How do you prove to any two simply-connected domains in the plane are homeomorphic without using the Riemann mapping theorem? An elementary proof would be nice.

  • $\begingroup$ Such a proof would certainly need the Jordan curve theorem (at least for smooth simple plane curves), and you might not call that elementary. $\endgroup$ May 26, 2011 at 13:12
  • $\begingroup$ @JohnPardon why does it necessarily need Jordan theorem? $\endgroup$ Jun 25, 2015 at 18:15
  • $\begingroup$ The Schoenflies theorem isn't particularly hard to prove. There's a nice cut-and-paste style argument given in Hatcher's 3-manifolds notes. $\endgroup$ Jun 25, 2015 at 23:34

3 Answers 3


Here's how one proof goes. I'm omitting some details, at least for now.

If $U$ is a simply-connected domain in the plane $\mathbb C$, then define the function $r: U \to \mathbb R_+$ to be the maximum radius of a (round) open disk in the standard Euclidean metric whose interior is contained in $U$. If $r(z) = \infty$ for any $z$, then $U$ is the entire plane, so there is nothing to prove. Otherwise, define a new Riemannian metric using the conformal factor $1/r$, that is, $ds' = 1/r\ ds$ where $ds$ is arc length in the standard metric.

Note that if $U$ is the upper half-plane, the metric coincides with the hyperbolic metric. If $U = \mathbb C \setminus \{0\}$, then all complex linear automorphisms of $\mathbb C$ preserve the metric, and the metric is that of a cylinder that is parametrized isometrically by $\log(z) / \left \langle 2 \pi i \right \rangle$. If $U$ is a round disk, then $ds'$ is a smooth negatively-curved metric except at the center of the disk, where there is a non-smooth point, but no cone angle: if the disk has radius 1, then a circle about the center of radius $\epsilon$ has length $2 \pi \epsilon / (1-\epsilon)$

In general, although the metric $ds'$ need not be smooth, it always has non-positive curvature. Intuitively: the $1/r$ factor means it takes infinite arc length to reach the boundary, and shortest $ds'$ geodesics try to thread their way into any bays and inlets of $U$ keeping far from the shoreline, since the speed limit is drastically reduced near the shore. In particular, there is a unique $ds'$ geodesic between any two points in $U$, and geodesics have the unique continuation property, they are determined by the tangent vector at the beginning and the length.

To parametrize $U$ by $\mathbb R^2$, choose any point $z_0$ in $U$. The tangent space to $U$ at $z_0$ parametrizes $U$, by $V$ goes to the geodesic through $V$ whose length is the length of $V$.

Sorry for leaving off details. Somewhere else on MO I believe I posted an alternate way to do this, using the convex hull of $S^2 \setminus U'$, where $U'$ is the stereographic image of $U$ on a sphere; in the projective hyperbolic metric, this boundary of the convex hull always is isometric to the hyperbolic plane, from which a proof is easy.


Since you ask for an elementary argument, let's assume that the domains have polygonal boundary. Then, the solution to the "carpenter's rule problem" (by Connelly-Demaine-Rote and Streinu) gives an actual algorithm for transfroming the domains into convex domains, for which any number of arguments work. (the carpenter's rule problem is described here:



Let me try to solve it completely elementary. My aim is to represent simply-connected domain $\Omega$ as a union $\cup_{i=1}^{\infty} P_i$, where $P_1,P_2,\dots$ are (closed) polygons and $P_i\subset {\rm Int}(P_{i+1})$ ($\rm Int$ is for interior). If such representation is obtained, then we may construct homeomorphisms $f_k:P_k\rightarrow \{z:|z|\leq 1-2^{-k}\}$, with $f_k(\partial P_k)=\{z:|z|=1-2^{-k}\}$, and $f_{k+1}$ extends $f_k$. For $x\in \Omega$ define $f(x)=f_k(x)$ for all $k$ such that $x\in P_k$, we get a homeomorphism from $\Omega$ to the open unit disc $\{z<1\}$.

How to construct our system of polygons? Enumerate all diadic squares $[a\cdot 2^n,(a+1)\cdot 2^n]\times [b\cdot 2^n,(b+1)\cdot 2^n]$, $a,b,n$ are integer, containing in $\Omega$: $S_1,S_2,\dots$. Define $P_1=S_1$. Next, if polygons $P_1,\dots,P_{n-1}$ are constructed, consider the square $S_n$ and denote $F=P_{n-1}\cup S_n$. If necessary, add to $F$ finite number of dyadic squares so that it becomes connected. It may contain holes, but since $\Omega$ is simply connected everything in holes belongs to $\Omega$, add all this to $F$. Finally, add several small dyadic cells to $F$ so that $P_{n-1}\subset {\rm Int}(F)$ and put $P_n:=F$. Our construction garantees that $\cup P_i=\cup S_i=\Omega$ as desired.


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