Consider the unit disc, complex plane, and Riemann sphere. For each ordered pair of these surfaces, is there a surjective holomorphic map? An injective holomorphic map?
Maps between simply-connected Riemann surfaces
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complex-analysis
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1Surely you can answer most of this question on your own. Could you please elaborate on what parts you are stuck on? – 2011-04-02
1 Answers
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Unit disk to complex plane:
Surjective? Consider $\sin((z+1)^{-3})$. The cubing sends a $120^\circ$ wedge at $-1$ included in the disk onto a punctured neighborhood of $-1$, and $\sin(1/z)$ sends each punctured disk at its essential singularity onto the plane. Easier would be to use a linear fractional transformation to map the open disk onto the half plane with real part greater than $-1$, then square.
Injective? Yes.
Unit disk to Riemann sphere:
- Surjective? There is a surjective holomorphic map from the disk to the plane (see above) and from the plane to the sphere (see below), so yes.
- Injective? Yes.
Complex plane to unit disk:
- There are no nonconstant holomorphic functions from the plane to the disk.
Complex plane to the Riemann sphere:
- Surjective? Consider $\frac{z^2}{1-z}$.
- Injective? Yes.
Riemann sphere to the complex plane or unit disk:
- There are no nonconstant holomorphic functions from the Riemann sphere to the complex plane. Removing $\infty$ from the domain would give an entire function with a removable singularity at $\infty$, which must be constant by Liouville's theorem. (Or, note that the image of the sphere would be compact, hence bounded, hence the function must be constant by Liouville.)
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0@elgeorges: Thank you very much for the compliment and for the informative comment. – 2011-04-03