the image of a morphism Let $X$ be an affine algebraic variety in $\mathbf{A}^n_k$ and $f\in A(X)$ be a non-constant morphism on $X$, how can we prove that $f(X)$ contains an open subset of $\mathbf{A}^1_k$?
 A: HINT 1: The statement is not true if $X$ is of dimension zero. For example, then $X$ could be a finite subset of $\mathbb A^1$, and $f$ the restriction of the identity map.
HINT 2:
A non-constant morphism is the same as a non-constant map $f:X \to \mathbb A^1$.
Note that the closed subsets of $\mathbb A^1$ are just the finite ones. Then I claim that the image cannot be closed. Reason: every closed subset of $\mathbb A^1$ is disconnected.
Can you go from here?
ADDED
Since the closed sets in $\mathbb A^1$ are just the finite sets, then as soon as a subset of $\mathbb A^1$ contains an open, it is open. Because if $U \subset Z$ is open, then $\mathbb A^1 \backslash Z \subset \mathbb A^1 \backslash U$, and the last one is a finite set. So we just have to show that $f(X)$ is open, or equivalently, that $f$ misses just a finite set of values.
COMMENT
The statement is clearly not true without restrictions. For example if $k=\mathbb R$, let $X=\{y - x^2 = 0 \} \subset \mathbb A^2$ and let $f$ be projection to the $y$ axis. Then the image is just the positive values, but by the above comment, this set does not contain any open. 
