Intersection of Simple Closed Curves in $\mathbb{R}P^2$

Question

Given a (smooth) simple closed curve $C$ in $\mathbb{R}P^2$, we either have that $\mathbb{R}P^2 \setminus C$ is the disjoint union of a disk and Möbius strip, or that $\mathbb{R}P^2 \setminus C$ is a single disk. In the first case, $C$ must have an even number of (transverse) intersections with any other curve $D$ because $D$ must enter and exit the disk the same number of times. How can I show that two curves of the second type must intersect an odd number of times? It's tempting to talk about a "left" and "right" side of the curve as the linked picture suggests, but in reality there is only one component.

"Sides" of the curve

Answer 1

If I have two transversely intersecting n-dimensional submanifolds $A,B$ of $M^{2n}$, then then (everything here is mod 2) the parity of the number of geometric intersections is given by the cup product $P([A])P([B])$ evaluated on the fundamental class $[M]$, where $P$ denotes the Poincare isomorphism and $[A],[B],[M]$ are the fundamental classes.

It is easy enough to argue (either via standard surgery arguments or ad hoc methods) that a curve of the second type represents the nontrivial loop in $\pi_1(\mathbb{R}P^2)$, hence under the mod 2 Poincare Isomorphism it represents the nontrivial class in $H^1 (\mathbb{R}P^2)$. Since we know the ring structure of $\mathbb{R}P^2$ is truncated polynomial on this class, we deduce that the number of intersections is $1$ mod 2, as desired.