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?  
I apologize if this has already been asked.
 A: 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.
