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Proof of Open Mapping Theorem? (Ash & Novinger)

  1. Jun 22, 2011 #1

    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]
  2. jcsd
  3. Jun 22, 2011 #2
    Hi mr. vodka! :smile:

    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.

    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].
  4. Jun 22, 2011 #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 :)
  5. Jun 22, 2011 #4
    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].

    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...
  6. Jun 22, 2011 #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 :)
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