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I'm trying to solve exercise I.6.7 in Hartshorne, stated (in part) here:

Let $P_1,\ldots,P_r, Q_1,\ldots,Q_s$ be distinct points of $\mathbb{A}^1$. If $\mathbb{A}^1-\{P_1,\ldots,P_r\}$ is isomorphic to $\mathbb{A}^1-\{Q_1,\ldots,Q_s\}$, show that $r=s$.

Here is my thinking so far. Let $U=\mathbb{A}^1-\{P_1,\ldots,P_r\}$ and $V=\mathbb{A}^1-\{Q_1,\ldots,Q_s\}$, and view $U$ and $V$ as open subsets of $\mathbb{P}^1$. Let $\varphi:U\rightarrow V$ be an isomorphism, and $i:V\rightarrow \mathbb{P}^1$ be inclusion. Then the morphism $i\circ\varphi:U\rightarrow \mathbb{P}^1$ can be uniquely extended (Hartshorne I.6.8) to a map $\overline{\varphi}:\mathbb{P}^1\rightarrow \mathbb{P}^1$. Must $\overline{\varphi}$ be an isomorphism? If so, the result follows because then $\overline{\varphi}$ must map the set $\{P_1,\ldots,P_r,\infty\}$ bijectively to the set $\{Q_1,\ldots,Q_s,\infty\}$.

I understand that any birational map between two nonsingular projective curves must 'induce' an isomorphism, as up to isomorphism, there is only one nonsingular projective curve in every birational equivalence class of curves. Where I lack understanding is exactly how a birational map induces an isomorphism. In my question, $\langle U,i\circ\varphi\rangle$ is a birational map from $\mathbb{P}^1\rightarrow\mathbb{P}^1$, but how do I know that extending it as I've done gives an isomorphism $\overline{\varphi}:\mathbb{P}^1\rightarrow\mathbb{P}^1$?

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  • $\begingroup$ Georges's answer is great for curves in general. In your case, you can also compute the group of units in the ring of regular functions of $\mathbb A^1\setminus \{P_1, \dots, P_r\}$. It is $k^* \times $ a free group of rank $r$. $\endgroup$ – user18119 Mar 23 '13 at 23:56
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Any non-constant morphism $\Phi\colon X\to Y$ between projective smooth curves has a degree $d$ and the morphism $\Phi$ will be an isomorphism if and only if that degree is $1$.

The incredibly good news is that you can calculate that degree by just looking at the fibre $\Phi^{-1}(y)$ of $\Phi$ at just one point ( any point!) of $y\in Y$: the degree is the dimension $d=dim_k \Gamma (\Phi^{-1}(y),\mathcal O)$ .
It is not terribly difficult to explain what the right-hand side means, but this is not even necessary here: we have $d=1$ as soon as for some non-empty $V\subset Y$ the restricted morphism $\Phi^{-1}(V)\to V$ is an isomorphism.
Since this is true in your case, we are done.

Edit
A more elementary proof (maybe the one Hartshorne had in mind at the level of Chapter 1, before "degree" is introduced) would be to consider the inverse isomorphism $\psi:V\to U$, to complete it to a morphism $\bar \psi:\mathbb P^1\to \mathbb P^1 $ (just as you did for $\phi$) and realize that $\bar \psi $ is an inverse to $\bar \phi $, which is thus an isomorphism.

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  • $\begingroup$ Thanks for this response. I'm not exactly sure what $\Gamma(\Phi^{-1}(y),\mathcal{O})$ represents, because $\Phi^{-1}(y)$ is closed. Is there anyway to answer my question without invoking the results you have used about the degree of a map of smooth projective curves? Or are these results fairly easy to see? $\endgroup$ – Jared Mar 22 '13 at 19:31
  • $\begingroup$ Dear Jared, I have added a small edit sketching a proof avoiding the notion of "degree". $\endgroup$ – Georges Elencwajg Mar 23 '13 at 12:50
  • $\begingroup$ Yep, that was the idea I needed. I had tried it before, but didn't think of it in the right way. I think to show that the composition of the two maps is the identity, you need to use uniqueness of extension. Thanks! $\endgroup$ – Jared Mar 23 '13 at 21:24

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