3Covering spaces
II Algebraic Topology
3.4 The Galois correspondence
Recall that at the beginning, we wanted to establish a correspondence between
covering spaces and fundamental groups. We have already established the result
that covering maps are injective on
π
1
. Therefore, given a (based) covering
space
p
: (
˜
X, ˜x
0
)
→
(
X, x
0
), we can give a subgroup
p
∗
π
1
(
˜
X, ˜x
0
)
≤ π
1
(
X, x
0
). It
turns out that as long as we define carefully what we mean for based covering
spaces to be “the same”, this is a one-to-one correspondence — each subgroup
corresponds to a covering space.
We can have the following table of correspondences:
Covering spaces Fundamental group
(Based) covering spaces ←→ Subgroups of π
1
Number of sheets ←→ Index
Universal covers ←→ Trivial subgroup
We now want to look at some of these correspondences.
Recall that we have shown that
π
1
(
X, x
0
) acts on
p
−1
(
x
0
). However, this is
not too interesting an action, since
p
−1
(
x
0
) is a discrete group with no structure.
Having groups acting on a cube is fun because the cube has some structure. So
we want something more “rich” for π
1
(X, x
0
) to act on.
We note that we can make
π
1
(
X, x
0
) “act on” the universal cover. How?
Recall that in the torus example, each item in the fundamental group corresponds
to translating the whole universal covering by some amount. In general, a point
on
˜
X
can be thought of as a path
α
on
X
starting from
x
0
. Then it is easy to
make a loop
γ
:
x
0
x
0
act on this: use the concatenation
γ · α
: [
γ
]
·
[
α
] = [
γ · α
].
˜x
0
˜x
0
0
˜γ
α
˜
X
X
p
x
0
γ
α
We will use this idea and return to the initial issue of making subgroups corre-
spond to covering spaces. We want to show that this is surjective — every sub-
group arises from some cover. We want to say “For any subgroup
H ≤ π
1
(
X, x
0
),
there is a based covering map
p
: (
˜
X, ˜x
0
)
→
(
X, x
0
) such that
p
∗
π
1
(
˜
X, ˜x
0
) =
H
”.
Except, this cannot possibly be true, since by taking the trivial subgroup, this
would imply that there is a universal covering for every space. So we need some
additional assumptions.
Proposition. Let
X
be a path connected, locally path connected and semi-
locally simply connected space. For any subgroup
H ≤ π
1
(
X, x
0
), there is a
based covering map p : (
˜
X, ˜x
0
) → (X, x
0
) such that p
∗
π
1
(
˜
X, ˜x
0
) = H.
Proof.
Since
X
is a path connected, locally path connected and semi-locally
simply connected space, let
¯
X
be a universal covering. We have an intermediate
group
H
such that
π
1
(
˜
X, ˜x
0
) = 1
≤ H ≤ π
1
(
X, x
0
). How can we obtain a
corresponding covering space?
Note that if we have
¯
X
and we want to recover
X
, we can quotient
¯
X
by the
action of
π
1
(
X, x
0
). Since
π
1
(
X, x
0
) acts on
¯
X
, so does
H ≤ π
1
(
X, x
0
). Now we
can define our covering space by taking quotients. We define
∼
H
on
¯
X
to be
the orbit relation for the action of
H
, i.e.
˜x ∼
H
˜y
if there is some
h ∈ H
such
that ˜y = h˜x. We then let
˜
X be the quotient space
¯
X/∼
H
.
We can now do the messy algebra to show that this is the covering space we
want.
We have just showed that every subgroup comes from some covering space,
i.e. the map from the set of covering spaces to the subgroups of
π
1
is surjective.
Now we want to prove injectivity. To do so, we need a generalization of the
homotopy lifting lemma.
Suppose we have path-connected spaces (
Y, y
0
), (
X, x
0
) and (
˜
X, ˜x
0
), with
f
: (
Y, y
0
)
→
(
X, x
0
) a continuous map,
p
: (
˜
X, ˜x
0
)
→
(
X, x
0
) a covering map.
When does a lift of
f
to
˜
f
: (
Y, y
0
)
→
(
˜
X, ˜x
0
) exist? The answer is given by the
lifting criterion.
Lemma (Lifting criterion). Let
p
: (
˜
X, ˜x
0
)
→
(
X, x
0
) be a covering map of path-
connected based spaces, and (
Y, y
0
) a path-connected, locally path connected
based space. If
f
: (
Y, y
0
)
→
(
X, x
0
) is a continuous map, then there is a (unique)
lift
˜
f
: (
Y, y
0
)
→
(
˜
X, ˜x
0
) such that the diagram below commutes (i.e.
p ◦
˜
f
=
f
):
(
˜
X, ˜x
0
)
(Y, y
0
) (X, x
0
)
p
f
˜
f
if and only if the following condition holds:
f
∗
π
1
(Y, y
0
) ≤ p
∗
π
1
(
˜
X, ˜x
0
).
Note that uniqueness comes from the uniqueness of lifts. So this lemma is
really about existence.
