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)/
_______ Ö1-x2-y2
g(u,v)
=
(u,v)/
_______ Ö1+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/
_____ Öx2+y2
,y/
_____ Öx2+y2
,z)
g(x,y,z)
=
(x/
____ Ö1-z2
,y/
____ Ö1-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/
_____ Öu2+v2
,v/
_____ Öu2+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.