II Algebraic Topology - Covering spaces

3Covering spaces

II Algebraic Topology

3.1 Covering space

Intuitively, a covering space of

is a pair (

X, p

X → X

), such that if we take

any

∈ X

, there is some neighbourhood

such that the pre-image of

the neighbourhood is “many copies” of U .

Definition (Covering space). A covering space of

is a pair (

X, p

X → X

such that each x ∈ X has a neighbourhood U which is evenly covered.

Whether we require

to be surjective is a matter of taste. However, it will

not matter if X is path-connected, which is the case we really care about.

Definition (Evenly covered). U ⊆ X is evenly covered by p :

X → X if

−1

(U)

∼

α∈Λ

where p|

: V

→ U is a homeomorphism, and each of the V

⊆

X is open.

Example. Homeomorphisms are covering maps. Duh.

Example. Consider

R → S

⊆ C

defined by

t 7→ e

2πit

. This is a covering

space.

−1

(1)

Here we have p

−1

(1) = Z.

As we said, we are going to use covering spaces to determine non-trivial

fundamental groups. Before we do that, we can guess what the fundamental

group of

is. If we have a loop on

, it could be nothing, it can loop around

the circle once, or loop around many times. So we can characterize each loop by

the number of times it loops around the circle, and it should not be difficult to

convince ourselves that two loops that loop around the same number of times

can be continuously deformed to one another. So it is not unreasonable to guess

that

(

∼

. However, we also know that

−1

(1) =

. In fact, for any

point

z ∈ S

−1

(

) is just “

many copies” of

. We will show that this is not

a coincidence.

Example. Consider

→ S

(for any

n ∈ Z \ {

}

) defined by

z 7→ z

. We

can consider this as “winding” the circle

times, or as the following covering

map:

where we join the two red dots together.

This time the pre-image of 1 would be

copies of 1, instead of

copies of 1.

Example. Consider

, which is the real projective plane. This is defined

/∼

, where we identify every

x ∼ −x

, i.e. every pair of antipodal points.

We can also think of this as the space of lines in

. This is actually really

difficult to draw (and in fact impossible to fully embed in

), so we will not

attempt. There is, however, a more convenient way of thinking about

. We

just use the definition — we imagine an

, but this time, instead of a point

being a “point” on the sphere, it is a pair of antipodal points.

We define

→ RP

to be the quotient map. Then this is a covering map,

since the pre-image of a small neighbourhood of any

is just two copies of the

neighbourhood.

As we mentioned, a covering space of

is (locally) like many “copies” of

. Hence, given any function

Y → X

, a natural question is whether we can

“lift” it to a function f : Y →

X, by mapping to one of the copies of X in

This is not always possible, but when it is, we call this a lift.

Definition (Lifting). Let

Y → X

be a map, and

X → X

a covering

space. A lift of

is a map

Y →

such that

p ◦

, i.e. the following

diagram commutes:

Y X

We can visualize this in the case where

is the unit interval

and the map

is just a path in X.

f(Y )

It feels that if we know which “copy” of

we lifted our map to, then we

already know everything about

, since we are just moving our

from

that particular copy of X. This is made precise by the following lemma:

Lemma. Let

X → X

be a covering map,

Y → X

be a map, and

be both lifts of f. Then

S = {y ∈ Y :

(y) =

(y)}

is both open and closed. In particular, if

is connected,

and

agree either

everywhere or nowhere.

This is sort of a “uniqueness statement” for a lift. If we know a point in the

lift, then we know the whole path. This is since once we’ve decided our starting

point, i.e. which “copy” of

we work in, the rest of

has to follow what

does.

Proof.

First we show it is open. Let

be such that

(

) =

(

). Then

there is an evenly covered open neighbourhood

U ⊆ X

(

). Let

such that

(

)

∈

(

) =

and

U → U

is a homeomorphism. Let

V =

−1

(

U) ∩

−1

(

U). We will show that

on V .

Indeed, by construction

◦

= p|

◦

Since p|

is a homeomorphism, it follows that

Now we show

is closed. Suppose not. Then there is some

y ∈

S \ S

. So

(

)

(

). Let

be an evenly covered neighbourhood of

(

). Let

−1

(

) =

Let

(

)

∈ U

and

(

)

∈ U

, where

β 6

. Then

−1

(

)

∩

−1

(

) is

an open neighbourhood of

, and hence intersects

by definition of closure. So

there is some

x ∈ V

such that

(

) =

(

). But

(

)

∈ U

and

(

)

∈ U

and hence

and

have a non-trivial intersection. This is a contradiction. So

S is closed.

We just had a uniqueness statement. How about existence? Given a map,

is there guarantee that we can lift it to something? Moreover, if I have fixed a

“copy” of

I like, can I also lift my map to that copy? We will later come up

with a general criterion for when lifts exist. However, it turns out homotopies

can always be lifted.

