Do Conformal Maps of Open Sets of the Complex Plane Always Take Boundary to Boundary?

For instance I'm trying to create a conformal map which takes the slit open unit disk in the complex plane to the open unit disk, and takes the boundary, that is, the set $(-1,0]\cup e^{i\theta}$ for $0\leq\theta<2\pi$ to the set $e^{i\phi}$ for $0\leq\phi<2\pi$.

I currently have a conformal map which takes the slit open unit disk to the open unit disk, can I be confident that it takes boundary to boundary?


Well, the relevant line in Ahlfohrs just says "For two-sided arcs the same will be true with obvious modifications. This is the second edition (1966), chapter 6, section 1.3 "Use of the Reflection principle," pages 225-226, just after the Riemann Mapping Theorem.

So, take your domain (you need to type in something about $r e^{i \theta}$ with $0 \leq r < 1$). First, take the principal branch of the square root. That maps your domain to the semicircle $|z| < 1, \; \mbox{Re}\; z > 0.$ Then we have a domain bounded by exactly two free one-sided analytic arcs. So, Theorem 4, section 1.4, page 227, both the boundary line segment and the boundary semicircle are mapped 1-1 and analytically to the boundary circle of the unit disk.

However, this says to me that the original slit does not have a single-valued map to the unit circle, it is evidently double-valued, just as the square root. Maybe it's just me.

EDIT: I have it correct. See pages 18-19 in Boundary Behavior of Conformal Maps by Christian Pommerenke. What he does is switch the order, the conformal map starts in the unit disk, denoted $$f : \mathbb D \rightarrow G.$$ Note: Falcao of Atletico Madrid just scored on Real Madrid, drawn 1-1 at minute 56. OK, the boundary of $\mathbb D$ is the unit circle, denoted $\mathbb T.$

Continuity Theorem. The function $f$ has a continuous extension to $\mathbb D \cup \mathbb T$ if and only if $\partial G$ is locally connected.

Caratheodory Theorem. The function $f$ has a continuous and injective extension to $\mathbb D \cup \mathbb T$ if and only if $\partial G$ is a Jordan curve.

If $\partial G$ is locally connected but not a Jordan curve then parts of $\partial G$ are run through several times. Some of the possibilities are indicated in Fig. 2.1 where points with the same letters correspond to each other. Note that e.g. the arcs $a_2 d_1$ and $a_1 d_2$ are both mapped onto the segment $ad.$

  • $\begingroup$ Hmm this is interesting, your description of the conformal map from the slit open unit disk to the open unit disk is basically what I have done with mine. I would be inclined to say that upon taking the slit open unit disk to the half open unit disk, that if boundary must go to boundary then the slit must go to the straight part of the boundary, is this consistent with what you are getting at? $\endgroup$ – Thoth Apr 11 '12 at 20:05
  • $\begingroup$ @Thoth, yes, that's about it. I would say that a maximal analytic arc, just continuous at the endpoints, goes to an analytic arc. So the slit becomes two segments that make up the straight part of the half-disk boundary, analytic except for continuous at the origin...I wish Ahlfohrs had done this example, it seems natural to wonder about a single slit. Meanwhile, the Riemann mapping is essentially unique. Page 136, exercise 5, Schwarz's lemma implies that any one to one conformal mmapping of the disk onto itself is linear fractional. So I do not see that we can demand better on the slit. $\endgroup$ – Will Jagy Apr 11 '12 at 20:42
  • $\begingroup$ The function $f(z) = -\frac{z^2 +2z -1}{z^2 -2z - 1}$ will then take the half open unit disk to the whole open unit disk. Have we decided what happens to the straight part of the boundary then? (and if you have hopefully you could explain it to me =]) $\endgroup$ – Thoth Apr 11 '12 at 21:07
  • $\begingroup$ The inverse of the full mapping, from the open disk to the disk minus the slit, extends continuously to the unit circle, but covers the slit twice. If you have appropriate software, have it draw pictures of the map your way and the inverse direction. It has been some 35 years since I wrote a specific conformal mapping, so do what you can to check that the map does what you think. $\endgroup$ – Will Jagy Apr 11 '12 at 22:17
  • $\begingroup$ I'm still processing this very good answer, thanks. $\endgroup$ – Thoth Apr 12 '12 at 18:57

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