# Does a conformal map take boundaries to boundaries?

I think it is a well-known result that conformal maps between sets in $\mathbb{C}$ take boundaries to boundaries. However, I looked around a little and I had trouble finding this result. Is it true? Also, is there a quick proof of it?

• In some sense, yes, but you need to be careful. A conformal map from a domain extends continuously to the boundary if the boundary is a Jordan curve. But if your open set is, say, the unit disc minus the nonnegative real axis, there's no continuous extension. There was an old question about this with a good answer, let me try to find it.
– user98602
Commented Dec 31, 2014 at 19:34
• Thank you. And when there is a continuous extension, this preserves boundaries, correct? Commented Dec 31, 2014 at 19:37
• With some minor assumptions (compactness of the closure, say), yes. I can write a proof in a minute.
– user98602
Commented Dec 31, 2014 at 19:38
• This seems to be harder than I remembered. For simply connected Jordan domains what you want is Carathéodory's theorem. For some discussion on the case I mentioned in my first comment see my comment and the first bit of Behaviour's answer here. I believe it to be true that a conformal map between bounded Jordan domains (no requirement of simple connectivity) extends to a homeomorphism of the closure, but I do not have a proof.
– user98602
Commented Dec 31, 2014 at 20:05
• Please clarify what you mean by "take boundaries to boundaries". Presumably, it's a statement involving some limits; make it precise.
– user147263
Commented Dec 31, 2014 at 20:36

Given that a conformal map $f: \Omega_1 \to \Omega_2$ extends continuously to $\bar f: \overline{\Omega}_1 \to \overline{\Omega}_2$, it is true that $\bar f(\partial \Omega_1) \subset \partial \Omega_2$. For pick a sequence of points $(z_i)$ converging to $z$ in the boundary; then $\bar f(z) = \lim f(z_i)$. Call $\lim f(z_i) = z'$, and suppose $z' \in \Omega_2$ itself. Since $f$ is a homeomorphism, let $g = f^{-1}$. Because $g$ is continuous, $g(z') = \lim g(f(z_i)) = \lim z_i$; but the former is in $\Omega_1$, so $\lim z_i \in \Omega_1$, contradictory to our assumption.
It is not a given that a conformal map should extend continuously to the boundary. For instance, the Riemannm map taking $\Omega = D^2 \setminus \Bbb R_{\geq 0}$ to the open unit disc doesn't have a continuous extention to its boundary. (The inverse map taking $D^2$ conformally to $\Omega$ does, though. It maps two points to each element of the strip.) Carathéodory's theorem says that the Riemann map from any (bounded) Jordan domain to the unit disc extends continuously to the boundary (inducing a homeomorphism). I suspect a similar theorem is true for multiply connected Jordan domains, but I don't have a proof.
• Is it true that a conformal equivalence $\Omega_1 \to \Omega_2$ extends continuously to $\overline{\Omega}_1$ (though not to a homeomorphism) if $\Omega_1$ is a Jordan domain but $\Omega_2$ is not?