# Triangulation of interior of piecewise smooth, simple, closed plane curve

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:

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 $$C \cup I$$.

• 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. May 27, 2022 at 1:19
• Continuous triangulating curves also work for me. Thanks a lot! May 27, 2022 at 1:32