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The proof of Theorem 13.5 in "Lectures on Riemann Surfaces" by Otto Forster begins by saying

Set $U_1:={\mathbb P}^1 \backslash \infty$ and $U_2:={\mathbb P}^1 \backslash 0$. Since $U_1 = {\mathbb C}$ and $U_2$ is biholomorphic to ${\mathbb C}$, it follows from (13.4) that $H^1(U_i, {\cal O})=0$.

In "1.5 Examples of Riemann Surfaces" in the book the maps $\phi_i:U_i \rightarrow {\mathbb C}, i=1,2$ are defined as follows:

$\phi_1$ is the identity map and $$ \phi_2(z) := \left\{ \begin{array}{ll} 1/z & \mbox{for} \; z \in {\mathbb C}^*\\ 0 & \mbox{for} \; z = \infty \end{array} \right. $$

But, It seems to me that $\phi_2$ cannot be biholomorphic at $\infty$, because since $\phi_2'(z)=-1/z^2$,

$$ \lim_{z\rightarrow \infty} \phi_2'(z) = 0. $$

Could someone point out where I made a mistake ?

  • 1
    Yes, the limit is correct. But how does this produce any contradiction?2012-01-13
  • 1
    I thought that $\phi_2'(\infty)\neq 0$ is a necessary condition for the existense of the holomorphic inverse mapping of$\phi_2$ at $\infty$.2012-01-13

1 Answers 1