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!!


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.

  • $\begingroup$ 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

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.