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I'm doing some exercise in which I need to prove that every morphism of prevarieties (As in Gathmann's, in fact it's an exercise from Gathmann's notes) $\mathbb{A}^1-0\to \mathbb{P}^1$ can be extended to a morphism $\mathbb{A}^1\to \mathbb{P}^1$.

My attempt started by taking the open cover $$U_0=\{[x:y]\mid x\neq 0\},U_1=\{[x:y]\mid y\neq 0\}\subseteq \mathbb{P}^1.$$ Then, from what I have done, it seems that a morphism $f:\mathbb{A}^1-0\to \mathbb{P}^1$ is just a map such that $f|_{f^{-1}(U_i)}:f^{-1}(U_i)\to U_i$ is a morphism of affine varieties, thus, after replacing $U_0,U_1\cong \mathbb{A}^1$ by a very simple map and $\mathbb{A}^1-0$ by $V(xy-1)\subseteq \mathbb{A}^2$ the induced maps must be polynomials.

I think up to the part in italic, I'm in the right track (but I think it's just the beginning). I'm not sure the part in italic can really help.

Now what I think follows is, consider the morphism $f:\mathbb{A}^1-0\to \mathbb{P}^1$. Then we consider the two restrictions $f_0:V_0\to U_0$, $f_1:V_1\to U_1$. I want to use this information to find some place to send $0$, but I'm not sure how I should do that.

I found a question about a map $\mathbb{A}^2-0\to \mathbb{P}^1$ which cannot be extended, and I see what fails there, but I'm not sure if I can use that as an idea to find the value of $f(0)$. It'd be more like that if I find the appropiate value for $f(0)$ then that might help me prove continuity. In a comment a suggestion for my exercise is given but I'm not sure it fits the way I'm trying to do it.

So, is there any idea about how can I continue with my attempt? If not, is there any idea about how can the attempt in that comment continue? Or an idea about a better approach for it?

For (a), I think the idea should be (I'm being very loose, but this works in general) to factor. In the sense that if $t$ is a coordinate on $\mathbb{A}^1$ then near $0$ your map looks something like $t↦[t^2+t,t^3−t^2]$. As it stands this formula does not determine a map at $0$. But away from $0$ it's the same as $[t+1,t^2−t]$, and that formula does say something at $0$.

Nell
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    Hi Nell, how much do you know about homogenous coordinates? A hint I can provide is that not all maps from A^1 minus a point to A^1 extend (for instance 1/x) so this is not something that can easily be proved by picking a single chart on A^1. It is easy to see, however, if we work with homogenous coordinates. – Sempliner Apr 29 '20 at 07:12
  • @Sempliner There shouldn't be a problem about using homogeneous coordinates. Yeah, one of my ideas was thinking to extend one of the $f_i$ but I figured they cannot necessarily be extended. I still think the information of the $f_i$ is key to find an image for $0$ in $\mathbb{P}^2$. – Nell Apr 29 '20 at 17:48
  • Take $f: A^1 - {0} \to \mathbb{P}^1$ then $f$ is given by $f(t) = [p(t), q(t)]$ with $p,q$ polynomials which have no common zeroes, except perhaps at $0$. So we can rewrite our morphism as $[t^a x(t), t^b y(t)]$ with $x,y$ coprime polynomials which do not vanish at $0$. Do you see how to extend the map now? – Sempliner Apr 29 '20 at 22:29
  • I think so. Since for $t\neq 0$, $[t^a x(t), t^b y(t)]=[x(t),y(t)]$ the map can just be extended to $\tilde{f}$ by letting $\tilde{f}(t)=[x(t),y(t)]$. It takes the same value as $f$ outside $0$ because of what I just wrote, its components are still polynomials, and it has a value at $0$. – Nell Apr 30 '20 at 01:51
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    Not quite, I was not assuming $a = b$! – Sempliner Apr 30 '20 at 01:57
  • Oh, right. So, if we assume $a\leq b$ we define it as $[x(t), t^{b-a} y(t)]$? – Nell Apr 30 '20 at 18:04
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    Yes. If you think about it this argument works equally well for any projective regular one dimensional scheme mapping to a projective variety. – Sempliner Apr 30 '20 at 21:16
  • @Sempliner: how do we see that every morphism $f\colon\mathbb{A}^1\setminus{0}\to\mathbb{P}^1$ is of the form $f(t)=[p(t),q(t)]$ (for $p,q\in k[t]$ without common zeros)? – Oskar Henriksson May 14 '20 at 17:41
  • @OskarHenriksson You can do it formally by seeing both $\mathbb{A}^1$ and $\mathbb{P}^1$ as ringed spaces and see how morphisms must behave, but more informally, for a map $[p(t):q(t)]$ to be well defined with codomain $\mathbb{P}^1$ the polynomials $p(t),q(t)$ cannot have common zeroes, because a point $[a:b]\in \mathbb{P}^1$ cannot have both of its coordinates being $0$. – Nell Jun 01 '20 at 21:25
  • @Nell Thanks! Would you mind outlining the more formal argument you have in mind (in terms of the ringed space structures on $\mathbb{A}^1$ and $\mathbb{P}^1$)? – Oskar Henriksson Jun 02 '20 at 08:29

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