I have stumbled onto the following fact and I am quite helpless in seeing why this is true (although I can agree intuitively).
Let $M$ be a surface of genus zero (open or closed, with or without boundary). The claim is now that there can't be two closed, smooth and transversal loops $\alpha, \beta : S^1 \rightarrow M$ with non-zero intersection number. (One loop is required to be simple closed. I am however not sure if this does really matter.)
The intersection number is meant to be the oriented intersection number of transversal loops (cf. chapter 3 of Differential topology by Guillemin and Pollack). It basically adds up the signs of each of the intersections. The sign of an intersection is set to $+1$ if loop $\alpha$ crosses loop $\beta$ from left to right and to $-1$ if $\alpha$ crosses $\beta$ from right to left. Transversality and smoothness of the loops is required to define this intersection number.
Intuitively it it clear that I can't draw two closed loops on a genus zero surface where $\alpha$ enters the right side of $\beta$ more often than it enters the left side.
Does anyone know about a reference where I can read up on this?
Many thanks! :)
If $M$ is $\ne S^2$ we can realize it as a "domain" in the plane ${\mathbb R}^2$. Both loops then appear as smooth closed curves in the plane, intersecting transversally. Since $\alpha$ is supposed simple, by Jordan's curve theorem it separates the plane into two open regions, one of them, call it $\Omega$, compact, and $\alpha=\pm\partial\Omega$. The sign here depends on the orientation of $\alpha$, and is $+1$ when $\Omega$ lies to the left of $\alpha$. It is obvious that $\beta$ enters $\Omega$ as often as it leaves $\Omega$, and this implies that the intersection number $\alpha\wedge\beta=0$.
When $M=S^2$ remove a point $p\in M$ lying neither on $\alpha$ nor on $\beta$, and apply the foregoing argument.