Every manifold can be cut into cubes

Question

I saw the following statement in my advanced calculus text, which was presented without proof:

If $\bar{D}$ is a compact domain in the plane (that is, closure of an open, connected and bounded subset of $\mathbb{R}^2$) whose boundary consists of piecewise smooth curves, then it is possible to cut $\bar{D}$ into a finite number of sets $C_n$, s.t. each $C_n$ is a smooth image of the closed unit cube $I = \{ (x, y) \in \mathbb{R}^2: 0 \leq x \leq 1, 0 \leq y \leq 1 \}$ and they intersect only at boundaries.

Since the environment of this theorem is Stokes theorem and integration on manifolds (Greens theorem and planar regions in this specific case), I would also expect the condition that the smooth maps of the closed unit cube onto $C_n$ restrict to orientation preserving diffeomorphisms on the open unit cube.

Does anyone know how one would prove this theorem or at least how one would approach such a problem?

Answer 1

The first step is to triangulate the compact domain $\bar{D}$. Triangulation means dividing $\bar{D}$ into a finite number of triangles such that any two triangles intersect only at a common vertex or along a complete common edge. Since $\bar{D}$ has a piecewise smooth boundary, such a triangulation always exists. There are various algorithms to achieve this, and the details can be found in computational geometry literature. The key is that the triangulation respects the boundary of $\bar{D}$, meaning that the boundary of each triangle is either completely inside $\bar{D}$ or forms part of the boundary of $\bar{D}$.

The next step is to show that each triangle can be smoothly mapped from the closed unit cube. This can be done by constructing explicit smooth maps. Consider a triangle in the plane with vertices $v_1$, $v_2$, and $v_3$. A point in the triangle can be represented as a convex combination of these vertices: $p = \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3$, where $\lambda_1 + \lambda_2 + \lambda_3 = 1$ and $\lambda_i \ge 0$. We can map the unit cube $[0, 1]^2$ onto this triangle using a smooth function. One way to do this is to divide the unit square into three regions and map each region smoothly onto a quadrilateral formed by two vertices of the triangle and two points on the medians. These quadrilaterals can then be mapped smoothly onto the triangle. The maps can be constructed in a way that they are orientation-preserving diffeomorphisms on the interior of the cube and the triangle.

Since the triangles in the triangulation only intersect along edges or vertices, the smooth maps from the cube to each triangle can be combined to form a piecewise smooth map from a finite collection of cubes to the domain $\bar{D}$. The condition that the maps restrict to orientation-preserving diffeomorphisms on the open unit cube ensures that the orientation is consistent across the triangles.