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I am reading the book From holomorphic functions to complex manifolds by Klaus Fritzsche and Hans Grauert. I have a question about a fiber bundle. On page 186, the last line. How to show that $$ \Gamma(U, \mathcal{O}^*_{X}) \cong \mathcal{O}^*(U):=\{f\in \mathcal{O}(U) : f(x) \neq 0 \text{ for every } x\in U\}? $$ Thank you very much.

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This follows almost immediately from the definitions. Note that (1) a section of a trivial bundle $U\to U\times F$ is the same as a map $U\to F$, and (2) there is a natural inclusion $\mathbb C^* \to \mathbb C$. – Aaron Oct 11 '11 at 3:40
@Aaron, thank you very much. – LJR Oct 11 '11 at 17:56
up vote 1 down vote accepted

To get this off the Unanswered list, I'll restate Aaron's explanation as an answer:

The notation is $\mathbb C^*=\mathbb C\setminus \{0\}$ and $\mathcal{O}_X^* = X\times \mathbb C^*$, a trivial fibre bundle. By definition of a section, an element of $\Gamma(U,\mathcal{O}_X^*)$ is a holomorphic map $s:U\to \mathcal{O}_X^*$ such that $\pi\circ s = \mathrm{id}_U$. The fibres being $\mathbb C^*$, what we have is a nonvanishing holomorphic function.

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