Just to be clear, the puzzle 2. that I proposed was only about subsets of K that are simple closed curves (without distinguishing anything about parametrizations). From that point of view the punctured plane has just two inequivalent families of simple closed curves. —Dan
On Saturday/28November/2020, at 4:35 PM, Andy Latto <andy.latto@pobox.com> wrote:
They do not form a group, because of the requirement that these be *simple* closed curves, while the fundamental group looks at all maps from the circle to the space in question. For example, the fundamental group of the punctured plane is Z, but the number of inequivalent simple closed curves is 4; clockwise and counterclockwise circles that don't go around the removed point, and clockwise and counterclockwise circles that do go around the removed point. Any map of the circle to the punctured plane that goes twice around the missing point must self-intersect.
Note that the requirement that the homotopy consist of simple closed curves means that the small clockwise and counterclockwise circle, both trivial as elements of the fundamental group, are not equivalent.