Proof of Open Mapping Theorem? (Ash & Novinger)

In summary, the conversation is about using a free course on complex analysis by Ash & Novinger. The person is stuck on the proof of the open mapping theorem and is specifically asking about proposition (d) and how it follows from propositions (a) and (b). The expert responds by summarizing the definitions and assumptions in propositions (a) and (b) and explains that proposition (d) follows from them. They also address a side question about the introduction of W_0 and explain its importance in the proof. The conversation ends with the person thanking the expert for clarifying and the expert pointing out the importance of assumptions in mathematical proofs.
  • #1
nonequilibrium
1,439
2
Hello,

In class we're using the free course on complexe analysis by Ash & Novinger, legally downloadable online. I'm stuck on their proof of the open mapping theorem. More specifically:
http://www.math.uiuc.edu/~r-ash/CV/CV4.pdf (page 15)

proposition (d) is [tex]f(\Omega) \textrm{ is open}[/tex] and its proof is that it follows out of proposition (a) and (b), so I'll give you (a) and (b) in case you don't want to click the link:
[tex]\textrm{Let $f$ be a non-constant analytic function on an open connected set $\Omega$. Let $z_0 \in \Omega$ and $w_0 = f(z_0)$, }[/tex]
[tex]\textrm{and let $k = m(f-w_0,z_0)$ be the order of zero which $f-w_0$ has at $z_0$. }[/tex]
[tex]\textrm{(a) There exists $\epsilon>0$ such that $\overline D(z_0, \epsilon) \subset \Omega$ and such that neither $f-w_0$ nor $f'$ has a zero in $\overline D(z_0, \epsilon)\backslash {z_0}$}. [/tex]
[tex]\textrm{(b) Let $\gamma$ be the positively oriented boundary of $\overline D(z_0, \epsilon)$, let $W_0$ be the component of $\mathbb C \backslash (f\circ \gamma)^*$ that contains $w_0$, }[/tex]
[tex]\textrm{and let $\Omega_1 = D(z_0,\epsilon) \cap f^{-1}(W_0)$. Then $f$ is a $k$-to-one map of $\Omega_1 \backslash {z_0}$ onto $W_0 \backslash {w_0}$ }.[/tex]

Proving (a) and (b) is not a problem, but I don't see how (d) follows out of (a) & (b)... It would follow if I knew that W_0 were open, but do I know that?

Side-Q: why is W_0 introduced? It seems obvious that [tex]W_0 = f(D(z_0,\epsilon)),[/tex]due to a continuous function sending connected sets to connected sets and [tex]w_0 \in f(D(z_0,\epsilon))[/tex]
 
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  • #2
Hi mr. vodka! :smile:

mr. vodka;3369909 [B said:
Proving (a) and (b) is not a problem, but I don't see how (d) follows out of (a) & (b)... It would follow if I knew that W_0 were open, but do I know that?[/B]

We know that W0 is open because [itex](f\circ \gamma)^*[/itex] is compact and thus closed. Indeed, take z in W0. By closedness of [itex](f\circ \gamma)*[/itex], we know that there exists a ball around z that does not meet [itex](f\circ \gamma)*[/itex]. I claim that this ball stays in W0. Indeed, if there was a point a outside W0 in the ball, then there would exist a path from a to z. But then W0 is not a component.

Side-Q: why is W_0 introduced? It seems obvious that [tex]W_0 = f(D(z_0,\epsilon)),[/tex]due to a continuous function sending connected sets to connected sets and [tex]w_0 \in f(D(z_0,\epsilon))[/tex]

It seems obvious, but it's not true. Who says that f doesn't send a point in [itex]D(z_0,\epsilon)[/itex] to [itex](f\circ \gamma)^*[/itex]? For example, we can send a circle to a figure 8, and in this case, some of the points inside [itex]D(Z_0,\epsilon)[/itex] get sent to [itex](f\circ \gamma)^*[/itex].
 
  • #3
Thank you micromass :)

Regarding the main question:
thank you for proving W_0 is open, but I'm still not convinced! In my OP I stated that I would be done if I knew W_0 were open, but now I take those words back. I feel like I also need to know that W_0 is a subset of the image of Omega_1. Is that true? If not, do you know how you can get (d) out of (a) and (b)?

Regarding the side-question:
Aha, interesting. But is it really possible to send a circle onto an "8"? After all, if there is a z inside the circle that gets projected onto the waist of the "8" (which there must be, due to connectedness(?)), then you can't draw a circle around that point which stays inside the "8"; in other words the image wouldn't be open... (?)
but anyway it wouldn't matter if that specific example wouldn't work; you've shown me that what I thought to be obvious, is not, and hence the usefulness of defining W_0 :)
 
  • #4
mr. vodka said:
Thank you micromass :)

Regarding the main question:
thank you for proving W_0 is open, but I'm still not convinced! In my OP I stated that I would be done if I knew W_0 were open, but now I take those words back. I feel like I also need to know that W_0 is a subset of the image of Omega_1. Is that true? If not, do you know how you can get (d) out of (a) and (b)?

That's exactly what (b) says. Be says that [itex]f(\Omega_1\setminus \{z_0\})=W_0\setminus \{w_0\}[/itex]. Thus [itex]f(\Omega_1)=W_0[/itex].

Regarding the side-question:
Aha, interesting. But is it really possible to send a circle onto an "8"? After all, if there is a z inside the circle that gets projected onto the waist of the "8" (which there must be, due to connectedness(?)), then you can't draw a circle around that point which stays inside the "8"; in other words the image wouldn't be open... (?)
but anyway it wouldn't matter if that specific example wouldn't work; you've shown me that what I thought to be obvious, is not, and hence the usefulness of defining W_0 :)

I agree that it's probably not possible to send a circle to an 8. But this is not a priori obvious. That is, before proving the open mapping theorem, you don't know that you can't do that. Thus an assumption must be made...
 
  • #5
Oh, of course, I had missed the "k-to-one"-segment, which implies /isthe whole set W_0\{z_0} was/is used!

Thanks a lot :)
 

FAQ: Proof of Open Mapping Theorem? (Ash & Novinger)

1. What is the Open Mapping Theorem?

The Open Mapping Theorem is a fundamental result in complex analysis that states if a holomorphic function maps a connected open set to another open set, then it is an open map.

2. What is the significance of the Open Mapping Theorem?

The Open Mapping Theorem has many important consequences, including the fact that a non-constant holomorphic function cannot have any isolated zeros. It also allows us to solve certain partial differential equations by transforming them into simpler ones.

3. Who first proved the Open Mapping Theorem?

The Open Mapping Theorem was first proved by German mathematician Edmund Landau in 1906. It was later independently proved by Hans Hahn and Stefan Banach in the 1920s.

4. What are the key assumptions of the Open Mapping Theorem?

The Open Mapping Theorem assumes that the function in question is holomorphic (analytic) and maps a connected open set to another open set. It also assumes that the function is not constant, meaning it has at least two distinct values.

5. Can the Open Mapping Theorem be generalized to other contexts?

Yes, the Open Mapping Theorem has been generalized to topological vector spaces, Banach spaces, and other types of analytic functions such as meromorphic functions. These generalizations are known as the Open Mapping Theorem in Functional Analysis and the Open Mapping Theorem in Complex Analysis.

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