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.
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2The natural inclusion is such a function. (You didn't specify it needs to be onto.) – Cheerful Parsnip Jun 20 '16 at 19:03
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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 $z$.
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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). – Jun 20 '16 at 19:03
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yes i got the point..... can there be Non constant analytic function from ${z\in\mathbb{C}:z\neq 0}$ to ${z\in\mathbb{C}:|z|>1}.$? – neelkanth Jun 20 '16 at 19:04
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http://math.stackexchange.com/questions/1833688/non-constant-analytic-function-from-z-in-mathbbcz-neq-0-to-z-in-math – neelkanth Jun 20 '16 at 19:13
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A slightly simpler map is $$ f(z)=\frac{z}{\lvert z\rvert}(\lvert z\rvert-1) $$ still a radial rescaling.
lhf
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