Extending complex functions f:C->C into f^:C^->C^

[Bacle2]

Extending complex functions f:C-->C into f^:C^-->C^

Hi, All:

I have a function f: ℂ-->ℂ' , i.e., a complex-valued function (use ℂ,ℂ' to make a distinction between the complexes as domain and codomain respectively), and i want to extend it into
the Riemann Sphere ℂ^, i.e., I am looking for f^ such that f^|_ℂ=f. If I remember correctly, a necessary and sufficient condition for extensibility is
that f must be a proper function, i.e., that for every K compact in ℂ' is sent to a compact
set in ℂ, i.e., f^-1(K) is compact in ℂ? I think this somehow had to see with
ℂ^ being the 1-pt compactification of ℂ; is this correct? If so, anyone have a ref. for the
proof, if not, could someone please let me know what the correct result is?

Thanks.

 

[micromass]

Indeed, the topology on the Riemann sphere is

$$\{U~\vert~U\cap\mathbb{C}~\text{is open in}~\mathbb{C}~\text{and if}~\infty\in U~\text{then}~U^c~\text{is compact}\}$$

Proving that the inverse of an open set is open under a proper map is now very easily done.

 

[Bacle2]

Thanks, Micromass, but I was hoping for a holomorphic extension. I was thinking of using some form of Riemann Removable singularity near ∞ , to avoid having to use manifold charts, or some heavy-handed methods. I know the functions that extend continuously in the real case are those that go to ∞ as x→∞ itself, but I don't see clearly how to extend holomorphically into C^.

 

[micromass]

Ah, so you want to extend a holomorphic function to a meromorphic function??

Well, then I need to say that the only meromorphic functions $\hat{f}:\hat{\mathbb{C}}\rightarrow \hat{\mathbb{C}}$ are the rational functions. So the condition you need to put on f is that it is a rational function.

 

[Bacle2]

I think rational functions f: C^-->C^ are holomorphic on C^ --tho I am not clear on how we define holomorphic on maps C^-->C^ without manifold charts of some sort, or maybe some function fields , since ∞ is not a problem
value anymore in C^. If we have, e.g.,az+b in the denominator, then z=-b/a is not
a singularity anymore. Maybe this is an issue for Algebraic Geometry.

 

[Bacle2]

Never mind, Micromass, I got it; thanks.

 

[Bacle2]

Sorry, just in case anyone else reads and is interested in the same issue: outside of ℝⁿ we cannot just differentiate; in manifolds,we need to pull back functions by chart maps and _then_ differentiate. In our case of the Riemann sphere, which is an analytic variety (i.e., a complex manifold), we use the stereographic projection , using the fact that the transition map in the stereographic projection is given by 1/z , we need to pre- and post- compose with 1/z (which will take us into ℂ) , where we can use the standard notion of differentiability. This way we can decide when a function is analytic in the Riemann sphere.