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Just based on some reading, I know that every Möbius transformation is a bijection from the Riemann sphere to itself.

I'm curious about the converse. For any holomorphic bijection on the sphere, why is it necessarily a Möbius transformation? Is there a proof or reference of why this converse is true? Thanks.

(I would appreciate an explanation at the level of someone whose just self-studying complex analysis for the first time.)

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If you added 'holomorphic' to your title, your post would be more precise. – Mariano Suárez-Alvarez Mar 9 '12 at 1:52
up vote 5 down vote accepted

Suppose $f$ is an holomorphic bijection of the sphere to itself. There is a Moebius transformation $g$ which maps $f(\infty)$ to $\infty$. Let $h=g\circ f$, which is again an holomorphic bijective map of the sphere to itself, and which maps $\infty$ to $\infty$. It follows that $h(\mathbb C)\subseteq\mathbb C$, because of injectivity, and the restriction $h|_{\mathbb C}:\mathbb C\to\mathbb C$ is a injective entire function.

Now, the big theorem of Picard, or several others, imply that $h|_\mathbb C$ is necessarily linear. It follows that $h$ itself is linear, and then $f=g^{-1}\circ h$ is a Moebius transformation.

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In the last line, I think you mean "$f=g^{-1}\circ h$". – Alex Becker Mar 9 '12 at 2:08
Indeed. Thanks! – Mariano Suárez-Alvarez Mar 9 '12 at 2:12
Thanks Mariano. Is there a more elementary way to see that $h|_\mathbb{C}$ is linear from the fact that $h|_\mathbb{C}:\mathbb{C}\to\mathbb{C}$ is injective and entire? I looked up a proof of Picard's theorem, and it seems to rely on Montel's theorem. If not, that's fine. – Dedede Mar 10 '12 at 1:01
Nevermind, there is some good discussion about that here: – Dedede Mar 10 '12 at 1:18

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