2Homotopy and the fundamental group

II Algebraic Topology



2.3 The fundamental group
The idea is to take spaces and turn them into groups. We want to try to do
this using paths. We’ve seen that if we want to do this, we should not work
directly with paths, but paths up to homotopy. According to our proposition,
this operation satisfies associativity, inverses and identity. The last issue we have
to resolve is that we can’t actually put two paths together unless they have the
same start and end points.
The idea is to fix one of the points
x
0
in our space, and only think about
loops that start and end at x
0
. So we can always join two paths together.
This tells us that we aren’t going to just think about spaces, but spaces with
basepoints. We’ve now got everything we need.
Definition (Fundamental group). Let
X
be a space and
x
0
X
. The fun-
damental group of
X
(based at
x
0
), denoted
π
1
(
X, x
0
), is the set of homotopy
classes of loops in
X
based at
x
0
(i.e.
γ
(0) =
γ
(1) =
x
0
). The group operations
are defined as follows:
We define an operation by [
γ
0
][
γ
1
] = [
γ
0
· γ
1
]; inverses by [
γ
]
1
= [
γ
1
]; and
the identity as the constant path e = [c
x
0
].
Often, when we write the homotopy classes of paths [
γ
], we just get lazy and
write γ.
Theorem. The fundamental group is a group.
Proof. Immediate from our previous lemmas.
Often in mathematics, after defining a term, we give lots of examples of
it. Unfortunately, it is rather difficult to prove that a space has a non-trivial
fundamental group, until we have developed some relevant machinery. Hence we
will have to wait for a while before we have some concrete examples. Instead,
we will look at some properties of the fundamental group first.
Definition (Based space). A based space is a pair (
X, x
0
) of a space
X
and a
point x
0
X, the basepoint. A map of based spaces
f : (X, x
0
) (Y, y
0
)
is a continuous map
f
:
X Y
such that
f
(
x
0
) =
y
0
. A based homotopy is a
homotopy rel {x
0
}.
Recall that for
π
0
, to every map
f
:
X Y
, we can associate a function
π
0
(f) : π
0
(X) π
0
(Y ). We can do the same for π
1
.
Proposition. To a based map
f : (X, x
0
) (Y, y
0
),
there is an associated function
f
= π
1
(f) : π
1
(X, x
0
) π
1
(Y, y
0
),
defined by [γ] 7→ [f γ]. Moreover, it satisfies
(i) π
1
(f) is a homomorphism of groups.
(ii) If f ' f
0
, then π
1
(f) = π
1
(f
0
).
(iii)
For any maps
(A, a) (B, b) (C, c)
h k
, we have
π
1
(
k h
) =
π
1
(k) π
1
(h).
(iv) π
1
(id
X
) = id
π
1
(X,x
0
)
Proof. Exercise.
In category-theoretic language, we say that π
1
is a functor.
So far so good. However, to define the fundamental group, we had to make a
compromise and pick a basepoint. But we just care about the space. We don’t
want a basepoint! Hence, we should look carefully at what happens when we
change the basepoint, and see if we can live without it.
The first observation is that
π
1
(
X, x
0
) only “sees” the path component of
x
0
,
since all loops based at
x
0
can only live inside the path component of
x
0
. Hence,
the first importance of picking a basepoint is picking a path component.
For all practical purposes, we just assume that
X
is path connected, since if
we weren’t, the fundamental group just describes a particular path component
of the original space.
Now we want to compare fundamental groups with different basepoints.
Suppose we have two basepoints
x
0
and
x
1
. Suppose we have a loop
γ
at
x
0
.
How can we turn this into a loop based at
x
1
? This is easy. We first pick a path
u
:
x
0
x
1
. Then we can produce a new loop at
x
1
by going along
u
1
to
x
0
,
take the path γ, and then return to x
0
by u, i.e. consider u
1
· γ · u.
x
0
x
1
u
Proposition. A path u : x
0
x
1
induces a group isomorphism
u
#
: π
1
(X, x
0
) π
1
(X, x
1
)
by
[γ] 7→ [u
1
· γ · u].
