Does there exist a regular map $\mathbb{A}^1\to\mathbb{P}^1$ which is surjective? Suppose $\mathbb{A}^1$ and $\mathbb{P}^1$ are affine space and projective space respectively. I'm not sure if it matters, but I don't mind if we assume that we're working over algebraically closed fields.
I'm curious, is it possible to find a surjective, regular mapping $\mathbb{A}^1\to\mathbb{P}^1$? 
 A: Yes it is possible to find a surjective, regular mapping $\mathbb{A}^1\to\mathbb{P}^1$!    
By algebra
$\mathbb A^1\to \mathbb P^1:z\mapsto[z:z^2+1] $
By geometry (better !)
Take a ramified $2$-covering $f:\mathbb P^1\to \mathbb P^1$, consider a non-critical point $a\in \mathbb P^1$ (easy: there are only two critical points for $f$ !) and the required surjective morphism is the restriction $$\text{res}(f):\mathbb P^1\setminus \{a\}=\mathbb A^1\to \mathbb P^1$$ By concrete geometry (best !)
Consider the projection  $p$ of a smooth conic  $C\subset \mathbb P^2$ from a point $Q$ outside $C$ onto a line $L\subset \mathbb P^2$,  take a point $a\in C$ such that the tangent line $\Theta_aC\subset \mathbb P^2$ at $a$ does not pass through $Q$ (easy: there are only two such undesirable points!) and restrict the projection $p$ to the complement of $a$ to obtain the required surjective morphism $$\text{res}(p):C\setminus \{a\}=\mathbb A^1 \to L=\mathbb P^1$$Confused? Make a drawing and just LOOK!
