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This is an exercise from a book called "theory of complex functions" I want to solve:

Let $c\in G$ and $B\subset G$ a disc around c. When is the $\mathbb{C}$-algebra homomorphism $\mathcal{O}(G) \rightarrow \mathcal{O} (B)$ defined by $f\mapsto f_{|B}$ one-to-one? When is it onto?

I am not sure what this exercise asks. The $\mathbb{C}$-algebra maps a function to a function with its ensemble of definition restricted to the map of its disc? Or a domain is mapped to a disc in the domain?

If the second is the case, then I would say that it is never one-to-one, because the ensemble of definition has a higher cardinality than the ensemble of the map, but always onto.

Merci for any explaination.

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    I thought of $1/(1-z)$, for example. There is no holomorphic $\log$ on the disc, you need to cut the domain along a line or curve from the origin to the boundary.2011-12-07

1 Answers 1

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  • It is one-to-one if and only if $G$ is connected. If $G$ is connected, you can apply the identity theorem. If $G$ is not connected, then two functions can be equal on the component containing $B$ while being different on a component not containing $B$.
  • It is onto if and only if the component of $G$ containing $B$ is equal to $B$. If the latter condition holds, then each $f\in \mathcal O(B)$ extends to a holomorphic function on $G$ by setting the values to $0$ (for example) on $G\setminus B$. If the condition does not hold, then there exists a $b\in G\cap\partial B$, and $z\mapsto\frac{1}{z-b}$ is an element of $\mathcal O(B)$ that has no extension in $\mathcal O(G)$.