Elements of the fundamental group of a space X can be any oriented closed curve that starts and ends at the basepoint, which is just some point chosen for the purpose. As long as X is path-connected, the resulting group π_1(X) will be independent of the choice. The fundamental group of the Klein bottle can be given the presentation π_1(K) = < a, b | a b a^(-1) b = 1 > which when made abelian must become the group H_1(K) = Z + Z_2. Since K is a surface instead of being higher-dimensional, lots of these elements of π_1(K) cannot be represented by *simple* closed curves. But Yes — each of the 5 classes of simple closed curves on K can be given an orientation ... and then as long as each curve contains the basepoint, they can then be thought of as elements of π_1(K) that have been identified to their inverses. *However*, there is a subtlety: The continuous deformation that can make two closed curves in a space X equivalent in π_1(X) must be through curves that all start and end at the same basepoint. This means that *more curves might be equivalent* in the present situation: Any two simple closed curves on K are equivalent (in this question) if there is a continuous map H : S^1 x [0, 1] —> K with one curve being H(( , 0)) : S^1 —> K and the other H(( , 1)) S^1 —> K. —Dan Cris Moore wrote: ----- On Dec 25, 2018, at 2:16 PM, Dan Asimov <dasimov@earthlink.net> wrote: PS For more fun with the Klein bottle, identify the exactly 5 classes of (unoriented) simple closed curves on it that cannot be deformed one to the other. I take it these are the elements of the fundamental group, with inverses identified? -----