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Oct 13, 2010 at 14:26 vote accept D. Savitt
Oct 13, 2010 at 8:36 comment added Greg Kuperberg In the compact case, every continuous bijection is a homeomorphism.
Oct 13, 2010 at 8:32 answer added Greg Kuperberg timeline score: 23
Oct 13, 2010 at 8:09 comment added D. Savitt (and of course a smooth homeomorphism would be even better, but I'd be happy with just a smooth bijection)
Oct 13, 2010 at 8:07 comment added D. Savitt Hi Theo, yes, "manifold" with (what I think of as) the default meaning -- second countable, without boundary, etc....
Oct 13, 2010 at 8:01 comment added Theo Johnson-Freyd Alternately, maybe you don't want to rest that much on the word "manifold", and instead mean to ask for smooth homeomorphisms that are not diffeomorphisms?
Oct 13, 2010 at 8:00 comment added Theo Johnson-Freyd I guess the word "manifold" forbids me from just rolling the half-open interval around the circle, and perhaps also from sticking continuum-many discrete points on the line....
Oct 13, 2010 at 6:15 history asked D. Savitt CC BY-SA 2.5