I have a question about a certain property of regular maps into $\mathbb{A}^1(k)$. This is my notation for the affine space over $k$, algebraically closed.
Suppose $f:X\to \mathbb{A}^n(k)$ is a regular map, where $X$ is some algebraic set/subvariety in some affine space. I read that if $n=1$, then $f$ is sometimes called a regular function, and in this case, the image $f(X)$ is necessarily open or closed in $\mathbb{A}^1(k)$. [Edit: From the current answer, why is $f(X)$ necessarily open?]
Why does this property hold in particular for regular functions? Thanks.
If the morphism $f$ is non-constant on at least one irreducible component of $X$, then its image $f(X)\subset \mathbb A^1(k)$ contains an open subset of $\mathbb A^1(k)$ (Chevalley's theorem) and is thus itself open, because the non-empty open subsets of $\mathbb A^1(k)$ are the complements of finite subsets.
The result is false for maps into higher dimensional affine spaces, as witnessed by the morphism $$\mathbb A^2(k) \to \mathbb A^2(k): (x,y)\mapsto (xy, y)$$ whose image is neither open nor closed.