Show that an analytic mapping is an open mapping

  • Thread starter Jamin2112
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In summary: That's part of it. Rouche's theorem is probably the main ingredient.It makes sense to me this way: An open set in f(z) (say an open ball B(f(z),r), to simplify) is one in which every point in B(f(z),r) is in f(z), i.e., every point in B(z,r) is in the image of f(z). This makes every point in B(f(z),r) a solution to , or a zero of, a certain equation. In summary, this student is trying to figure out how to continue an analytic function but doesn't seem to understand the concept of an open set.
  • #1
Jamin2112
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Homework Statement



As in in title.

Homework Equations



Open mapping: maps open sets to open sets.

The Attempt at a Solution



Not sure.
 
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  • #2
Jamin2112 said:

Homework Statement



As in in title.

Homework Equations



Open mapping: maps open sets to open sets.

The Attempt at a Solution



Not sure.

That's a pretty poor attempt. If you don't know how to prove it can you at least give us a few thoughts on why you think it might (or might not) be true?
 
  • #3
Dick said:
That's a pretty poor attempt. If you don't know how to prove it can you at least give us a few thoughts on why you think it might (or might not) be true?

If a function f(z) is differentiable on a set G, then it is continuous on G. If it continuous, the image of every open set is an open set. Done(?).
 
  • #4
Jamin2112 said:
If a function f(z) is differentiable on a set G, then it is continuous on G. If it continuous, the image of every open set is an open set. Done(?).

Nice that you are trying to think about it. But that's not true in the real numbers. f(x)=x^2 is continuous and differentiable, but it's not open. The proof is going to have to involve special properties of analytic functions.
 
  • #5
Dick said:
Nice that you are trying to think about it. But that's not true in the real numbers. f(x)=x^2 is continuous and differentiable, but it's not open. The proof is going to have to involve special properties of analytic functions.

Anything to do with "Analytic continuation" that I see on Wikipedia?
 
  • #6
Jamin2112 said:
Anything to do with "Analytic continuation" that I see on Wikipedia?

That's part of it. Rouche's theorem is probably the main ingredient.
 
  • #7
It makes sense to me this way:

An open set in f(z) (say an open ball B(f(z),r), to simplify) is one in which every point in B(f(z),r) is in f(z), i.e., every point in B(z,r) is in the image of f(z). This makes every point in B(f(z),r) a solution to , or a zero of, a certain equation.
 

Related to Show that an analytic mapping is an open mapping

1. What is an analytic mapping?

An analytic mapping is a function between two complex manifolds that is infinitely differentiable. This means that the function has derivatives of all orders at every point in its domain.

2. What does it mean for a mapping to be open?

An open mapping is a function that preserves the open sets of a topological space. In other words, if an open set is mapped to another space, the image of the open set will also be open.

3. How do you prove that an analytic mapping is an open mapping?

To show that an analytic mapping is an open mapping, you must prove that the function preserves open sets. This can be done by showing that the image of any open set in the domain is also open in the codomain.

4. What is the significance of an analytic mapping being an open mapping?

An analytic mapping being an open mapping means that the function is well-behaved and has nice properties. It also allows for the use of complex analysis techniques to study the function.

5. Can a mapping be analytic but not open?

Yes, it is possible for a mapping to be analytic but not open. This can occur if the function fails to preserve open sets in its domain or if the function is not defined on a complex manifold.

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