>Kempe's method was accepted as proof of the FCT until >Heawood created his counter-example. > >Suppose that there was a simple way to 4-color Heawood's graph >without worrying about the problem of "tangled chains"? Would >that be sufficient for a proof?
Heawood's graph is a counterexample to Kempe's proposed coloring strategy. According to Kempe's claimed proof, Heawood's graph can be 4-colored by a specific strategy used in the proof. Heawood identifies a specific planar graph which, if one follows Kempe's coloring strategy, then two adjacent vertices will be forced to have the same color. The result is to show that Kempe's proof is invalid as a proof of 4-colorability for planar graphs.
However, Heawood's graph _is_ a planar graph, hence it _can_ be 4-colored (and probably easily so). So if you show a 4-coloring of Heawood's graph, that reveals nothing we don't already know.