Don't the path components form a group, in general, with the operation being "component containing the composition of representative homeomorphisms"? So it's incurious to ask anything less than "what group?" On Thu, Dec 20, 2018, 8:32 PM Dan Asimov <dasimov@earthlink.net wrote:
Here's a puzzle I think I know the answer to, but I don't have a proof:
Let K denote the Klein bottle, the ideal surface you get when you identify the top and bottom edges of the unit square [0,1] x [0,1] normally, by
(x,0) ~ (x,1),
but identify the left and right edges by a flip:
(0,y) ~ (1,1-y).
Puzzle: ----------------------------------------------------------------------- Consider the space Homeo(K) of self-homeomorphisms of the Klein bottle. I.e., Homeo(K) consists of all continuous bijections
h : K —> K
having a continuous inverse.
Two self-homeomorphisms h_0, h_1 of K are *in the same path component* of Homeo(K) if there is a continuous *family*
{h(t) | 0 <= t <= 1}
of homeomorphisms h(t) in Homeo(K) such that h(e) = h_e for e = 0, 1.
The continuity of this family just amounts to there being a continuous map
H : K x [0,1] —> K
such that the restriction of H to any time-slice K x {t}:
H | K x {t} —> K
is the homeomorphism h(t) : K —> K.
QUESTION: ————————— How many path components does Homeo(K) have? -----------------------------------------------------------------------
—Dan
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