Suppose we have a regular map - so that each vertex has the same order - with
faces on a surface with Euler characteristic
Suppose also that the surface can be coloured with at most
colours where
(1)
Then at least one face must have a boundary with less than
edges.
Proof:
(2)
Since the map is regular every vertex has the same order (number of edges entering or exiting the vertex) so if the order of each vertex is
then number of edges is![]()
We must have k>=3 for a regular map so 2E >= 3
Rearrange (2) to give![]()
Now use (2) to give![]()
If
is the number of faces with
edges, then
and
so
![]()
Hence at least one face must have a boundary consisting of less than six edges.