2Homotopy and the fundamental group
II Algebraic Topology
2.0 Motivation
Recall that we wanted to prove that
R
n
6
∼
=
R
m
for
n 6
=
m
. Let’s first do the
simple case, where m = 1, n = 2. We want to show that R 6
∼
=
R
2
.
This is not hard. We know that
R
is a line, while
R
2
is a plane. Let’s try to
remove a point from each of them. If we remove a point from
R
, the space stops
being path connected. However, removing a point does not have this effect on
R
2
. Since being path connected is a topological property, we have now showed
that R and R
2
are not homeomorphic.
Unfortunately, this does not extend very far. We cannot use this to show
that R
2
and R
3
are not homeomorphic. What else can we do?
Notice that when we remove a point from
R
2
, sure it is still connected, but
something has changed.
Consider a circle containing the origin in
R
2
\ {
0
}
. If the origin were there,
we can keep shrinking the circle down until it becomes a point. However, we
cannot do this if the origin is missing.
The strategy now is to exploit the fact that
R
2
\ {
0
}
has circles which cannot be
deformed to points.