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#1 2010-10-29 00:39:29

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Homeomorphism of Euclidean plane to sphere with point removed

It is clear that
is not homeomorphic to a sphere (by which I mean the surface only, not the interior). The latter is compact whereas the former isn’t. However if you remove a single point from the sphere, then the two spaces are homeomorphic.

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

[align=center]

[/align]
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. tongue

Last edited by JaneFairfax (2010-10-29 01:50:05)


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#2 2011-01-14 23:58:13

scientia
Member
Registered: 2009-11-13
Posts: 222

Re: Homeomorphism of Euclidean plane to sphere with point removed

The surface of the sphere is homeomorphic to the extended complex plane
.

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#3 2011-01-15 07:58:17

coprime
Member
Registered: 2011-01-15
Posts: 1

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 (2011-01-15 08:14:11)

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