This satisfies
(i) If u ' u
0
, then u
#
= u
0
#
.
(ii) (c
x
0
)
#
= id
π
1
(X,x
0
)
(iii) If v : x
1
x
2
. Then (u · v)
#
= v
#
u
#
.
(iv) If f : X Y with f(x
0
) = y
0
, f(x
1
) = y
1
, then
(f u)
#
f
= f
u
#
: π
1
(X, x
0
) π
1
(Y, y
1
).
A nicer way of writing this is
π
1
(X, x
0
) π
1
(Y, y
0
)
π
1
(X, x
1
) π
1
(Y, y
1
)
f
u
#
(fu)
#
f
The property says that the composition is the same no matter which way
we go from
π
1
(
X, x
0
) to
π
1
(
Y, y
1
). We say that the square is a commutative
diagram. These diagrams will appear all of the time in this course.
(v)
If
x
1
=
x
0
, then
u
#
is an automorphism of
π
1
(
X, x
0
) given by conjugation
by u.
It is important (yet difficult) to get the order of concatenation and composition
right. Path concatenation is from left to right, while function composition is
from right to left.
Proof.
Yet another exercise. Note that (
u
1
)
#
= (
u
#
)
1
, which is why we have
an isomorphism.
The main takeaway is that if
x
0
and
x
1
are in the same path component,
then
π
1
(X, x
0
)
=
π
1
(X, x
1
).
So the basepoint isn’t really too important. However, we have to be careful.
While the two groups are isomorphic, the actual isomorphism depends on which
path
u
:
x
0
x
1
we pick. So there is no natural isomorphism between the two
groups. In particular, we cannot say “let
α π
1
(
X, x
0
). Now let
α
0
be the
corresponding element in π
1
(X, x
1
)”.
We can also see that if (
X, x
0
) and (
Y, y
0
) are based homotopy equivalent,
then
π
1
(
X, x
0
)
=
π
1
(
Y, y
0
). Indeed, if they are homotopy equivalent, then there
are some f : X Y , g : Y X such that
f g ' id
Y
, g f ' id
X
.
So
f
g
= id
π
1
(Y,y
0
)
, g
f
= id
π
1
(X,x
0
)
,
and f
and g
are the isomorphisms we need.
However, can we do this with non-based homotopies? Suppose we have
a space
X
, and a space
Y
, and functions
f, g
:
X Y
, with a homotopy
H : X × I Y from f to g. If this were a based homotopy, we know
f
= g
: π
1
(X, x
0
) π
1
(Y, f(x
0
)).
Now we don’t insist that this homotopy fixes
x
0
, and we could have
f
(
x
0
)
6
=
g
(
x
0
). How can we relate
f
:
π
1
(
X, x
0
)
π
1
(
Y, f
(
x
0
)) and
g
:
π
1
(
X, x
0
)
π
1
(Y, g(x
0
))?
First of all, we need to relate the groups
π
1
(
Y, f
(
x
0
)) and
π
1
(
Y, g
(
x
0
)). To
do so, we need to find a path from
f
(
x
0
) to
g
(
x
0
). To produce this, we can use
the homotopy
H
. We let
u
:
f
(
x
0
)
g
(
x
0
) with
u
=
H
(
x
0
, ·
). Now we have
three maps f
, g
and u
#
. Fortunately, these do fit together well.
x
0
X Y
f(x
0
)
g(x
0
)
f
g
Lemma. The following diagram commutes:
π
1
(Y, f(x
0
))
π
1
(X, x
0
)
π
1
(Y, g(x
0
))
u
#
f
g
In algebra, we say
g
= u
#
f
.
Proof. Suppose we have a loop γ : I X based at x
0
.
We need to check that
g
([γ]) = u
#
f
([γ]).
