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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.2012-03-09

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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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    Nevermind, there is some good discussion about that here: http://math.stackexchange.com/questions/29758/entire-1-1-function.2012-03-10