0

If there is a conformal map $f:X \to Y$ where $X,Y \subset \mathbb{C}$ are closed (and bounded), will $f$ necessarily map $\partial X \mapsto \partial Y$ (boundary of $X$ to boundary of $Y$)? Does this hold with weaker conditions also, if so what can be said?

My intuition is that it should do since conformal maps are continuous, so points not on the boundary of $X$ are inside a neighbourhood mapping to a neighbourhood in $Y$. And then since the boundary of $Y$ is mapped to (since image is closed), it must be mapped to from a subset of $\partial X$. Is this true? My doubt comes from not knowing if $f$ will map some points in $\partial X$ to $Y \backslash (\partial Y)$ since conformal does not imply biholomorphic.

Edit: Is it true that the following holds:

A map $f: D \to \mathbb{C}$ is conformal where $D \subset \mathbb{C}$ is closed $\implies$ $D$ is the continuous extension of an open set, $U \subset \mathbb{C}$ so that $D = \overline U$.

Jamie
  • 117
  • 1
    Possibly helpful: https://math.stackexchange.com/q/1825101/42969. – Martin R Dec 08 '21 at 14:56
  • 1
    What do you mean by a conformal map of closed subsets? For instance, suppose that $X$ is a finite subset, what would a conformal map mean? – Moishe Kohan Dec 08 '21 at 15:06
  • Thank you, this is helpful. So if I understand their answer correctly, would this imply $f(\partial X) \supseteq \partial Y$ with equality when $f$ is surjective? – Jamie Dec 08 '21 at 15:07
  • @MoisheKohan Ah yes, I understand your criticism. Perhaps what I meant was a closed and simple connected set? Although I am not particularly familiar with this terminology so correct me if this still does not make sense. – Jamie Dec 08 '21 at 15:12
  • 1
    Also related: https://en.wikipedia.org/wiki/Carathéodory%27s_theorem_(conformal_mapping). – Martin R Dec 08 '21 at 15:39
  • 1
    No, simply-connected will not be very helpful. Normally, conformal maps are defined between open subsets. Then the question becomes if such a map extends continuously to the boundary (not always) and if it maps boundary points to boundary points (yes, for those points where the extension exists). – Moishe Kohan Dec 08 '21 at 16:44
  • Okay thank you, so if I now understand correctly, when a conformal map from an open set can be extended continuously to also map the boundary, then it's necessarily true that said boundary will map to the boundary of the image? – Jamie Dec 08 '21 at 17:14
  • To clarify something in my head, it is still valid to say that a map is conformal from a closed set if and only if the closed set can be expressed as the continuous extension of an open set? For example a conformal map could map the closed unit disk but couldn't map the real line? or is it only an if statement and not an iff statement? – Jamie Dec 08 '21 at 17:17
  • 1
    I know complex analysis reasonably well and I never saw anybody using the notion of conformal maps between non-open subsets of the complex plane (or, more generally, of Riemann surfaces). I do not think there is a satisfactory notion for such a concept. (I can formulate one, but I doubt anybody uses it.) Thus, I suggest you stick to the concept of conformal maps of open subsets. As for your other question asked in a comment, please, edit your original post. Then I might answer it. I do not like answering questions from comments. – Moishe Kohan Dec 10 '21 at 22:57
  • @MoisheKohan Thanks for your comment, my question was initially motivated by a problem asking to show there does not exist a conformal map from the closed upper unit half disk to the closed unit disk, this wasn't a long answer but I couldn't help notice with the result I presumed may exist in my question that it would be a one liner. I have updated my question with an edit. Thanks – Jamie Dec 11 '21 at 19:21
  • Your edit does not make things any clearer. I suggest, you pick up a textbook in complex analysis and read the relevant definitions. In the current form, your question is un-answerable since you did not provide a definition of a conformal map from a non-open subset of the complex plane. – Moishe Kohan Dec 11 '21 at 19:23
  • The definition I have learnt of a conformal map, $f: D \to \mathbb{C}$ is on any subset $D \in \mathbb{C}$ where the map is real differentiable and the conformal condition is that the map preserves orientation and angle of any two tangent vectors at any point $z_0 \in D$. – Jamie Dec 11 '21 at 19:29
  • @MoisheKohan Does the edit I have made now make sense as a question to ask? Apologies for lack of clarity in original edit, I have only been studying complex maps for one term and was just curious if (my original question) was a result as my textbook did not mention it but it seemed intuitive to me. – Jamie Dec 11 '21 at 19:38
  • @JamieReason This definition is (almost) standard if the domain is open. One also usually requires invertible derivative and, frequently, that the map itself is invertible. Try to write down the detailed meaning of your definition if the domain is not open. – Moishe Kohan Dec 11 '21 at 19:56
  • @MoisheKohan Thank you for your input, I will find another textbook to find a more rigorous definition that is still compatible with one in the notes I am provided, and then use this an exercise. Thanks – Jamie Dec 12 '21 at 22:52

0 Answers0