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Alexandre Eremenko
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An elementary proof is the following: first you prove that every holomorphic map $f:S^2\to S^2$ is a rational function. Indeed, such map must be a meromorphic function, it has finitely many zeros and poles and is either regular or has a pole at $\infty$. So the degree of $f$ is defined: the equation $f(z)=a$ has $d$ solutions, counting multiplicity, for every $a\in S^2$. Therefore, if the map is bijective, we must have $d=1$ that is $f\in PSL(2,C)$.

This proof is based onessentially elementary algebra, no analysisonly one fact from Analysis is used: the "Fundamental theorem of algebra".

An elementary proof is the following: first you prove that every holomorphic map $f:S^2\to S^2$ is a rational function. Indeed, such map must be a meromorphic function, it has finitely many zeros and poles and is either regular or has a pole at $\infty$. So the degree of $f$ is defined: the equation $f(z)=a$ has $d$ solutions, counting multiplicity, for every $a\in S^2$. Therefore, if the map is bijective, we must have $d=1$ that is $f\in PSL(2,C)$.

This proof is based on elementary algebra, no analysis is used.

An elementary proof is the following: first you prove that every holomorphic map $f:S^2\to S^2$ is a rational function. Indeed, such map must be a meromorphic function, it has finitely many zeros and poles and is either regular or has a pole at $\infty$. So the degree of $f$ is defined: the equation $f(z)=a$ has $d$ solutions, counting multiplicity, for every $a\in S^2$. Therefore, if the map is bijective, we must have $d=1$ that is $f\in PSL(2,C)$.

This proof is essentially elementary algebra, only one fact from Analysis is used: the "Fundamental theorem of algebra".

Source Link
Alexandre Eremenko
  • 91.8k
  • 9
  • 259
  • 429

An elementary proof is the following: first you prove that every holomorphic map $f:S^2\to S^2$ is a rational function. Indeed, such map must be a meromorphic function, it has finitely many zeros and poles and is either regular or has a pole at $\infty$. So the degree of $f$ is defined: the equation $f(z)=a$ has $d$ solutions, counting multiplicity, for every $a\in S^2$. Therefore, if the map is bijective, we must have $d=1$ that is $f\in PSL(2,C)$.

This proof is based on elementary algebra, no analysis is used.