3Covering spaces
II Algebraic Topology
3.3 Universal covers
We have defined universal covers mysteriously as covers that are simply connected.
We have just shown that
p
:
R → S
1
is a universal cover. In general, what do
universal covers look like?
Let’s consider a slightly more complicated example. What would be a
universal cover of the torus
S
1
× S
1
? An obvious guess would be
p × p
:
R × R →
S
1
× S
1
. How can we visualize this?
First of all, how can we visualize a torus? Often, we just picture it as the
surface of a doughnut. Alternatively, we can see it as a quotient of the square,
where we identify the following edges:
Then what does it feel like to live in the torus? If you live in a torus and look
around, you don’t see a boundary. The space just extends indefinitely for ever,
somewhat like
R
2
. The difference is that in the torus, you aren’t actually seeing
free space out there, but just seeing copies of the same space over and over again.
If you live inside the square, the universe actually looks like this:
As we said, this looks somewhat likes
R
2
, but we know that this is not
R
2
,
since we can see some symmetry in this space. Whenever we move one unit
horizontally or vertically, we get back to “the same place”. In fact, we can move
horizontally by
n
units and vertically by
m
units, for any
n, m ∈ Z
, and still get
back to the same place. This space has a huge translation symmetry. What is
this symmetry? It is exactly Z × Z.
We see that if we live inside the torus
S
1
× S
1
, it feels like we are actually
living in the universal covering space
R × R
, except that we have an additional
symmetry given by the fundamental group Z × Z.
Hopefully, you are convinced that universal covers are nice. We would like to
say that universal covers always exist. However, this is not always true.
Firstly, we should think — what would having a universal cover imply?
Suppose
X
has a universal cover
˜
X
. Pick any point
x
0
∈ X
, and pick an
evenly covered neighbourhood
U
in
X
. This lifts to some
˜
U ⊆
˜
X
. If we draw a
teeny-tiny loop
γ
around
x
0
inside
U
, we can lift this
γ
to
˜γ
in
˜
U
. But we know
that
˜
X
is simply connected. So
˜γ
is homotopic to the constant path. Hence
γ
is also homotopic to the constant path. So all loops (contained in
U
) at
x
0
are
homotopic to the constant path.
It seems like for every
x
0
∈ X
, there is some neighbourhood of
x
0
that is
simply connected. Except that’s not what we just showed above. The homotopy
from
˜γ
to the constant path is a homotopy in
˜
X
, and can pass through anything
in
˜
X
, not just
˜
U
. Hence the homotopy induced in
X
is also a homotopy in
X
,
not a homotopy in
U
. So
U
itself need not be simply connected. What we have
is a slightly weaker notion.
Definition (Locally simply connected).
X
is locally simply connected if for all
x
0
∈ X, there is some neighbourhood U of x
0
such that U is simply connected.
As we mentioned, what we actually want is a weaker condition.
Definition (Semi-locally simply connected).
X
is semi-locally simply connected
if for all
x
0
∈ X
, there is some neighbourhood
U
of
x
0
such that any loop
γ
based at x
0
is homotopic to c
x
0
as paths in X.
We have just argued that if a universal cover
p
:
˜
X → X
exists, then
X
is
semi-locally simply connected. This is really not interesting, since we don’t care
if a space is semi-locally simply connected. What is important is that this is the
other direction. We still need one more additional condition:
Definition (Locally path connected). A space
X
is locally path connected if for
any point
x
and any neighbourhood
V
of
x
, there is some open path connected
U ⊆ V such that x ∈ U.
It is important to note that a path connected space need not be locally path
connected. It is an exercise for the reader to come up with a counterexample.
Theorem. If
X
is path connected, locally path connected and semi-locally
simply connected, then X has a universal covering.
Note that we can alternatively define a universal covering as a covering space
of
X
that is also a covering space of all other covers of
X
. If we use this definition,
then we can prove this result easily using Zorn’s lemma. However, that proof
is not too helpful since it does not tell us where the universal covering comes
from. Instead, we will provide a constructive proof (sketch) that will hopefully
be more indicative of what universal coverings are like.
Proof.
(idea) We pick a basepoint
x
0
∈ X
for ourselves. Suppose we have a
universal covering
˜
X
. Then this lifts to some
˜x
0
in
˜
X
. If we have any other point
˜x ∈
˜
X
, since
˜
X
should be path connected, there is a path
˜α
:
˜x
0
˜x
. If we
have another path, then since
˜
X
is simply connected, the paths are homotopic.
Hence, we can identify each point in
˜
X with a path from ˜x
0
, i.e.
{points of
˜
X} ←→ {paths ˜α from ˜x
0
∈
˜
X}/'.
This is not too helpful though, since we are defining
˜
X
in terms of things in
˜
X
.
However, by path lifting, we know that paths
˜α
from
˜x
0
in
˜
X
biject with paths
α
from
x
0
in
X
. Also, by homotopy lifting, homotopies of paths in
X
can be
lifted to homotopies of paths in
˜
X. So we have
{points of
˜
X} ←→ {paths α from x
0
∈ X}/'.
So we can produce our
˜
X by picking a basepoint x
0
∈ X, and defining
˜
X = {paths α : I → X such that α(0) = x
0
}/'.
The covering map p :
˜
X → X is given by [α] 7→ α(1).
One then has to work hard to define the topology, and then show this is
simply connected.