The way I was trying to do things, I think it works easier going from the outside in, though. (Uh oh, this is that “worrying about embedding” problem, isn’t it.)
But on a graph, don’t the outermost regions of the map correspond to those nodes on the graph from which you can draw infinite rays that do not cross any of the lines on the graph?
Maybe not, because of something I just realized I don’t know: Is the graph of a triangle plus a node over the triangle and connected to just one other node considered equivalent, for graph-theoretical purposes, to a graph of a triangle plus a node in the middle of the triangle and connected to just one other node?
Yes, they’re graph-theoretically equivalent; worrying about differences between them is worrying about embedding, which is essentially irrelevant to the four-coloring problem.
(A graph, for our purposes, is just a set of nodes along with an irreflexive symmetric relation upon them (adjacency). Two graphs are graph-theoretically equivalent if there’s some correspondence between their nodes which preserves the adjacency structure. The graphs we’re concerned with satisfy a further constraint (planarity) which can be expressed purely in terms of this adjacency structure, and which thus respects graph-theoretic equivalence.)
Okay, I’ve got it now. And on further thought, I was wrong to think the approach I was taking is easier to deal with going “from the outside in.”
(I know I’m being mysterious here since I haven’t said what that approach is, but I don’t want to burden people with further details of what is basically me just messing around).
Alright, time to do something important for a while.
Incidentally, it suddenly occurs to me that I don’t know if the four-color theorem applies to maps with infinitely many regions as well… I imagine it does. Is it easy to see that it must generalize in this way?
See, although I don’t know what your approach was, I think one of the advantages of the graph-theoretic perspective, ignoring the details of the embedding, is that this may well have then been clear from the start, via the fact that the node representing the outside has no special graph-theoretic properties.