^3 + a_2 x^2 + a_1 x + a_0) + (b_3 x^3 + b_2 x^2 + b_1 x + b_0)) = (a_3 + b_3,a_2 + b_2,a_1 + b_1,a_0 + b_0)$ 
= $ (a_3,a_2,a_1,a_0) + (b_3,b_2,b_1,b_0)$
= $\phi((a_3 x^3 + a_2 x^2 + a_1 x + a_0)$ + $\phi(b_3 x^3 + b_2 x^2 + b_1 x + b_0))$
And $\phi(\lambda(a_3 x^3 + a_2 x^2 + a_1 x + a_0)) = \lambda(a_3,a_2,a_1,a_0)$
= $\lambda \phi(a_3 x^3 + a_2 x^2 + a_1 x + a_0)$
So $\phi$ is a linear transformation.
Ker($\phi$) = {$a_3 x^3 + a_2 x^2 + a_1 x + a_0$ | $ (a_3,a_2,a_1,a_0)$ = (0,0,0,0)}
= {$a_3 x^3 + a_2 x^2 + a_1 x + a_0$ | $a_i = 0 \forall i$}
= {0}
So $\phi$ is injective.
And given any $ (a_3,a_2,a_1,a_0) \epsilon R^4$ it is clear that there exists $(a_3 x^3 + a_2 x^2 + a_1 x + a_0) \epsilon P_3$.
So $\phi$ is surjective.
Hence $\phi$ is a linear transformation between two vector spaces which is also bijective, hence $\phi$ is a vector space isomorphism.
