1Homotopy
III Algebraic Topology
1 Homotopy
In this course, the word “map” will always mean “continuous function”.
In topology, we study spaces up to “continuous deformation”. Famously, a
coffee mug can be continuously deformed into a doughnut, and thus they are
considered to be topologically the same. Now we also talk about maps between
topological spaces. So a natural question is if it makes sense to talk about the
continuous deformations of maps. It turns out it does, and the definition is
sort-of the obvious one:
Definition
(Homotopy)
.
Let
X, Y
be topological spaces. A homotopy between
f
0
, f
1
:
X → Y
is a map
F
: [0
,
1]
× X → Y
such that
F
(0
, x
) =
f
0
(
x
) and
F
(1
, x
) =
f
1
(
x
). If such an
F
exists, we say
f
0
is homotopic to
f
1
, and write
f
0
' f
1
.
This ' defines an equivalence relation on the set of maps from X to Y .
Just as concepts in topology are invariant under homeomorphisms, it turns
out the theories we develop in algebraic topology are invariant under homotopies,
i.e. homotopic maps “do the same thing”. This will be made more precise later.
Under this premise, if we view homotopic functions as “the same”, then we
have to enlarge our notion of isomorphism to take this into account. To do
so, we just write down the usual definition of isomorphism, but with equality
replaced with homotopy.
Definition
(Homotopy equivalence)
.
A map
f
:
X → Y
is a homotopy equiva-
lence if there is some
g
:
Y → X
such that
f ◦ g ' id
Y
and
g ◦ f ' id
X
. We call
g a homotopy inverse to f.
As always, homotopy equivalence is an equivalence relation. This relies on
the following straightforward property of homotopy:
Proposition.
If
f
0
' f
1
:
X → Y
and
g
0
' g
1
:
Y → Z
, then
g
0
◦ f
0
' g
1
◦ f
1
:
X → Z.
X Y Z
f
0
f
1
g
0
g
1
Example
(Stupid example)
.
If
f
:
X → Y
is a homeomorphism, then it is a
homotopy equivalence — we take the actual inverse for the homotopy inverse,
since equal functions are homotopic.
Example
(Interesting example)
.
Let
i
:
{
0
} → R
n
be the inclusion map. To
show this is a homotopy equivalence, we have to find a homotopy inverse.
Fortunately, there is only one map
R
n
→ {
0
}
, namely the constant function 0.
We call this
r
:
R
n
→ {
0
}
. The composition
r ◦ i
:
{
0
} → {
0
}
is exactly the
identity. So this is good.
In the other direction, the map
i ◦ r
:
R
n
→ R
n
sends everything to 0. We
need to produce a homotopy to the identity. We let F : [0, 1] × R
n
→ R
n
be
F (t, v) = tv.
We have
F
(0
, v
) = 0 and
F
(1
, v
) =
v
. So this is indeed a homotopy from
i ◦ r
to id
R
n
.
So from the point of view of homotopy, the one-point space
{
0
}
is the same
as R
n
! So dimension, or even cardinality, is now a meaningless concept.
Example
(Also interesting example)
.
Let
S
n
⊆ R
n+1
be the unit sphere, and
i
:
S
n
→ R
n+1
\ {
0
}
. We show that this is a homotopy equivalence. We define
r : R
n+1
\ {0} → S
n
by
r(v) =
v
kvk
.
Again, we have
r ◦ i
=
id
S
n
. In the other direction, we need to construct a path
from each v to
v
kvk
in a continuous way. We could do so by
H : [0, 1] × (R
n+1
\ {0}) → R
n+1
\ {0}
(t, v) 7→ (1 − t)v + t
v
kvk
.
We can easily check that this is a homotopy from id
R
n+1
\{0}
to i ◦ r.
Again, homotopy equivalence allowed us to squash down one dimension of
R
n+1
\ {
0
}
to get
S
n
. However, there is one thing we haven’t gotten rid of —
the hole. It turns out what is preserved by homotopy equivalence is exactly the
holes.
Now we might ask ourselves — can we distinguish holes of “different dimen-
sion”? For
n 6
=
m
, is
S
n
homotopy equivalent to
S
m
? If we try hard to construct
homotopies, we will always fail, and then we would start to think that maybe
they aren’t homotopy equivalent. However, at this point, we do not have any
tools we can use the prove this.
The solution is algebraic topology. The idea is that we assign algebraic
objects to each topological space in a way that is homotopy-invariant. We then
come up with tools to compute these algebraic invariants. Then if two spaces
are assigned different algebraic objects, then we know they cannot be homotopy
equivalent.
What is this good for? At this point, you might not be convinced that it
is a good idea to consider spaces up to homotopy equivalence only. However,
algebraic topology can also help us show that spaces are not homeomorphic.
After all, homeomorphic spaces are also homotopy equivalent. For example,
suppose we want to show that if
R
n
∼
=
R
m
, then
n
=
m
. Algebraic topology
doesn’t help us directly, because both
R
n
and
R
m
are homotopy equivalent to a
point. Instead, we do the slightly sneaky thing of removing a point. If
R
n
∼
=
R
m
,
then it must also be the case that
R
n
\{
0
}
∼
=
R
m
\{
0
}
. Since these are homotopy
equivalent to
S
n−1
and
S
m−1
, this implies that
S
n−1
' S
m−1
are homotopy
equivalent. By algebraic topology, we will show that that this can only happen
if
m
=
n
. So indeed we can recover the notion of dimension using algebraic
topology!