# Proof of Open Mapping Theorem? (Ash & Novinger)

1. Jun 22, 2011

### nonequilibrium

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 $$f(\Omega) \textrm{ is open}$$ 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:
$$\textrm{Let f be a non-constant analytic function on an open connected set \Omega. Let z_0 \in \Omega and w_0 = f(z_0), }$$
$$\textrm{and let k = m(f-w_0,z_0) be the order of zero which f-w_0 has at z_0. }$$
$$\textrm{(a) There exists \epsilon>0 such that \overline D(z_0, \epsilon) \subset \Omega and such that neither f-w_0 nor f' has a zero in \overline D(z_0, \epsilon)\backslash {z_0}}.$$
$$\textrm{(b) Let \gamma be the positively oriented boundary of \overline D(z_0, \epsilon), let W_0 be the component of \mathbb C \backslash (f\circ \gamma)^* that contains w_0, }$$
$$\textrm{and let \Omega_1 = D(z_0,\epsilon) \cap f^{-1}(W_0). Then f is a k-to-one map of \Omega_1 \backslash {z_0} onto W_0 \backslash {w_0} }.$$

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 $$W_0 = f(D(z_0,\epsilon)),$$due to a continuous function sending connected sets to connected sets and $$w_0 \in f(D(z_0,\epsilon))$$

2. Jun 22, 2011

### micromass

Hi mr. vodka!

We know that W0 is open because $(f\circ \gamma)^*$ is compact and thus closed. Indeed, take z in W0. By closedness of $(f\circ \gamma)*$, we know that there exists a ball around z that does not meet $(f\circ \gamma)*$. 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.

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

3. Jun 22, 2011

### nonequilibrium

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. Jun 22, 2011

### micromass

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

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. Jun 22, 2011

### nonequilibrium

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 :)