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I'm reading the notes where the following statement is assumed to be a well-known result:

Why is any birational map which maps the projective line $\mathbb{P}_{\mathbb{C}}^1$ onto itself belongs to $Aut(\mathbb{P}_{\mathbb{C}}^1)$?

I will appreciate any help!

Rainbow57
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    Briefly, a rational mapping from an algebraic curve to the projective line may be represented locally as $[p(z) : q(z)]$ for some one-variable polynomials $p$ and $q$. Any common factors ("potential points of indeterminacy") may be divided out, yielding a holomorphic extension. By contrast, a rational mapping in two or more variables can have genuine indeterminacy, e.g., of the type $f(z, w) = z/w$ at the origin. <> This is surely explained elsewhere on-site, hence the comment instead of an answer. – Andrew D. Hwang Jul 15 '21 at 16:30
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    Georges' answer here may be of interest. – Andrew D. Hwang Jul 15 '21 at 16:33
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    This is also available as the "curve-to-projective extension theorem" in section 16.5 of Vakil's Rising Sea. – KReiser Jul 15 '21 at 18:29

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