Does there exist an example

Hi there!

This is an interesting question.

The Riemannian mapping thm. actually gives a partial answer to that question - it is possible to find such a mapping, if the domain is not all of C, so the statement is seriously doubted - otherwise Riemann would have stated his thm in a more general way

However, I'll try to sketch a counter proof:

Assume that f: C -> D is holomorphic. It is evident that f is then an entire function. What is more, for all z in C it is true that |f(z)|<=1, i.e. f is bounded, because D is bounded. Now the Liouville's thm implies that f must be constant, implying the statement is incorrect.
(the Liouville's thm is that very nice tool also used for proving the Fundamental Theorem of Algebra in less that 5 lines)


regards,
marin

 

Holomorphic maps are continuous. The unit disk is compact. Continuous maps apply compact set to compact set. The complex plane is not compact. Hence, you can't map continuously (or holomorphically) D onto C.

 

Of course he means the open disk. The disk and the plane are homeomorphic. But not conformally equivalent (as noted, by Liouville's theorem). So what about mapping the open disk onto the complex plane in a many-to-one manner?

 

For some reason I didn't even consider that the OP might mean the open disk!