In other words, we want to show that
g γ ' u
1
· (f γ) · u.
To prove this result, we want to build a homotopy.
Consider the composition:
F : I × I X × I Y.
γ×id
I
H
Our plan is to exhibit two homotopic paths `
+
and `
in I × I such that
F `
+
= g γ, F `
= u
1
· (f γ) · u.
This is in general a good strategy
X
is a complicated and horrible space we
don’t understand. So to construct a homotopy, we make ourselves work in a
much nicer space I × I.
Our `
+
and `
are defined in a rather simple way.
I × I
`
`
+
F
Y
u
1
f
u
g
More precisely,
`
+
is the path
s 7→
(
s,
1), and
`
is the concatenation of the
paths s 7→ (0, 1 s), s 7→ (s, 0) and s 7→ (1, s).
Note that
`
+
and
`
are homotopic as paths. If this is not obvious, we can
manually check the homotopy
L(s, t) = t`
+
(s) + (1 t)`
(s).
This works because I × I is convex. Hence F `
+
'
F L
F `
as paths.
Now we check that the compositions
F `
±
are indeed what we want. We
have
F `
+
(s) = H(γ(s), 1) = g γ(s).
Similarly, we can show that
F `
(s) = u
1
· (f γ) · u(s).
So done.
It is worth looking at this proof hard and truly understand what is going
on, since this is a really good example of how we can construct interesting
homotopies.
With this lemma, we can show that fundamental groups really respect
homotopies.
Theorem. If
f
:
X Y
is a homotopy equivalence, and
x
0
X
, then the
induced map
f
: π
1
(X, x
0
) π
1
(Y, f(x
0
)).
is an isomorphism.
While this seems rather obvious, it is actually non-trivial if we want to do
it from scratch. While we are given a homotopy equivalence, we are given no
guarantee that the homotopy respects our basepoints. So this proof involves
some real work.
Proof.
Let
g
:
Y X
be a homotopy inverse. So
f g '
H
id
Y
and
gf '
H
0
id
X
.
x
0
X Y
f(x
0
)
g f(x
0
)
u
0
f
g
We have no guarantee that
g f
(
x
0
) =
x
0
, but we know that our homotopy
H
0
gives us u
0
= H
0
(x
0
, · ) : x
0
g f(x
0
).
Applying our previous lemma with id
X
for f and g f for g”, we get
u
0
#
(id
X
)
= (g f)
Using the properties of the
operation, we get that
g
f
= u
0
#
.
However, we know that
u
0
#
is an isomorphism. So
f
is injective and
g
is
surjective.
Doing it the other way round with
f g
instead of
g f
, we know that
g
is
injective and f
is surjective. So both of them are isomorphisms.
With this theorem, we can finally be sure that the fundamental group
is a property of the space, without regards to the basepoint (assuming path
connectedness), and is preserved by arbitrary homotopies.
We now use the fundamental group to define several simple notions.
Definition (Simply connected space). A space
X
is simply connected if it is
path connected and π
1
(X, x
0
)
=
1 for some (any) choice of x
0
X.
Example. Clearly, a point
is simply connected since there is only one path on
(the constant path). Hence, any contractible space is simply connected since
it is homotopic to . For example, R
n
is simply connected for all n.
There is a useful characterization of simply connected spaces:
Lemma. A path-connected space
X
is simply connected if and only if for any
x
0
, x
1
X, there exists a unique homotopy class of paths x
0
x
1
.
Proof.
Suppose
X
is simply connected, and let
u, v
:
x
0
x
1
be paths. Now
note that
u · v
1
is a loop based at
x
0
, it is homotopic to the constant path, and
v
1
· v is trivially homotopic to the constant path. So we have
u ' u · v
1
· v ' v.
On the other hand, suppose there is a unique homotopy class of paths
x
0
x
1
for all
x
0
, x
1
X
. Then in particular there is a unique homotopy class of loops
based at x
0
. So π
1
(X, x
0
) is trivial.