Q on Riemann Domains: Is P Injective on U?

In summary, P is a local homeomorphism on U and a global homeomorphism on U if B is simply connected. If P(U) is ring-shaped, P is still a local homeomorphism but not a global one. If P is proper and U is connected, then P is a homeomorphism on U.
  • #1
lark
163
0
If P is the projection map from a Riemann domain M [itex]\rightarrow C^n[/itex], and U is a connected subset of M with P(U)=B, where B is a ball in [itex]C^n[/itex], 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
 
Last edited:
Physics news on Phys.org
  • #2
,

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!
 

Related to Q on Riemann Domains: Is P Injective on U?

1. What is Q on Riemann Domains?

Q on Riemann Domains refers to a mathematical concept that involves the use of a function called "Q" on a specific type of mathematical structure called a Riemann Domain. A Riemann Domain is a complex mathematical space that has certain properties that allow for the application of Q in a meaningful way.

2. What does it mean for P to be injective on U?

When a function P is considered to be injective on a set U, it means that for every input value in U, there is a unique output value. In other words, no two input values in U will produce the same output value, making P a one-to-one function on U.

3. Why is the question of P being injective on U important in the context of Riemann Domains?

In the context of Riemann Domains, P being injective on U is important because it allows for the study and analysis of the behavior of the function Q. This is because Q is only well-defined on Riemann Domains where P is injective on U. Additionally, the injectivity of P on U can provide insights into the properties and structure of the Riemann Domain itself.

4. How is the injectivity of P on U determined?

The injectivity of P on U can be determined by analyzing the function P and the set U. One way to determine injectivity is by using the horizontal line test, where a horizontal line drawn through the graph of P will only intersect the graph at most once. Another method is to check the values of P for all elements in U and see if there are any repeated outputs.

5. What are the implications of Q being well-defined on a Riemann Domain?

If Q is well-defined on a Riemann Domain, it means that the function is valid and meaningful on that specific mathematical space. This allows for the application and study of Q on the Riemann Domain, and can also provide insights into the properties and structure of the Riemann Domain itself.

Similar threads

Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Differential Geometry
Replies
14
Views
3K
Replies
9
Views
2K
Replies
5
Views
2K
  • Differential Geometry
Replies
8
Views
2K
  • Differential Geometry
Replies
12
Views
2K
  • Differential Geometry
Replies
17
Views
2K
  • Special and General Relativity
Replies
25
Views
2K
  • Differential Geometry
Replies
6
Views
1K
Back
Top