The topology of the sphere

The sphere is homeomorphic to a cube.



The formulae are
f(x,y,z)
=
(x,y,z)/   ________
x2+y2+z2
 
g(x,y,z)
=
(x,y,z)/max(|x|,|y|,|z|)
If we remove the north pole from the sphere, then the remaining space is homeomorphic to the plane R2 by stereographic projection, as illustrated by the diagram below.



The formulae for the homeomorphisms f:S2\{N}->R2 and g:R2->S2\{N} shown here are
f(x,y,z) = (x/(1-z),y/(1-z))
g(u,v) = (2u,2v,u2+v2-1)/ (u2+v2+1)

It follows that S2\{N} is contractible. The diagram below illustrates a contraction.



Using a similar construction, we see that the open unit disk is homeomorphic to the lower half of the sphere and to the whole plane.



The formulae for the homeomorphisms f:U->R2 and g:R2->U (where U is the open unit disk) are
f(x,y)
=
(x,y)/   _______
1-x2-y2
 
g(u,v)
=
(u,v)/   _______
1+u2+v2
 

If we remove both poles from the sphere, then the remaining space is homeomorphic to the cylinder S1xR

The formulae are
f(x,y,z)
=
(x/   _____
x2+y2
 
,y/   _____
x2+y2
 
,z)
g(x,y,z)
=
(x/   ____
1-z2
 
,y/   ____
1-z2
 
,z)
It is thus homotopy equivalent to S1

The cylinder is also homeomorphic to the plane with the origin removed:

f(x,y,z)
=
(x ez,y ez)
g(u,v)
=
(x/   _____
u2+v2
 
,v/   _____
u2+v2
 
,log(u2+v2)/2)

If instead we remove the equatorial circle from S2, we are left a space homeomorphic to a union of two discs



It is thus homotopy equivalent to S0


Finally, this picture shows that the inclusion map of the equator is homotopic to a constant map.