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Let $A=\{z\in \Bbb C:|z|>1\}$; $B=\{z\in \Bbb C:z\neq 0\}$ Which of the following are true:

  • There is a continuous onto function $f:A\to B$
  • There is a continuous one-one function $f:B\to A$
  • There is a non-constant analytic function $f:A\to B$
  • There is a non-constant analytic function $f:B\to A$

My try:

  1. Unable to prove

  2. Consider $f:B\to A$ by $f(z)=\frac{1}{z};|z|<1$ and $f(z)=z;|z|\ge 1$

3 . Consider the function $f(z)=e^z$

4 Unable to show .

Please give some hints.Really clueless

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  • $\begingroup$ "3 unable to show" ? $\endgroup$
    – A.G
    Sep 3 '16 at 14:27
  • $\begingroup$ Any thing wrong in the problem @A.G $\endgroup$
    – Learnmore
    Sep 3 '16 at 14:49
  • $\begingroup$ I have edited that @A.G $\endgroup$
    – Learnmore
    Sep 3 '16 at 15:10
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Hint for 1: Consider the intersection of $A$ with the positive real axis. Now consider what happens if you subtract $1$ from each point. Now do the same thing for the upper imaginary axis but subtract $i$ from each point instead of $1$.

Can you generalize to the line $\arg(z)=\theta$?

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For 1, 2: Let $g(r) = r^2 - r.$ Then $g$ is a nice differentiable (in fact, real analytic) bijection of $(1,\infty)$ onto $(0,\infty)$ with $g'>0$ everywhere. The map $f(re^{it}) = g(r)e^{it}$ is thus a real analytic diffeomorphism from $A$ onto $B,$ and $f^{-1}(re^{it}) = g^{-1}(r)e^{it}$ is the same coming back.

Hint for 4: If $f:B\to A$ is analytic, then  $1/f$ is bounded and analytic in $B.$

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