Download Covering spaces

Covering spaces
John Terilla
Fall 2014
Contents
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The structure of the Galois group of a covering
2.1 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 The action of π1 (X, x) on the fibers p−1 (x) of a cover . . . . . . . .
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7
8
3
The existence of a universal cover
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1
Introduction
1.1 Covering Spaces . . . . . . .
1.2 The general lifting problem . .
1.3 Morphisms of covering spaces
1.4 The Galois group of a covering
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Introduction
I wanted to outline a presentation of covering spaces, trying to establish the overall
picture of a “Galois theory” of covering spaces quickly and clearly. It’s worth noting
that there are two occasions of a purely topological problem to be solved (lifting a map,
constructing a simply connected covering space). By applying π1 to a solution, one
discovers a necessary algebraic condition that in both cases turns out to be sufficient.
This is algebraic topology par excellence.
I found May’s book [3] helpful in separating the formal algebra from the topology, but I wanted to introduce the topic with less abstraction than his approach using
groupoid covering morphisms. So, I pick base points and the more familiar algebraic
language of group actions applies. I adopted Hatcher’s notation, but in order to streamline the presentation, I make several assumptions. Specifically, in these notes I assume
that all spaces are path connected and locally path connected and covering maps are
required to be surjective.
1
1.1
Covering Spaces
Definition 1. A surjection p : X̃ → X is a covering if X if every x ∈ X has a
neighborhood U , called a fundamental or elementary neighborhood, with the property
that each component of p−1 (U ) is mapped by p homeomorphically onto U . A based
covering is a based map p : (X̃, x̃) → (X, x) that is a covering.
The fundamental lemma is that covering spaces have the path lifting property:
p
The Path Lifting Property. Suppose X̃ → X is a covering.
1. Given a path γ : I → X and a lift of the initial point x̃0 ∈ X̃, there is a unique
lift γ̃ : I → X̃ with pγ̃ = γ.
2. If γ ' γ 0 are homotopic paths in X, then they lift to homotopic paths γ̃ ' γ˜0 in
X̃ (in particular, the endpoints agree).
Immediate consequences of the path lifting property include
p
Corollary 1. If X̃ → X is a covering then p∗ : π1 (X̃, x̃) → π1 (X, x) is injective.
Proof. Suppose γ̃ : I → X̃ satisfies p∗ [γ̃] = [cx ]. Then γ := pγ̃ ' cx . Thus, the lifts
are homotopic: γ
e'e
cx . The fact that c̃x = cx̃ concludes the proof.
So, we can and will view π1 (X̃, x̃0 ) as a subgroup of π1 (X, x0 )—I’ll just write
π1 (X̃, x̃) ⊂ π1 (X, x) instead of p∗ (π1 (X̃, x̃)) ⊂ π1 (X, x). A typical loop γ based
at x ∈ X lifts to a path begininng at x̃ ∈ X̃ and ending at another point in the fiber
p−1 (x). The geometric way to think of π1 (X̃, x̃) as a subgrpoup of π1 (X, x) is as the
homotopy classes of loops [γ] ∈ π1 (X, x) that lift to loops γ̃ based at x̃ ∈ X̃, not just
paths starting at x̃.
Problem 1. Let p : X̃ → X is a covering and x̃1 and x̃2 be two distinct points in the
fiber p−1 (x). Check that fundamental groups π1 (X̃, x̃1 ) and π1 (X̃, x̃2 ) are conjugate
subgroups in π1 (X, x).
It will be established that (for spaces X satisfying a reasonable local connectivity
condition) there is a bijective correspondence between isomorphism classes of based
covering spaces over (X, x) and subgroups of π1 (X, x0 ). In order to describe this correspondence, it’s necessary to say what an isomorphism of covering spaces is. Before
doing that, it will be convenient to consider the general lifting problem.
1.2
The general lifting problem
Suppose p : (X̃, x̃) → (X, x) is a covering and f : (Y, y) → (X, x) is a map from any
path connected space Y . The path lifting problem is to determine whether there exists
2
a lift f˜ : (Y, y) → (˜(X), x̃) so that pf˜ = f :
(X̃, x̃)
f˜
(Y, y)
f
p
(X, x)
Notice that the path lifting property implies that if the lift f˜ exists, it’s unique.
