- #1

nonequilibrium

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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]