Morphism of varieties from $\mathbb A^1 \to \mathbb A^1 \{0\}$? 
Suppose $f$ is a morphism of varieties from $\mathbb A^1 \to \mathbb A^1 \backslash \{0\}$. What can we say about $f$ ?

I am unable to deduce much. I know that $f$ will induce a $k$ algebra map $k[x] \to k[x]$ by pulling back the regular functions. What more can be said about $f$?

Are you aware that $\mathbb A^1\backslash \{ 0 \}$ is also affine? It is isomorphic to $V(xy - 1) \subset \mathbb A^2$ via the mapping $x \mapsto (x, 1/x)$. The coordinate ring of $V(xy - 1) \subset \mathbb A^2$ is
$$ k[x,y]/(xy - 1) \cong k[x,x^{-1}].$$
Therefore, following your line of reasoning, your geometric problem reduces to the algebraic problem of finding $k$-algebra morphisms from $k[x,x^{-1}] $ to $k[x]$. I'll leave this to you.

Alternatively, extend your morphism $\mathbb A^1 \to \mathbb A^1 \backslash \{ 0 \}$ to a morphism $\mathbb A^1 \to \mathbb A^1$ by composing with the inclusion $\mathbb A^1 \backslash \{ 0 \} \hookrightarrow \mathbb A^1$. The composition $\mathbb A^1 \to \mathbb A^1$ must be given by $x \mapsto f(x)$ for some polynomial $f(x) \in k[x]$. Assuming $k$ is algebraically closed, is it possible for the image of this map to avoid $0$ if $f(x)$ is a non-constant polynomial?