Also, note that the condition holds trivially when
Y
is simply connected, e.g.
when it is an interval (path lifting) or a square (homotopy lifting). So paths and
homotopies can always be lifted.
Proof.
One direction is easy: if
˜
f
exists, then
f
=
p ◦
˜
f
. So
f
∗
=
p
∗
◦
˜
f
∗
. So we
know that im f
∗
⊆ im p
∗
. So done.
In the other direction, uniqueness follows from the uniqueness of lifts. So we
only need to prove existence. We define
˜
f as follows:
Given a
y ∈ Y
, there is some path
α
y
:
y
0
y
. Then
f
maps this to
β
y
:
x
0
f
(
y
) in
X
. By path lifting, this path lifts uniquely to
˜
β
y
in
˜
X
. Then
we set
˜
f
(
y
) =
˜
β
y
(1). Note that if
˜
f
exists, then this must be what
˜
f
sends
y
to.
What we need to show is that this is well-defined.
Suppose we picked a different path
α
0
y
:
y
0
y
. Then this
α
0
y
would have
differed from α
y
by a loop γ in Y .
Our condition that
f
∗
π
1
(
Y, y
0
)
≤ p
∗
π
1
(
˜
X, ˜x
0
) says
f ◦ γ
is the image of a
loop in
˜
X
. So
˜
β
y
and
˜
β
0
y
also differ by a loop in
˜
X
, and hence have the same
end point. So this shows that
˜
f is well-defined.
Finally, we show that
˜
f
is continuous. First, observe that any open set
U ⊆
˜
X
can be written as a union of
˜
V
such that
p|
˜
V
:
˜
V → p
(
˜
V
) is a homeomorphism.
Thus, it suffices to show that if
p|
˜
V
:
˜
V → p
(
˜
V
) =
V
is a homeomorphism, then
˜
f
−1
(
˜
V ) is open.
Let
y ∈
˜
f
−1
(
˜
V
), and let
x
=
f
(
y
). Since
f
−1
(
V
) is open and
Y
is locally
path-connected, we can pick an open
W ⊆ f
−1
(
V
) such that
y ∈ W
and
W
is
path connected. We claim that W ⊆
˜
f
−1
(
˜
V ).
Indeed, if
z ∈ W
, then we can pick a path
γ
from
y
to
z
. Then
f
sends
this to a path from
x
to
f
(
z
). The lift of this path to
˜
X
is given by
p|
−1
˜
V
(
f
(
γ
)),
whose end point is p|
−1
˜
V
(f(z)) ∈
˜
V . So it follows that
˜
f(z) = p|
−1
˜
V
(f(z)) ∈
˜
V .
Now we prove that every subgroup of
π
1
comes from exactly one covering
space. What this statement properly means is made precise in the following
proposition:
Proposition. Let (
X, x
0
), (
˜
X
1
, ˜x
1
), (
˜
X
2
, ˜x
2
) be path-connected based spaced,
and p
i
: (
˜
X
i
, ˜x
i
) → (X, x
0
) be covering maps. Then we have
p
1∗
π
1
(
˜
X
1
, ˜x
1
) = p
2∗
π
1
(
˜
X
2
, ˜x
2
)
if and only if there is some homeomorphism
h
such that the following diagram
commutes:
(
˜
X
1
, ˜x
1
) (
˜
X
2
, ˜x
2
)
(X, x
0
)
h
p
1
p
2
i.e. p
1
= p
2
◦ h.
Note that this is a stronger statement than just saying the two covering
spaces are homeomorphic. We are saying that we can find a nice homeomorphism
that works well with the covering map p.
Proof.
If such a homeomorphism exists, then clearly the subgroups are equal. If
the subgroups are equal, we rotate our diagram a bit:
(
˜
X
2
, ˜x
2
)
(
˜
X
1
, ˜x
1
) (X, x
0
)
p
2
p
1
h=˜p
1
Then
h
=
˜p
1
exists by the lifting criterion. By symmetry, we can get
h
−1
=
˜p
2
.
To show
˜p
2
is indeed the inverse of
˜p
1
, note that
˜p
2
◦ ˜p
1
is a lift of
p
2
◦ ˜p
1
=
p
1
.
Since
id
˜
X
1
is also a lift, by the uniqueness of lifts, we know
˜p
2
◦ ˜p
1
is the identity
map. Similarly, ˜p
1
◦ ˜p
2
is also the identity.
(
˜
X
1
, ˜x
1
)
(
˜
X
1
, ˜x
1
) (
˜
X
2
, ˜x
2
) (X, x
0
)
p
1
p
1
˜p
1
˜p
2
p
2
Now what we would like to do is to forget about the basepoints. What
happens when we change base points? Recall that the effect of changing base
points is that we will conjugate our group. This doesn’t actually change the
group itself, but if we are talking about subgroups, conjugation can send a
subgroup into a different subgroup. Hence, if we do not specify the basepoint,
we don’t get a subgroup of π
1
, but a conjugacy class of subgroups.
Proposition. Unbased covering spaces correspond to conjugacy classes of
subgroups.