Coloring question

Consider any closed curve on the plane that does not repeat any segments, but possibly crossing itself at several points.

How do I prove that the faces are $2$-colourable.

For example:

Help appreciated!

Solution 1:

Take the curve as a simple graph, where the intersections are vertices. Then this graph is obviously planar and allows an Eulerian cycle.

Check, that on planar graphs, Eulerianess implies that its dual graph is bipartite. In the dual graph, the faces become vertices and a bipartite graph is certainly 2-colorable.

Solution 2:

Here is an answer that appeals to intuition:

Let the (sufficiently nice) curve be $\gamma$ and consider the winding number function $f(z)=n(\gamma,z)$. Notice that $f(z)$ is constant in any of the regions of the complex plane defined by $\gamma$. Also, $f(z)$ will change by $1$ or $-1$ when it moves from one region to the next. Now, consider the values of $f(z)$ mod $2$ to see why the checkerboard pattern occurs.

Solution 3:

Let's prove it for a finite union of finite-length closed curves, with finitely many intersections.

Proof by induction on that number of intersections. If 0, you have a finite union of disjoint loops; color each point by how many loops contain it (mod 2). If greater than 0, you may take any intersection. Because there are finitely many, there is some small neighborhood around it, where you may replace X by )(, removing the intersection and reconfiguring the curves (but they're still finite-length closed). Further, the result has fewer intersections, so we may apply induction to color the result. The three regions have to have color A-B-A, so restoring the intersection preserves the 2-colorability.