If we remove the north pole from the sphere, then the remaining
space is homeomorphic to the plane R^{2} by
stereographic projection, as illustrated by the diagram below.
The formulae for the homeomorphisms
f:S^{2}\{N}->R^{2} and
g:R^{2}->S^{2}\{N} shown here are
f(x,y,z) =
(x/(1-z),y/(1-z))
g(u,v) =
(2u,2v,u^{2}+v^{2}-1)/
(u^{2}+v^{2}+1)
It follows that S^{2}\{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->R^{2} and
g:R^{2}->U (where U is the open unit disk) are
f(x,y)
=
(x,y)/
_______ Ö1-x^{2}-y^{2}
g(u,v)
=
(u,v)/
_______ Ö1+u^{2}+v^{2}
If we remove both poles from the sphere, then the remaining space is
homeomorphic to the cylinder S^{1}xR
The formulae are
f(x,y,z)
=
(x/
_____ Öx^{2}+y^{2}
,y/
_____ Öx^{2}+y^{2}
,z)
g(x,y,z)
=
(x/
____ Ö1-z^{2}
,y/
____ Ö1-z^{2}
,z)
It is thus homotopy equivalent to S^{1}
The cylinder is also homeomorphic to the plane with the origin
removed:
f(x,y,z)
=
(x e^{z},y e^{z})
g(u,v)
=
(x/
_____ Öu^{2}+v^{2}
,v/
_____ Öu^{2}+v^{2}
,log(u^{2}+v^{2})/2)
If instead we remove the equatorial circle from S^{2}, we
are left a space homeomorphic to a union of two discs
It is thus homotopy equivalent to S^{0}
Finally, this picture shows that the inclusion map of the equator is
homotopic to a constant map.