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Let $A=\{z\in \mathbb C:|z|>1\}$ and $B=\{z\in \mathbb C:z\neq 0\}$. Then which of the following is/are true?

  1. There exists a continuous onto function $f:A\to B$.

  2. There exists a continuous one-one function $f:B\to A$.

  3. There exists a non-constant analytic function $f:A\to B$.

  4. There exists a non-constant analytic function $f:B\to A$.

3 is correct as $f(z)=e^z$ is one such function. What about others? I am confused.

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  • $\begingroup$ If you can find a bijection beween the two, you ar golden. It an it's inverse answer 1,2. Any analytic funtion (even the identity) on either can be composed with the bijection to get an analytic function from one set to the other. What about 1) f:z -> z(|z|-1)? $\endgroup$ – fleablood Oct 8 '16 at 4:24
  • $\begingroup$ For 2, $re^{i\theta}\mapsto(r+1)e^{i\theta}$ $\endgroup$ – Gerry Myerson Oct 8 '16 at 4:36
  • $\begingroup$ @GerryMyerson: That works for 1 & 2. $\endgroup$ – copper.hat Oct 8 '16 at 5:03
  • $\begingroup$ $f(z) = z$ works for 3. $\endgroup$ – copper.hat Oct 8 '16 at 5:03
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for (4) :

if $f : B \to A$ is analytic then $g(z) = (f(\exp z))^{-1}$ is analytic from $\Bbb C$ to $\{z \in \Bbb C \mid 0 < |z| < 1 \}$.

Since $g$ is entire and bounded it has to be a constant $c$. Then since $\exp$ is onto $B$, $f$ has to be constant too (to $c^{-1}$).

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  • $\begingroup$ +1: Great! This is much simpler than mine. My solution gives what form $f(B)$ must have. So, mine worked out like an overkill. $\endgroup$ – i707107 Oct 8 '16 at 18:07
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$4.$ The answer is false.

Consider the Laurant series of $f(z) = \sum_{n\in \mathbb{Z}} a_n z^n$. Suppose that there is an integer $M_0$ such that $a_n= 0$ for all $n\leq M_0$. If there is an integer $N_0$ and $a_n\neq 0$ for infinitely many $n\geq N_0$, then $f(1/z)$ has an essential singularity at $0$. Then by Casorati-Weierstrass Theorem, $f(B)$ is dense in $\mathbb{C}$.

Suppose that there is an integer $M_0$ such that $a_n= 0$ for all $n\leq M_0$. And, if there is an integer $N_0$ such that $a_n\neq 0$ for only finitely many $n\geq N_0$. Then $z^kf(z)$ is a polynomial for some $k\in \mathbb{Z}$. By the Fundamental Theorem of Algebra, the image of $f$ contains $\mathbb{C}-\{z_0\}$ for some $z_0\in\mathbb{C}$.

Suppose $f(z)$ has an essential singularity at $z=0$, then by Casorati-Weierstrass Theorem, $f(B)$ is dense in $\mathbb{C}$.

We have a stronger result if we apply the Great Picard Theorem instead of Casorati-Weierstrass. For some $z_0\in\mathbb{C}$,

$f(B)$ contains $\mathbb{C}-\{z_0\}$.

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