Cassidy Curtis sez: “I hate to break it to your readers, but that piece of playground equipment is not really a Moebius strip. It has two twists, not one, which means it’s an orientable surface (having two unique sides), topologically equivalent to an ordinary circular strip. A kid on one side of the strip could crawl around forever and not meet any kids on the other side. Real Moebius strips have an odd number of twists, and thus have only one side.”
Kevin McCarty sez:That’s not quite right either. A two-twist strip is not topologically
equivalent to a no-twist strip, nor to a one-twist strip (the Moebius
strip). They’re all topologically inequivalent. The two circular
edges of the 2-twist strip are linked, while with the 0-twist strip
they’re not. For all integers n, the n-twist strips are inequivalent
to each other. The ones with even number of twists are orientable
with two circular edges with varying linking number, and the
odd-numbered twists are non-orientable with a single circular edge
that links (knots) itself a varying number of times. Welcome to the
wonderful world of homotopy equivalence classes.
Tim sez:
I should add that Kevin McCarty’s comment is only true for
strips nested in euclidean space. Without such an embedding, there’s
no way to distinguish strips differing in an even number of twists.
See: on homotopy, where the space Y is in fact euclidean 3-space. Without
such a reference space we can only say whether our strips are
homeomorphic or not, which they indeed are whenever they differ
in an even number of twists.
I believe that any confusion here originates from the use of
homotopy in knot theory where an ignorance of the space in which
the knot is embedded would result in every knot being equivalent to
the trivial knot.