Birational Maps of Nonsingular Projective Curves 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 $\overline{\varphi}:\mathbb{P}^1\rightarrow\mathbb{P}^1$?
 A: 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.
