Let $\gamma: [0,1]\rightarrow \mathbb{R}^2$ a piecewise smooth, simple, closed plane curve with at least 4 vertices (points where $\gamma$ is not smooth). Let $C$ denote its trace and $I$ its interior. I am reading do Carmo's differential geometry book on curves and surfaces, and he states a topological theorem (without proof) on the existence of triangulations for regular regions on surfaces. A corollary of this is that $C\cup I$ can be triangulated. My question is whether this triangulation can be of the following type:

Let $v_1,...,v_k$ be the vertices $k\ge 4$. Can we choose smooth curves connecting $v_1$ to $v_3$, $v_1$ to $v_4$,..., $v_1$ to $v_{k-1}$, which a) lie in $I$ (excluding their beginnings and ends), b) together with the edges of $\gamma$ form a triangulation of $C\cup I$ like in the picture:

enter image description here


1 Answer 1


Yes, a triangulation of $C \cup I$ as you describe it certainly does exist.

The full theorem is that $C \cup I$ is homeomorphic to the closed unit disc; this is known as the Schönflies Theorem. And the closed unit disc is in turn homeomorphic to any convex polygon union its interior.

So, for example, $C \cup I$ is homeomorphic to a regular pentagon union its interior. The latter can be triangulated exactly as you drew it, except with straight lines; and then using the homeomorphism that triangulation can be transported to $C \cup I$.

  • 1
    $\begingroup$ Right, except OP cares about smoothness, so you will need a version of Schoenflies where the map is smooth on the interior. (Which does exist.) I am not sure if OP cares about smoothness at the end-points of the triangulating curves. $\endgroup$ Commented May 27, 2022 at 1:19
  • $\begingroup$ Continuous triangulating curves also work for me. Thanks a lot! $\endgroup$
    – Plemath
    Commented May 27, 2022 at 1:32

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