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Comment by dietrichepp | Hacker News Reader

The four-color theorem is typically formulated for planar figures. It also applies to the surface of anything homeomorphic to a sphere, because a stereographic projection will preserve the four-color property. It does not apply to other shapes -- a simple torus may need seven colors. Adding holes lets you increase the number of required colors without bound.

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Is there a statement for polyhedra similar to four-color theorem?

I could only google a book called "Map Coloring Polyhedra and the Four Color Problem". Unfortunately, it's not available to read online. From what I've found, it seems to me that there is only a handful of proven facts about colorings of some polyhedra.

[ADD]: four color theorem works for every planar graph and it looks like one can make such graph corresponding to any convex polyhedron, so it seems that original OP statement holds.

Just make a hole in the center of one side of the polyhedra. Then stretch out the hole, flattening the polyhedra in the process. Eventually you get a planar map, while retaining the same topology as the original polyhedra.

Or, in math-speak, the polyhedron is homeomorphic to the sphere, and a stereographic projection of the sphere will preserve the four-color property while projecting the sphere to a plane.

I first thought, "That's exactly what I said," and then I realized that "homeomorphic" isn't exactly a household word.

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