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I am working on following Question;

f is analytic on $D(0,1)\setminus\{0\}$ with an essential singularity at 0, show f is not one-to-one.

I am a bit stuck. All I have been able to figure out is the following;

  • $\lim_{z \rightarrow 0} f(z)$ does not exist since the singularity is not removable.
  • $f(D(0,1)\setminus\{0\})$ is open by the Open Mapping Theorem
  • $f(D(0,1)\setminus\{0\})$ is dense in $\mathbb{C}$, by Casorati-Weierstrass.

Now I suspect that we should try the proof by contradiction route and suppose that $f$ is one-to-one and show that then the singularity is maybe not essential, potentially removable. Or maybe we should construct two sequence converging to different limits while the images of the sequences converge to the same thing.

Cheers for the help in advance; If possible please do not give the whole answer but rather give hints and when I feel I have a solution I will post it as an answer (Of course I will accept the most useful hint as the answer to this question)

Jandré Snyman
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    Try considering a single point $a = f(b)$ and tiny disks around it - what happens then if you consider pre-images of such disks, $f^{-1}(D(f(b),\delta))$? – João Ramos Sep 11 '18 at 07:01
  • What can we say about pre-images of open sets other than the fact that they are not necessarily open and that there must in fact exist an open set that has a non-open pre-image? – Jandré Snyman Sep 11 '18 at 07:12
  • Last time I checked, pre-images of open sets were open if the function in question is continuous... if $f$ is one-to-one, then, by shrinking your disk, you have to have a ‘small’ preimage $f^{-1}(D(f(a),\delta))$. – João Ramos Sep 11 '18 at 07:17
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    I might Mark this question as a duplicate as this question is very similar to the ones that Martin posted above. – Jandré Snyman Sep 11 '18 at 07:34

1 Answers1

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Some useful ideas are:

  • As you mentioned, the Cassorati-Weierstrass theorem states that $f(D(0,1)\backslash\{0\})$ is dense in $\mathbb{C},$ but also that the same holds for any disk $D(0,\varepsilon)\backslash\{0\}.$

  • If $f$ is one-to-one, then, for some $a \in D(0,1)\backslash\{0\}$, if we consider the pre-images $f^{-1}(D(f(a),\delta)),$ they should be, by the (holomorphic) inverse function theorem, contained in a tiny neighborhood $D(a,\eta)$ of $a.$

  • By 'shrinking' the original disk to something like $D(0,\varepsilon)\backslash\{0\},$ for $\varepsilon$ small, what does the 'lack of a disk' in the image of $f$ say, in conjunction with the first point presented?

João Ramos
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