4Some group theory
II Algebraic Topology
4.2 Another view of free groups
Recall that we have not yet properly defined free groups, since we did not show
that multiplication is well-defined. We are now going to do this using topology.
Again let
S
be a set. For the following illustration, we will just assume
S = {a, b}, but what we will do works for any set S. We define X by
a
b
x
0
We call this a “rose with 2 petals”. This is a cell complex, with one 0-cell and
|S|
1-cells. For each
s ∈ S
, we have one 1-cell,
e
s
, and we fix a path
γ
s
: [0
,
1]
→ e
s
that goes around the 1-cell once. We will call the 0-cells and 1-cells vertices and
edges, and call the whole thing a graph.
What’s the universal cover of
X
? Since we are just lifting a 1-complex, the
result should be a 1-complex, i.e. a graph. Moreover, this graph is connected
and simply connected, i.e. it’s a tree. We also know that every vertex in the
universal cover is a copy of the vertex in our original graph. So it must have 4
edges attached to it. So it has to look like something this:
˜x
0
In
X
, we know that at each vertex, there should be an edge labeled
a
going in;
an edge labeled
a
going out; an edge labeled
b
going in; an edge labeled
b
going
out. This should be the case in
˜
X as well. So
˜
X looks like this:
˜x
0 aa
b
b
a
b
b
The projection map is then obvious — we send all the vertices in
˜
X
to
x
0
∈ X
,
and then the edges according to the labels they have, in a way that respects the
direction of the arrow. It is easy to show this is really a covering map.
We are now going to show that this tree “is” the free group. Notice that
every word
w ∈ S
∗
denotes a unique “edge path” in
˜
X
starting at
˜x
0
, where an
edge path is a sequence of oriented edges
˜e
1
, · · · , ˜e
n
such that the “origin” of
˜e
i+1
is equal to the “terminus” of ˜e
i
.
For example, the following path corresponds to w = abb
−1
b
−1
ba
−1
b
−1
.
˜x
0
We can note a few things:
(i)
˜
X is connected. So for all ˜x ∈ p
−1
(x
0
), there is an edge-path ˜γ : ˜x
0
˜x.
(ii)
If an edge-path
˜γ
fails to be locally injective, we can simplify it. How can
an edge fail to be locally injective? It is fine if we just walk along a path,
since we are just tracing out a line. So it fails to be locally injective if two
consecutive edges in the path are the same edge with opposite orientations:
˜e
i−1
˜e
i
˜e
i+1
˜e
i+2
We can just remove the redundant lines and get
˜e
i−1
˜e
i+2
This reminds us of two things — homotopy of paths and elementary
reduction of words.
(iii)
Each point
˜x ∈ p
−1
(
x
0
) is joined to
˜x
0
by a unique locally injective
edge-path.
(iv)
For any
w ∈ S
∗
, from (ii), we know that
˜γ
is locally injective if and only if
w is reduced.
We can thus conclude that there are bijections
F (S) p
−1
(x
0
) π
1
(X, x
0
)
that send
˜x
to the word
w ∈ F
(
S
) such that
˜γ
w
is a locally injective edge-path
to ˜x
0
˜x; and ˜x to [γ] ∈ π
1
(X, x
0
) such that ˜x
0
· [γ] = ˜x.
So there is a bijection between
F
(
S
) and
π
1
(
X, x
0
). It is easy to see that the
operations on
F
(
S
) and
π
1
(
X, x
0
) are the same, since they are just concatenating
words or paths. So this bijection identifies the two group structures. So this
induces an isomorphism F (S)
∼
=
π
1
(X, x
0
).