Lemma (Homotopy lifting lemma). Let

X → X

be a covering space,

Y × I → X

be a homotopy from

. Let

be a lift of

. Then there

exists a unique homotopy

H : Y × I →

X such that

(i)

H( · , 0) =

; and

(ii)

H is a lift of H, i.e. p ◦

H = H.

This lemma might be difficult to comprehend at first. We can look at the

special case where

∗

. Then a homotopy is just a path. So the lemma

specializes to

Lemma (Path lifting lemma). Let

X → X

be a covering space,

I → X

path, and

˜x

∈

such that

(

˜x

) =

(0). Then there exists a unique path

˜γ : I →

X such that

(i) ˜γ(0) = ˜x

; and

(ii) ˜γ is a lift of γ, i.e. p ◦ ˜γ = γ.

This is exactly the picture we were drawing before. We just have to start

at a point

˜x

, and then everything is determined because locally, everything

upstairs in

is just like

. Note that we have already proved uniqueness. So

we just need to prove existence.

In theory, it makes sense to prove homotopy lifting, and path lifting comes

immediately as a corollary. However, the proof of homotopy lifting is big and

scary. So instead, we will prove path lifting, which is something we can more

easily visualize and understand, and then use that to prove homotopy lifting.

Proof. Let

S = {s ∈ I : ˜γ exists on [0, s] ⊆ I}.

Observe that

(i) 0 ∈ S.

(ii) S

is open. If

s ∈ S

and

˜γ

(

)

∈ V

⊆ p

−1

(

), we can define

˜γ

on some

small neighbourhood of s by

˜γ(t) = (p|

)

−1

◦ γ(t)

(iii) S

is closed. If

s 6∈ S

, then pick an evenly covered neighbourhood

(

Suppose

((

s − ε, s

))

⊆ U

. So

s −

6∈ S

. So (

s −

∩ S

∅

. So

closed.

Since

is both open and closed, and is non-empty, we have

. So

˜γ

exists.

How can we promote this to a proof of the homotopy lifting lemma? At

every point

y ∈ Y

, we know what to do, since we have path lifting. So

(

y, ·

)

is defined. So the thing we have to do is to show that this is continuous. Steps

of the proof are

(i)

Use compactness of

to argue that the proof of path lifting works on small

neighbourhoods in Y .

(ii)

For each

, we pick an open neighbourhood

, and define a good path

lifting on U × I.

(iii)

By uniqueness of lifts, these path liftings agree when they overlap. So we

have one big continuous lifting.

With the homotopy lifting lemma in our toolkit, we can start to use it to

do stuff. So far, we have covering spaces and fundamental groups. We are now

going to build a bridge between these two, and show how covering spaces can be

used to reflect some structures of the fundamental group.

At least one payoff of this work is that we are going to exhibit some non-trivial

fundamental groups.

We have just showed that we are allowed to lift homotopies. However, what

we are really interested in is homotopy as paths. The homotopy lifting lemma

does not tell us that the lifted homotopy preserves basepoints. This is what we

are going to show.

Corollary. Suppose

γ, γ

I → X

are paths

and

˜γ, ˜γ

I →

are lifts

of γ and γ

respectively, both starting at ˜x

∈ p

−1

γ ' γ

as paths, then

˜γ

and

˜γ

are homotopic as paths. In particular,

˜γ(1) = ˜γ

(1).

Note that if we cover the words “as paths” and just talk about homotopies,

then this is just the homotopy lifting lemma. So we can view this as a stronger

form of the homotopy lifting lemma.

Proof.

The homotopy lifting lemma gives us an

, a lift of

with

(

· ,

0) =

˜γ

lift

˜γ

˜x

In this diagram, we by assumption know the bottom of the

square is

˜γ

. To

show that this is a path homotopy from

˜γ

, we need to show that the other

edges are c

˜x

, c

˜x

and ˜γ

respectively.

Now

(

· ,

1) is a lift of

(

· ,

1) =

, starting at

˜x

. Since lifts are unique,

we must have

(

· ,

1) =

˜γ

. So this is indeed a homotopy between

˜γ

and

˜γ

Now we need to check that this is a homotopy of paths.

We know that

, ·

) is a lift of

, ·

) =

. We are aware of one lift of

, namely

˜x

. By uniqueness of lifts, we must have

, ·

) =

˜x

. Similarly,

H(1, · ) = c

˜x

. So this is a homotopy of paths.

So far, our picture of covering spaces is like this:

Except. . . is it? Is it possible that we have four copies of

but just three copies

? This is obviously possible if

is not path connected — the component

containing

and the one containing

are completely unrelated. But what if

X is path connected?

Corollary. If

is a path connected space,

, x

∈ X

, then there is a bijection

−1

) → p

−1

Proof.

Let

be a path. We want to use this to construct a bijection

between each preimage of

and each preimage of

. The obvious thing to do

is to use lifts of the path γ.

