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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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What do we know about $G$? – Davide Giraudo Dec 6 '11 at 19:48
I assume $\cal O(B)$ is the set of holomorphic functions on $B$. The question(s) you are supposed to answer are: i) if $f, g$ are holomorphic on $G$ and $f=g$ on $B$, does it follow that $f=g$ on $G$? -- this is the one-to-one part -- and ii) if $f$ is a holomorphic function on $B$, is it true that there is a holomorphic $F$ defined on $G$ such that the restriction of $F$ to $B$ coincides with $f$ -- this is the onto part. – user20266 Dec 6 '11 at 19:49
Merci Thomas Andrews, i) no, because the radius of convergence of f and g can be different, so that from $f=g$ on $B$ it doesn't follow that $f=g$ on G. ii) Yes, because the radius of convergence of f and F can be different, but the one of f is always smaller, so they will still both converge in the same disc. ? – VVV Dec 6 '11 at 20:11
i) careful, the key words you want to look up are 'analytical continuation' and 'connectedness'. ii) assume the disc has radius 1 and $G$ is the disc of radius 2. Try to find a holomorphic function on the disc of radius 1 which cannot be extended beyond that radius. – user20266 Dec 6 '11 at 20:18
i) yes, because f-g is an analytic function which vanishes in the open, connected domain B , so it must vanish on its entire domain. ii) $log(z)$ is holomorphic, but it can't be extended beyond the radius 1, since 0 doesn't work with it. ? Is this what you meant? Merci – VVV Dec 6 '11 at 20:44
  • 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 $\mathcal O(G)$.
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