Consider a complex polynomial map $f: \mathbb{C}^n \rightarrow \mathbb{C}^n$. For $n = 1$, the fundamental theorem of algebra says that, for any $y \in \mathbb{C}$ there exists $x \in \mathbb{C}$ such that $y = f(x)$. Thus for $n = 1$, $f(\mathbb{C}) = \mathbb{C}$ if and only if $f$ is a non-constant polynomial. Can we generalize this statement to $n > 1$?

For arbitrary $n$, what is a necessary and sufficient condition to say that closure$\left(f(\mathbb{C}^n)\right)$ $ = \mathbb{C}^n$?

Suppose the polynomials $f_1,f_2,\cdots,f_n$ are algebraically dependent, i.e., there exists an annihilating polynomial $F$ such that $F(f_1,f_2,\cdots,f_n) = 0$, then the image $f(\mathbb{C}^n)$ is a subset of the affine variety $V(F)$ of dimension $n-1$. Hence a necessary condition is that the polynomials $f_1,f_2,\cdots,f_n$ must be algebraically independent. Can we show that it a sufficient condition as well?