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Does there exist a continuous one to one map from the set $A= \{z \in \Bbb C : z \neq 0 \}$ to the set $B = \{z \in \Bbb C : |z|>1 \}$?

I have tried to find this map though I have failed to do this. Please help me in this regard.

Thank you very much.

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$$f(z)=e^{\lvert z\rvert}\frac{z}{\lvert z\rvert}.$$

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  • $\begingroup$ This map is clearly continuous and since $z \neq 0$ so we have $|z|>0 \implies |f(z)|=|e^{|z|}|>1$ i.e. $f$ maps $A$ into $B$. But why is it one to one? $\endgroup$
    – D_C
    Jun 12 '18 at 6:10
  • $\begingroup$ @charMD No, on $(0,\infty)$ it's $e^x$, which is strictly increasing. $\endgroup$ Jun 12 '18 at 6:15
  • $\begingroup$ @D_C Try to write its inverse. $\endgroup$ Jun 12 '18 at 6:17
  • $\begingroup$ @SaucyO'Path do you mean that here $f$ is a bijection. $\endgroup$
    – D_C
    Jun 12 '18 at 6:19
  • $\begingroup$ @D_C Even better, it's a homeomorphism. $\endgroup$ Jun 12 '18 at 6:20
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$f(z)=z+\frac{z}{|z|}.$

The geometric meaning is: for a vector $z\neq 0$, we put it away from the origin $1$ unit in the direction of $z$, the inverse is pull it back $1$ unit.

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