If one could prove that for every 4-chromatic planar graph $X$ every color identical pair in $X$ is separated by a cycle, would that be a proof of the 4-color theorem?

Explanation:

A pair of vertices $\{u, v\}$ in a $k$-chromatic graph $G$ is a **color identical pair** iff the color of $u$ equals the color of $v$ in every $k$-coloring of $G$, and the supergraph of $G$ where $u$ and $v$ are adjacent, $(V(G), E(G) \cup \{uv\})$, then requires $k + 1$ colors.

For every $k$-critical graph $X$ and for every edge $xy$ in $X$, there is a $(k - 1)$-chromatic graph $Y = (V(X), E(X) \setminus \{xy\})$ where $\{x, y\}$ is a color identical pair (because, if there were a $(k - 1)$-coloring of $Y$ where $x$ and $y$ had different colors, that would also have been a $(k - 1)$-coloring of $X$).

E.g. removing an edge from $K_{5}$ gives a 4-chromatic graph of two tetrahedrons whose intersection is a triangle, and the two vertices that are not in the intersection are a color identical pair. On the plane these vertices are separated by the triangle.

x,y}, and if in each subgraphxandyare separated by a cycle there is no planar 5-critical graph. $\endgroup$ – Asbjørn Brændeland Mar 11 '14 at 22:09