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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 Jul 22 '13 at 16:20

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Think about $f(z)=z^2$. – nsrt Jul 22 '13 at 15:45
... or $f(z)=e^z$. – diverietti Jul 22 '13 at 15:52
@nsrt, hm... (way too trivial for you :-). Let's take $f(z):=z^5$ where domain is restricted $\Re(z) > 0$. – Włodzimierz Holsztyński 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
up vote 7 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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