2Homotopy and the fundamental group
II Algebraic Topology
2.2 Paths
Definition (Path). A path in a space
X
is a map
γ
:
I → X
. If
γ
(0) =
x
0
and
γ(1) = x
1
, we say γ is a path from x
0
to x
1
, and write γ : x
0
x
1
.
If γ(0) = γ(1), then γ is called a loop (based at x
0
).
x
0
x
1
Note that this map does not have to be injective. It can be self-intersecting
or do all sorts of weird stuff.
Recall that the basic idea of algebraic topology is to assign to each space
X
an (algebraic) object, which is (hopefully) easier to deal with. In homotopy
theory, we do so using the idea of paths.
To do so, we need to be able to perform operations on paths.
Definition (Concatenation of paths). If we have two paths
γ
1
from
x
0
to
x
1
;
and γ
2
from x
1
to x
2
, we define the concatenation to be
(γ
1
· γ
2
)(t) =
(
γ
1
(2t) 0 ≤ t ≤
1
2
γ
2
(2t − 1)
1
2
≤ t ≤ 1.
This is continuous by the gluing lemma.
Note that we concatenate left to right, but function composition goes from
right to left.
x
0
x
1
x
2
Definition (Inverse of path). The inverse of a path γ : I → X is defined by
γ
−1
(t) = γ(1 − t).
This is exactly the same path but going in the opposite direction.
What else do we need? We need an identity.
Definition (Constant path). The constant path at a point
x ∈ X
is given by
c
x
(t) = x.
We haven’t actually got a good algebraic system. We have
γ
and
γ
−1
, but
when we compose them, we get a path from
x
1
to
x
2
and back, and not the
identity. Also, we are not able to combine arbitrary paths in a space.
Before we make these into proper algebraic operations, we will talk about
something slightly different. We can view this as a first attempt at associating
things to topological spaces.
Definition (Path components). We can define a relation on
X
:
x
1
∼ x
2
if
there exists a path from
x
1
to
x
2
. By the concatenation, inverse and constant
paths,
∼
is an equivalence relation. The equivalence classes [
x
] are called path
components. We denote the quotient X/∼ by π
0
(X).
In the above space, we have three path components.
This isn’t really a very useful definition, since most spaces we care about are
path-connected, i.e. only have one path component. However, this is a first step
at associating something to spaces. We can view this as a “toy model” for the
more useful definitions we will later have.
One important property of this
π
0
is that not only does it associate a set to
each topological space, but also associates a function between the corresponding
sets to each continuous map.
Proposition. For any map f : X → Y , there is a well-defined function
π
0
(f) : π
0
(X) → π
0
(Y ),
defined by
π
0
(f)([x]) = [f(x)].
Furthermore,
(i) If f ' g, then π
0
(f) = π
0
(g).
(ii) For any maps A B C
h k
, we have π
0
(k ◦ h) = π
0
(k) ◦ π
0
(h).
(iii) π
0
(id
X
) = id
π
0
(X)
Proof.
To show this is well-defined, suppose [
x
] = [
y
]. Then let
γ
:
I → X
be a
path from x to y. Then f ◦ γ is a path from f (x) to f (y). So [f(x)] = [f(y)].
(i)
If
f ' g
, let
H
:
X × I → Y
be a homotopy from
f
to
g
. Let
x ∈ X
. Then
H
(
x, ·
) is a path from
f
(
x
) to
g
(
x
). So [
f
(
x
)] = [
g
(
x
)], i.e.
π
0
(
f
)([
x
]) =
π
0
(g)([x]). So π
0
(f) = π
0
(g).
(ii) π
0
(k ◦ h)([x]) = π
0
(k) ◦ π
0
(h)([x]) = [k(h(x))].
(iii) π
0
(id
X
)([x]) = [id
X
(x)] = [x]. So π
0
(id
X
) = id
π
0
(X)
.
Corollary. If f : X → Y is a homotopy equivalence, then π
0
(f) is a bijection.
Example. The two point space
X
=
{−
1
,
1
}
is not contractible, because
|π
0
(X)| = 2, but |π
0
(∗)| = 1.
This is a rather silly example, since we can easily prove it directly. However,
this is an example to show how we can use this machinery to prove topological
results.
Now let’s return to our operations on paths, and try to make them algebraic.
Definition (Homotopy of paths). Paths
γ, γ
0
:
I → X
are homotopic as paths if
they are homotopic rel
{
0
,
1
} ⊆ I
, i.e. the end points are fixed. We write
γ ' γ
0
.
x
0
x
1
Note that we would necessarily want to fix the two end points. Otherwise, if we
allow end points to move, we can shrink any path into a constant path, and our
definition of homotopy would be rather silly.
This homotopy works well with our previous operations on paths.
Proposition. Let
γ
1
, γ
2
:
I → X
be paths,
γ
1
(1) =
γ
2
(0). Then if
γ
1
' γ
0
1
and
γ
2
' γ
0
2
, then γ
1
· γ
2
' γ
0
1
· γ
0
2
.
x
0
x
1
x
2
γ
1
γ
2
γ
0
1
γ
0
2
Proof. Suppose that γ
1
'
H
1
γ
0
1
and γ
2
'
H
2
γ
0
2
. Then we have the diagram
γ
1
γ
2
γ
0
1
γ
0
2
x
0
x
1
x
2
H
1
H
2
We can thus construct a homotopy by
H(s, t) =
(
H
1
(s, 2t) 0 ≤ t ≤
1
2
H
2
(s, 2t − 1)
1
2
≤ t ≤ 1
.
To solve all our previous problems about operations on paths not behaving
well, we can look at paths up to homotopy.
Proposition. Let γ
0
: x
0
x
1
, γ
1
: x
1
x
2
, γ
2
: x
2
x
3
be paths. Then
(i) (γ
0
· γ
1
) · γ
2
' γ
0
· (γ
1
· γ
2
)
(ii) γ
0
· c
x
1
' γ
0
' c
x
0
· γ
0
.
(iii) γ
0
· γ
−1
0
' c
x
0
and γ
−1
0
· γ
0
' c
x
1
.
Proof.
(i) Consider the following diagram:
x
0
x
3
γ
0
γ
1
γ
2
γ
0
γ
1
γ
2
x
1
x
2
(ii) Consider the following diagram:
γ
0
c
x
1
γ
0
x
0
x
1
x
1
(iii) Consider the following diagram:
γ
0
γ
−1
0
c
x
0
x
0
x
0
Turning these into proper proofs is left as an exercise for the reader.