1
$\begingroup$

Does there is one one continuous function from $\{z\in\mathbb{C}:|z|>1\}$ to $\{z\in\mathbb{C}:z\neq 0\}?$ I tried many examples but did't found. Is there any concept about existence or non-existence of such a function? Please help. Thanks a lot.

$\endgroup$
1
  • 2
    $\begingroup$ The natural inclusion is such a function. (You didn't specify it needs to be onto.) $\endgroup$ Jun 20 '16 at 19:03
2
$\begingroup$

For instance, $$f(z)=\frac{z}{\lvert z\rvert}\ln \lvert z\rvert$$ The idea: $\ln x$ maps bijectively $(1,\infty)\to (0,\infty)$. So we just identify $\Bbb C=\Bbb R^2$ and rescale along the radii. In fact, $\dfrac z{\lvert z\rvert}$ is the norm-1 vector in the direction of $z$.

$\endgroup$
9
  • $\begingroup$ it is clearly continuous. Is it one one? $\endgroup$
    – neelkanth
    Jun 20 '16 at 18:57
  • $\begingroup$ Yes: its just a smooth radial rescaling via a bijective function. $\endgroup$
    – user228113
    Jun 20 '16 at 18:59
  • $\begingroup$ ok thanks a lot...i am looking at your solution. $\endgroup$
    – neelkanth
    Jun 20 '16 at 19:00
  • $\begingroup$ An important detail is that $\ln\lvert z\rvert$ is always positive on the domain (so you never send a vector in the opposite quadrant). $\endgroup$
    – user228113
    Jun 20 '16 at 19:03
  • 1
    $\begingroup$ @neelkanth, you should post it as a different question. $\endgroup$
    – Hmm.
    Jun 20 '16 at 19:07
1
$\begingroup$

A slightly simpler map is $$ f(z)=\frac{z}{\lvert z\rvert}(\lvert z\rvert-1) $$ still a radial rescaling.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.