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 consideritself. 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 defineby and for
I was thinking about this last night. Thinking about math problems is a great way to pass the time when youre having insomnia.
Last edited by JaneFairfax (2010-10-29 01:50:05)
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 (2011-01-15 08:14:11)