# one one continuous function from $\{z\in\mathbb{C}:|z|>1\}$ to $\{z\in\mathbb{C}:z\neq 0\}$

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.

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

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 $z$.

• it is clearly continuous. Is it one one? Jun 20 '16 at 18:57
• Yes: its just a smooth radial rescaling via a bijective function.
– user228113
Jun 20 '16 at 18:59
• ok thanks a lot...i am looking at your solution. Jun 20 '16 at 19:00
• 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).
– user228113
Jun 20 '16 at 19:03
• @neelkanth, you should post it as a different question.
– Hmm.
Jun 20 '16 at 19:07

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