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Suppose $f$ is analytic, say, in $\mathbb{C}$, and suppose $\Omega$ is a bounded simple connected open domain whose boundary we denote as $\Gamma$, then is $f(\Omega)$ also a simple connected domain whose boundary is $f(\Gamma)$?

I think $f(\Omega)$ is also connected becasue the continuity of $f$ suffices, but I'm not sure whether $f(\Omega)$ is simple connected and whether $f(\Gamma)$ will be the boundary of the domain.

Sorry for the above too simple question...

Now I put an additional condition on $f$, assuming that $f$ maps $\Gamma$ injectively into $f(\Gamma)$, then what can we say about $f(\Omega)$ ? Or, what if $f$ is injective on $\overline{\Omega}$ ?

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migrated from mathoverflow.net Jul 22 '13 at 16:20

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Think about $f(z)=z^2$. –  nsrt Jul 22 '13 at 15:45
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... or $f(z)=e^z$. –  diverietti Jul 22 '13 at 15:52
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@nsrt, hm... (way too trivial for you :-). Let's take $f(z):=z^5$ where domain is restricted $\Re(z) > 0$. –  wlod Jul 22 '13 at 15:52
    
sorry, I forgot to say that we assume $\Omega$ a bounded domain –  booksee Jul 22 '13 at 16:03
    
How about the disk $\lvert z \rvert < 2\pi$ under the function $f(z) = e^z$? –  MartianInvader Jul 22 '13 at 16:31

1 Answer 1

up vote 5 down vote accepted

As @nsrt indicates, the set $\{ z \mid \Im(z) \gt 0, 1 \lt |z| \lt 2 \}$ maps under $z\mapsto z^3$ to $\{|z| \lt 8\}\setminus \{|z|\le 1\}$, which is not simply connected.

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What number $z$ with positive imaginary part has square $2$? –  GEdgar Jul 22 '13 at 20:31
    
Thank you for pointing out my mistake! Should be $\geq$, not $>$. –  Eric Auld Jul 22 '13 at 23:34
    
With $\ge$ is it not "open". A better fix would be to use $z^3$ with your original set. Since it is already CW I will do that for you. –  GEdgar Jul 23 '13 at 12:20
    
Good point; thanks for the fix. –  Eric Auld Jul 23 '13 at 12:21

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