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I am trying to see if it is true that if we are given f:C-->C analytic ; C complex plane,

then, to extend f to a function defined on C^ (Riemann Sphere) , i.e, to get:

f^: C^ -->C^

with f^|_C =f

i.e., the restriction of f^ to the complex plane agrees with f ,

If we need to define f(oo) =oo .

I think the answer is yes. Here's what I have:

We consider a 'hood ('hood:=neighborhood.) W of oo in C^ , which are complements

of compact 'hoods K in C , together with {oo}, i.e., W=C\K U {oo} , for K compact in C.

By Liouville's thm., |f|-->oo on balls B(0;r) , as r-->oo . And then by continuity,

it would seem that we need f(oo)=oo, since W= C-B(0;r) is a 'hood of oo.

Alternatively, if we had an analytic map f on C , f would go to oo on balls B(0;r)

as r->oo . Then if we used the stereo projection S: C-->C^ , and push f

forward by this projection, we would have Sof (composition)-->oo .

But this is still not rigorous-enough.

Any Ideas?.

Thanks.

P.S: If it bothers people to use regular ASCII, please let me know. I use ASCII

as a way to force myself to keep things clear . But I can change if neccessary.

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# Extending Analytic Functions f:C->C , to f^:C^->C^; C^=Riemann Sphere

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