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#1 2010-10-29 23:39:29
- JaneFairfax
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Homeomorphism of Euclidean plane to sphere with point removed
Let us take the unit sphere centred at , given by , and let us remove the north pole . Call this “pointless” sphere . Then is the disjoint union of circles formed by the intersection of with the plane as t varies from 0 to 2 (including 0 but not 2).
Now consider itself. This is the disjoint union of origin-centred circles of all possible non-negative radii (counting as a circle of radius 0). Let be any continuous bijection (e.g. or ).
Then if we define by and for
where , we have a homeomorphism!
I was thinking about this last night. Thinking about math problems is a great way to pass the time when you’re having insomnia.
Last edited by JaneFairfax (2010-10-30 00:50:05)
#2 2011-01-15 22:58:13
- scientia
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Re: Homeomorphism of Euclidean plane to sphere with point removed
#3 2011-01-16 06:58:17
- coprime
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Re: Homeomorphism of Euclidean plane to sphere with point removed
since homeomorphism is defined by continuity but does not require smoothness we can map the punctured sphere to sharpened pencil shape: a truncated half-cone with apex at the south pole intersecting the sphere at the equator, glued to a half infinite cylinder. then the cylinder is easily mapped to the remaining (infinite) portion of the cone. finally flatten out the cone.
use the unit sphere centred at the origin:
for the map of lower hemisphere to truncated half-cone we send
(x,y,z) to (px,py,z) where p=√ (1-z²)
for the map of the upper punctuated hemisphere to the half- cylinder we send
(x,y,z) to (x/r,y/r,zq/r) where r=√ (x²+y²) and q=√(1-r²)
to map the half-cyclinder to the remainder of the half-cone
send (x,y,z) to (-ipx,-ipy,z) where i=√(-1)
to flatten the half-cone send (x,y,z) to (x,y)
(NB the use of imaginaries is merely a notational convenience)
Last edited by coprime (2011-01-16 07:14:11)