2Homotopy and the fundamental group
II Algebraic Topology
2.1 Homotopy
We have just talked about the notion of “deforming” circles to a point. We can
think of a circle in
X
as a map
S
1
→ X
, and we want to “deform” this map to
a point. This process of deformation is known as homotopy. Here we are going
to use the interval [0, 1] ⊆ R a lot, and we will just call it I.
Notation.
I = [0, 1] ⊆ R.
Definition (Homotopy). Let
f, g
:
X → Y
be maps. A homotopy from
f
to
g
is a map
H : X × I → Y
such that
H(x, 0) = f (x), H(x, 1) = g(x).
We think of the interval
I
as time. For each time
t
,
H
(
· , t
) defines a map
X → Y
.
So we want to start from f, move with time, and eventually reach g.
If such an
H
exists, we say
f
is homotopic to
g
, and write
f ' g
. If we want
to make it explicit that the homotopy is H, we write f '
H
g.
As mentioned at the beginning, by calling
H
a map, we are requiring it to
be continuous.
Sometimes, we don’t want a general homotopy. We might want to make sure
that when we are deforming from a path
f
to
g
, the end points of the path don’t
move. In general, we have
Definition (Homotopy
rel
A). We say
f
is homotopic to
g rel A
, written
f ' g rel A, if for all a ∈ A ⊆ X, we have
H(a, t) = f (a) = g(a).
This notion will find itself useful later, but we don’t have to pay too much
attention to this yet.
Our notation suggests that homotopy is an equivalence relation. Indeed, we
have
Proposition. For spaces
X, Y
, and
A ⊆ X
, the “homotopic
rel A
” relation is
an equivalence relation. In particular, when
A
=
∅
, homotopy is an equivalence
relation.
Proof.
(i) Reflexivity: f ' f since H(x, t) = f (x) is a homotopy.
(ii)
Symmetry: if
H
(
x, t
) is a homotopy from
f
to
g
, then
H
(
x,
1
− t
) is a
homotopy from g to f .
(iii)
Transitivity: Suppose
f, g, h
:
X → Y
and
f '
H
g rel A
,
g '
H
0
h rel A
.
We want to show that
f ' h rel A
. The idea is to “glue” the two maps
together.
We know how to continuously deform
f
to
g
, and from
g
to
h
. So we just
do these one after another. We define H
00
: X × I → Y by
H
00
(x, t) =
(
H(x, 2t) 0 ≤ t ≤
1
2
H
0
(x, 2t − 1)
1
2
≤ t ≤ 1
This is well-defined since
H
(
x,
1) =
g
(
x
) =
H
0
(
x,
0). This is also continuous
by the gluing lemma. It is easy to check that
H
00
is a homotopy
rel A
.
We now have a notion of equivalence of maps — two maps are equivalent if
they are homotopic. We can extend the notion of homotopy to spaces as well.
Recall that when we defined homeomorphism, we required that there be some
f, g such that f ◦ g = id, g ◦ f = id. Here, we replace equality by homotopy.
Definition (Homotopy equivalence). A map
f
:
X → Y
is a homotopy equiva-
lence if there exists a
g
:
Y → X
such that
f ◦ g ' id
Y
and
g ◦ f ' id
X
. We call
g a homotopy inverse for f .
If a homotopy equivalence
f
:
X → Y
exists, we say that
X
and
Y
are
homotopy equivalent and write X ' Y .
We are soon going to prove that this is indeed an equivalence relation on
spaces, but we first look at some examples of homotopy equivalent spaces. Clearly,
homeomorphic spaces are homotopy equivalent. However, we will see that we
can do much more “violent” things to a space and still be homotopy equivalent.
Example. Let
X
=
S
1
,
Y
=
R
2
\ {
0
}
. We have a natural inclusion map
i
:
X → Y
. To obtain a map
Y → X
, we can project each point onto the circle.
r(y)
y
In particular, we define r : Y → X by
r(y) =
y
kyk
.
We immediately have
r ◦ i
=
id
X
. We will now see that
i ◦ r ' id
Y
. The
composition
i ◦ r
first projects each object into
S
1
, and then includes it back
into
R
2
\ {
0
}
. So this is just the projection map. We can define a homotopy
H : Y × I → Y by
H(y, t) =
y
t + (1 − t)kyk
.
