If you consider *oriented* simple closed curves on K, then the one that bounds a disk (5.) and the one that goes around once in the "y-direction" are each equivalent to their mirror images; the other three equivalence classes are not. They don't quite all correspond to different elements of the fundamental group of K, since they don't a priori share a basepoint. But they can be shown inequivalent by the use of Z_2 intersection number of homology classes.* The two classes that go once around horizontally each Z_2-intersect themselves once but not each other, so they're distinct. The horizontal class going twice around has 0 Z_2-intersection with each of these, so it's distinct from them. The other two classes are easy to distinguish from the above three classes and from each other using their homology class. To show there are no other classes is trickier and I'm not prepared to explain it here; it requires understanding covering spaces and the fundamental group of K, which has the presentation π_1(K,x_0) = 〈a,b | a^2 b^2〉. —Dan _____ * To get the Z_2 intersection number of two closed loops C and C', represent them by smooth curves that have a finite number of intersection points, at each of which they're not tangent to each other. Then just count them mod 2.
On Saturday/28November/2020, at 4:42 PM, Andy Latto <andy.latto@pobox.com> wrote:
How many different equivalence classes of oriented simple closed curves are there? The fifth of your curve (the one that represents the trivial element of the fundamental group) is equivalent to its reverse, because the Klein bottle is nonorientable. Are any of the others equivalent to their reverses?
Showing that the five curves you exhibit are inequivalent is straightforward, since they all correspond to different elements of the fundamental group. Can you give a hint as to how you would go about showing that these 5 are all that there are?