When you divide the real projective plane into two subsets, does it always have exactly one non-orientable component?

Question

Let's say you divide the real projective plane into two subsets, are exactly one these subsets non-orientable?

In particular, we will require that each subset $S$ is "nice" in the sense their common boundary consists of a finite number of non-intersecting polygons (in particular, this excludes 1 dimensional regions).

Note the answer when divide it into three subsets is "no". You can divide the real projective plane into three orientable components.

(The motivation for this question is that, if the answer to this question is yes, you can play a variant of the Hex board game where you win if you form a non-orientable subset of the projective plane.)

EDIT: There actually is a game (called Projex) based on this concept. The website seems to imply that the answer to this question is yes. It is based on a discrete version of this problem though, and includes no proof.

Answer 1

I, too, have thought about (and played) this $\mathbb{R}\mathrm{P}^2$ Hex game, and came up with the following sort of "hands-on" approach.

Start with a decomposition into two closed submanifolds of $\mathbb{R}\mathrm{P}^2$ with $1$-manifold shared boundary, and call the regions (connected components) "red" or "blue" based on which of the two submanifolds it is from. Suppose one of the regions is a disk. Since color changes when crossing a boundary, one can see that the regular neighborhood of each $1$-manifold in the intersection is orientable. So, switching the color of the disk does not change which of the two submanifolds is orientable or non-orientable.

Hence, we may assume none of the regions are disks. In particular, every boundary curve is essential (that is, the lift to $S^2$ is connected). There can only be one by the Jordan curve theorem, since a second such curve contains a path between the two disks the first such curve bounds in lifts to $S^2$. If there is at most one boundary curve and it is essential, then there are no boundary curves at all, since both sides are the same side, so the same color.

Thus, one of the two submanifolds is empty (hence orientable), and the other is all of $\mathbb{R}\mathrm{P}^2$ (hence non-orientable).

Looking through the argument, one can see that the submanifolds are disjoint unions of punctured spheres (with at least one puncture) and a projective plane (with any number of punctures), with at most one such punctured projective plane between them.