I'm reading Pearls in Graph Theory by Nora Hartsfield and Gerhard Ringel, and the proof of theorem 8.2.5 confused me:
Theorem 8.2.5. If in a normal map $M$, the vertices can be labeled by black and white so that around each country in $M$ the number of black vertices minus the number of white vertices is a multiple of three, then the edges of $M$ can be properly colored by three colors.
Proof. Given a normal map $M$ and a labeling of the vertices by black and white with the above property, we construct the normal $M’$ by blowing up every white vertex into a triangle with three black vertices. So we have a map $M'$ such that all the vertices are black, and all the countries are triangles, hexagons, nonagons, etc. We now color the edges of $M'$ by three colors as follows. We begin at any country and walk counterclockwise around the inside of the country and color the edges by $a, b, c, a, b, c, ...$ as we come to them. The number of edges around each country is a multiple of three, so the coloring is proper so far. Now we color a region adjoining the first region by the same recipe; one edge will already be colored, so the coloring is determined. To see that the coloring is proper, consider Figure 8.2.10.
Now we return to the map $M$ from $M′$. As in Figure 8.2.8, when we restore a white vertex, we shrink a triangle with three black vertices, and we keep the coloring of the remaining edges. This is a proper coloring, since we shrink a triangle with edges colored $a$, $b$, and $c$, and the edges remaining around the white vertex must be colored $a$, $b$, and $c$.
Here normal means connected, bridgeless, planar multigraph that is cubic.
I’m lost on the bold part -- is it guaranteed that the coloring will always be proper?
Consider a country surrounded by others, if coloring of the surrounding countries are determined, then for the country in the centre, will the coloring be automatically proper, i.e. in the good $a$,$b$,$c$ sequence?
I don't see this straightforward. Pls enlighten me.