I know this question has been asked a couple of times on this site, but both of the times $\mathbb{P}^1$ was regarded as a space of homogeneous coordinates and not as $\mathbb{A}^1\cup \infty$ and I want to use this definition.
I'll try to be as detailed as possible, because I'd like to know where I'm wrong.
We have that $\mathbb{P}^1=\mathbb{A}^1\cup \mathbb{A}^1_\infty$ where $\mathbb{A}^1_\infty$ is just the affine line $\mathbb{P}^1-\{0\}$. So a morphism $f:\mathbb{A}^1-\{0\}\to \mathbb{P}^1$ can be decomposed in two morphisms: $$ f_1:f^{-1}(\mathbb{A}^1)\to \mathbb{A}^1 $$ $$ f_2:f^{-1}(\mathbb{A}^1_{\infty})\to \mathbb{A}^1_\infty $$ Using the standard identification of $\mathbb{A}^1_\infty$ with $\mathbb{A}^1$ (i.e. $x\mapsto 1/x$), we get that $f_2$ is the reciprocal of a morphism $\mathbb{A}^1-\{0\}\to \mathbb{A}^1$. We know that: $$\mathscr{O}_{\mathbb{A}^1}(\mathbb{A}^1-\{0\})=\mathbb{K}[x]_x=\left\{\frac{g}{x^n}:g\in\mathbb{K}[x]\right\}.$$ So: can be decomposed in two morphisms: $$ f_1:f^{-1}(\mathbb{A}^1)\to \mathbb{A}^1,\ \ \ x\mapsto g/x^n, $$ $$ f_2:f^{-1}(\mathbb{A}^1_\infty)\to \mathbb{A}^1_\infty,\ \ \ x\mapsto x^m/h. $$ Clearly I can assume that both $g$ and $h$ aren't divided by $x$, because otherwise I can simplify $x$, since $0$ is not in the domain. So $g(0),h(0)\neq 0$. Clearly $f_1$ and $f_2$ have to agree on the intersection of their domains i.e. $D(g)\cap D(h)=D(gh)$ . So we have that: $$gh=x^{n+m} \text{ on $D(gh)$}$$ If $D(gh)$ is empty then $g=0$ or $h=0$ and it's trivial. If $D(gh)$ is not empty then we have two polynomials that agree on a open set and this implies that they are equal, but since $x$ doesn't divide $g$ nor $h$ this would imply $n=m=0$ and $gh=1$ and we get constant maps!
Basically, by this (surely wrong) reasoning every morphism $\mathbb{A}^1-\{0\}\to \mathbb{P}^1$ is constant.
Where's my mistake?
EDIT: I figured out what was my mistake. Now I'd like to know how to effectively solve this exercise.