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Let $A=\mathbb{C}- B_{1}[0]$, where $B_{1}[0]$ refers to the closed unit disc. Also, Let B be the complex plane punctured at the origin, i.e $B=\mathbb{C}-\{\ 0\}\ $. Then which of the following statements are correct?

(a) there exist continuos onto function $f:A\to B$.

(b) there exists continuos one-one function $f:B\to A$.

(c) there exists non-constant analytic function $f:B\to A$.

(d) there exists non-constant analytic function $f:A\to B$.

My attempt: Clearly (a) and (d) are true. $f(z)=e^{z}$ does the job for these statements. Now the answers say (b) is correct and (c) is not. I can't figure it out. I think Liouville's theorem also rule out (c) by considering the function $1/f$. But what about the existence of a one-one continuous function?

Thanks in advance!!

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For b), you can simply use the map $g : z \mapsto (|z|-1)z$.

For c), let's take $f : B \to A$. If $f$ extends to $\Bbb C$ this contradicts Liouville theorem. If $f$ has a pole at $0$, your argument with Liouville applied to $1/f$ works. If $f$ has an essential singularity, the image has to be dense so can't be contained in $B$. So such $f$ doesn't exist.

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  • $\begingroup$ I don't think you need to consider 3 cases. Just if we look at 1/f beforehand without doing any analysis, we have that 1/f has a removable singularity at 0 and hence 1/f is analytic on the whole of $\mathbb{C}$, and is also bounded and hence the result. $\endgroup$ – Riju Jun 17 '17 at 6:18
  • $\begingroup$ Indeed, you are totally right ! I $\endgroup$ – user171326 Jun 17 '17 at 6:19

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