This is continuous, and H( · , 0) = i ◦ r, H( · , 1) = id
Y
.
As we have said, homotopy equivalence can do really “violent” things to a
space. We started with a 2-dimensional
R
2
\ {
0
}
space, and squashed it into a
one-dimensional sphere.
Hence we see that homotopy equivalence doesn’t care about dimensions.
Dimensions seem to be a rather fundamental thing in geometry, and we are
discarding it here. So what is left? What does homotopy equivalence preserve?
While
S
1
and
R
2
\{
0
}
seem rather different, they have something in common —
they both have a “hole”. We will later see that this is what homotopy equivalence
preserves.
Example. Let
Y
=
R
n
,
X
=
{
0
}
=
∗
. Let
X → Y
be the inclusion map, and
r
:
Y → X
be the unique map that sends everything to
{
0
}
. Again, we have
r ◦ i = id
X
. We can also obtain a homotopy from i ◦ r to id
Y
by
H(y, t) = ty.
Again, from the point of view of homotopy theory,
R
n
is just the same as a
point! You might think that this is something crazy to do — we have just given
up a lot of structure of topological spaces. However, by giving up these structures,
it is easier to focus on what we really want to care about — holes. For example,
it is often much easier to argue about the one-point space
∗
than the whole of
R
2
! By studying properties that are preserved by homotopy equivalence, and
not just homeomorphism, we can simplify our problems by reducing complicated
spaces to simpler ones via homotopy equivalence.
In general things homotopy equivalent to a single point are known as con-
tractible spaces.
Notation. ∗ denotes the one-point space {0}.
Definition (Contractible space). If X ' ∗, then X is contractible.
We now show that homotopy equivalence of spaces is an equivalence relation.
To do this, we first need a lemma.
Lemma. Consider the spaces and arrows
X Y Z
f
0
f
1
g
0
g
1
If f
0
'
H
f
1
and g
0
'
H
0
g
1
, then g
0
◦ f
0
' g
1
◦ f
1
.
Proof.
We will show that
g
0
◦ f
0
' g
0
◦ f
1
' g
1
◦ f
1
. Then we are done since
homotopy between maps is an equivalence relation. So we need to write down
two homotopies.
(i) Consider the following composition:
X × I Y Z
H
g
0
It is easy to check that this is the first homotopy we need to show
g
0
◦ f
0
'
g
0
◦ f
1
.
(ii) The following composition is a homotopy from g
0
◦ f
1
to g
1
◦ f
1
:
X × I Y × I Z
f
1
×id
I
H
0
Proposition. Homotopy equivalence of spaces is an equivalence relation.
Proof.
Symmetry and reflexivity are trivial. To show transitivity, let
f
:
X → Y
and
h
:
Y → Z
be homotopy equivalences, and
g
:
Y → X
and
k
:
Z → Y
be their homotopy inverses. We will show that
h ◦ f
:
X → Z
is a homotopy
equivalence with homotopy inverse g ◦ k. We have
(h ◦ f ) ◦ (g ◦ k) = h ◦ (f ◦ g) ◦ k ' h ◦ id
Y
◦ k = h ◦ k ' id
Z
.
Similarly,
(g ◦ k) ◦ (h ◦ f ) = g ◦ (k ◦ h) ◦ f ' g ◦ id
Y
◦ f = g ◦ f ' id
X
.
So done.
Definition (Retraction). Let
A ⊆ X
be a subspace. A retraction
r
:
X → A
is a map such that
r ◦ i
=
id
A
, where
i
:
A → X
is the inclusion. If such an
r
exists, we call A a retract of X.
This map sends everything in
X
to
A
without moving things in
A
. Roughly
speaking, if such a retraction exists, then A is no more complicated than X.
Definition (Deformation retraction). The retraction
r
is a deformation retrac-
tion if
i ◦ r ' id
X
. A deformation retraction is strong if we require this homotopy
to be a homotopy rel A.
Roughly, this says that A is as complicated as X.
Example. Take
X
any space, and
A
=
{x} ⊆ X
. Then the constant map
r
:
X → A
is a retraction. If
X
is contractible, then
A
is a deformation retract
of X.