Q on Riemann Domains: Is P Injective on U?

[lark]

If P is the projection map from a Riemann domain M $\rightarrow C^n$, and U is a connected subset of M with P(U)=B, where B is a ball in $C^n$, then is P injective on U, so it's a homeomorphism on U?
P is locally a homeomorphism by definition.
It would be related to B being simply connected. WOuldn't be true if P(U) were ring-shaped.
I read something saying that in complex analysis, local homeomorphisms being global homeomorphisms relates to connectivity.
If P is proper, meaning if K is a compact subset of B, the inverse image in U of K is compact, it would be true by a theorem of Ho, apparently.
Laura

 

[jvicens]

,

Thank you for your post. I can confirm that your understanding is correct. The statement that P is locally a homeomorphism means that it is a continuous and bijective map between topological spaces, and its inverse is also continuous. This implies that P is injective on U, meaning that distinct points in U map to distinct points in B. Therefore, P is a homeomorphism on U.

You are also correct in saying that this is related to the connectivity of B. If B is simply connected, then P(U) will also be simply connected, and this implies that P is a global homeomorphism. This is because, in simply connected spaces, any loop can be continuously deformed into a point. Therefore, if P is injective on U, any loop in P(U) can be continuously deformed into a point, which means that P(U) is simply connected.

However, as you mentioned, this would not be true if P(U) were ring-shaped, as it would have holes and would not be simply connected. In this case, P would not be a global homeomorphism, but it would still be a local homeomorphism.

The statement about P being proper is also correct. If P is proper, then it satisfies certain conditions that guarantee that it is a homeomorphism. This is known as the proper mapping theorem, and it is a generalization of the Inverse Function Theorem. So, if P is proper and U is connected, then P is a homeomorphism on U.

I hope this helps clarify your understanding. Keep up the good work in your studies of complex analysis!