Define a map

−1

(

)

→ p

−1

(

) that sends

˜x

to the end point of the

unique lift of γ at ˜x

The inverse map is obtained by replacing

with

−1

, i.e.

−1

. To show this

is an inverse, suppose we have some lift

˜γ

˜x

, so that

(

˜x

) =

˜x

. Now

notice that

˜γ

−1

is a lift of

−1

starting at

˜x

and ending at

˜x

. So

−1

(

˜x

) =

˜x

So f

−1

is an inverse to f

, and hence f

is bijective.

Definition (

-sheeted). A covering space

X → X

of a path-connected space

X is n-sheeted if |p

−1

(x)| = n for any (and hence all) x ∈ X.

Each covering space has a number associated to it, namely the number of

sheets. Is there any number we can assign to fundamental groups? Well, the

index of a subgroup might be a good candidate. We’ll later see if this is the case.

One important property of covering spaces is the following:

Lemma. If p :

X → X is a covering map and ˜x

∈

X, then

∗

: π

(

X, ˜x

) → π

(X, x

)

is injective.

Proof.

To show that a group homomorphism

∗

is injective, we have to show

that if p

∗

(x) is trivial, then x must be trivial.

Consider a based loop

˜γ

. We let

p ◦ ˜γ

. If

is trivial, i.e.

γ ' c

paths, the homotopy lifting lemma then gives us a homotopy upstairs between

˜γ

and c

˜x

. So ˜γ is trivial.

As we have originally said, our objective is to make our fundamental group

act on something. We are almost there already.

Let’s look again at the proof that there is a bijection between

−1

(

) and

−1

(

). What happens if

is a loop? For any

˜x

∈ p

−1

(

), we can look at

the end point of the lift. This end point may or may not be our original

˜x

. So

each loop γ “moves” our ˜x

to another ˜x

However, we are not really interested in paths themselves. We are interested

in equivalence classes of paths under homotopy of paths. However, this is fine.

is homotopic to

, then this homotopy can be lifted to get a homotopy

between

˜γ

and

˜γ

. In particular, these have the same end points. So each (based)

homotopy class gives a well-defined endpoint.

˜γ

˜x

Now this gives an action of

(

X, x

) on

−1

(

)! Note, however, that this will

not be the sort of actions we are familiar with. We usually work with left-actions,

where the group acts on the left, but now we will have right-actions, which

may confuse you a lot. To see this, we have to consider what happens when we

perform two operations one after another, which you shall check yourself. We

write this action as ˜x

· [γ].

When we have an action, we are interested in two things — the orbits, and

the stabilizers. This is what the next lemma tells us about.

Lemma. Suppose X is path connected and x

∈ X.

(i)

The action of

(

X, x

) on

−1

(

) is transitive if and only if

is path

connected. Alternatively, we can say that the orbits of the action correspond

to the path components.

(ii) The stabilizer of ˜x

∈ p

−1

) is p

∗

(π

(

X, ˜x

)) ⊆ π

(X, x

(iii) If

X is path connected, then there is a bijection

∗

(π

(

X, ˜x

))\π

(X, x

) → p

−1

Note that

∗

(

X, ˜x

))

\π

(

X, x

) is not a quotient, but simply the set of

cosets. We write it the “wrong way round” because we have right cosets

instead of left cosets.

Note that this is great! If we can find a covering space

and a point

such that

−1

(

) is non-trivial, then we immediately know that

(

X, x

) is

non-trivial!

Proof.

(i)

˜x

, ˜x

∈ p

−1

(

), then since

is path connected, we know that there

is some

˜γ

˜x

. Then we can project this to

p ◦ ˜γ

. Then

a path from

, i.e. a loop. Then by the definition of the action,

˜x

· [γ] = ˜γ(1) = ˜x

(ii)

Suppose [

]

∈ stab

(

˜x

). Then

˜γ

is a loop based at

˜x

. So

˜γ

defines

[˜γ] ∈ π

(

X, ˜x

) and γ = p ◦ ˜γ.

(iii) This follows directly from the orbit-stabilizer theorem.

We now want to use this to determine that the fundamental group of a space

is non-trivial. We can be more ambitious, and try to actually find

(

X, x

In the best possible scenario, we would have

(

X, ˜x

) trivial. Then we have a

bijection between

(

X, x

) and

−1

(

). In other words, we want our covering

space

X to be simply connected.

Definition (Universal cover). A covering map

X → X

is a universal cover

X is simply connected.

We will look into universal covers in depth later and see what they really are.

Corollary. If

X → X

is a universal cover, then there is a bijection

(X, x

) → p

−1

Note that the orbit-stabilizer theorem does not provide a canonical bijection

between

−1

(

) and

∗

(

X, ˜x

)

\π

(

X, x

). To obtain a bijection, we need to

pick a starting point

˜x

∈ p

−1

(

). So the above bijection

depends on a choice

of ˜x