This is because for any y 0 ∈ Y , one can take a path γ : I → Y from y to y 0 . Then
f γ : I → X is a path from x to f (y), which lifts uniquely to a path that must agree
with f˜γ, hence the value f˜(y 0 ) is determined as the lift of the endpoint of f γ.
A necessary condition for the path lifting problem to have a solution is determined
by assuming a solution f˜ exists and applying π1 to get
π1 (X̃, x̃)
f˜∗
π1 (Y, y)
p∗
f∗
π1 (X, x)
It follows that f∗ (π1 (Y, y)) ⊂ p∗ (π1 (X̃, x̃)). It turns out that this condition is also sufficient. But before proving it, here’s an example showing that this necessary condition
can be powerful.
Example 1. The exponential map exp : C → C∗ is a covering. It’s not injective so it
cannot have a left inverse. It’s surjective, so it has a set-theoretic right inverse, but does
exp have a continuous right inverse? That is, does there exists a continuous function
g
exp
g : C∗ → C whose composition C∗ → C → C∗ is the identity? This amounts to
asking whether there is a lift
C
g
C∗
exp
id
C∗
The answer, is “no” since the necessary condition is not satisfied: for any base points
z and ez , id∗ (π1 (Z, ez )) = Z is not contained in exp∗ (π1 (C), z) = ∗.
Theorem 1. Suppose p : (X̃, x̃) → (X, x) is a covering and consider f : (Y, y) →
(X, x). There exists a unique lift f˜ : (Y, y) → (X̃, x̃) if and only if f∗ (π1 (Y, y)) ⊂
p∗ (π1 (X̃, x̃)).
Proof. Necessity and uniqueness have already been established. The only thing that
remains is to construct a lift. For any y 0 ∈ Y , take a path γ : I → Y connecting y to
y 0 . Lift f γ to a path I → X̃ and define f˜(y 0 ) to be the endpoint of the lifted path. The
hypothesis that f∗ (π1 (Y, y)) ⊂ p∗ (π1 (X̃, x̃)) implies that this definition of f˜ is well
defined—that the value f˜(y 0 ) does not depend on the path γ chosen. The check that f˜
is continuous is left as an exercise.
3
Note that we obtain unique lifts by keeping track of base points, which makes sense
since we are studying the lifting problem by applying the functor π1 . One can consider
a non-based lifting problem
X̃
f˜
Y
f
p
X
To obtain the necessary and sufficient condition on the map f , one should choose a
point y ∈ Y and take π1 . If there is any solution to the lifting problem, there may exist
many solutions depending on which point in the fiber p−1 (f (y)) the lift maps y to. If
two lifts agree at any point of y they must be equal.
1.3
Morphisms of covering spaces
We will consider covering spaces and based covering spaces. Fix (X, x) as a base
space.
p1
Definition 2. A morphism from a based covering (X̃1 , x̃1 ) → (X, x) to a based covp2
ering (X̃2 , x̃2 ) → (X, x) is a continuous function f : (X̃1 , x̃1 ) → (X̃2 , x̃2 ) satisfying
p1 = f p2 .
f
(X̃1 , x̃1 )
(X̃2 , x̃2 )
p1
p2
(X, x0 )
Based covering spaces over (X, x0 ) form a category. The definition of morphism
p1
determines the notion of isomorphism: two based covering spaces (X̃1 , x̃1 ) → (X, x)
p2
and (X̃2 , x̃2 ) → (X, x) are isomorphic iff there are covering maps f : X̃1 → X̃2 and
g : X̃2 → X̃1 so that gf = idX̃1 and f g = idX̃2 . Notice that isomorphic covering
spaces are homeomorphic topological spaces.
p1
Now, notice that a morphism f from a based covering (X̃1 , x̃1 ) → (X, x0 ) to a
p2
based covering (X̃2 , x̃2 ) → (X, x0 ) is a solution to a lifting problem:
(X̃2 , x̃2 )
f
(X̃1 , x̃1 )
p1
p2
(X, x0 )
Thus, there exists a morphism of based covering spaces if and only if π1 (X̃1 , x̃1 ) ⊂
π1 (X̃2 , x̃2 ), viewed as subgroups of π1 (X, x0 ). Therefore, if p1 : (X̃1 , x̃1 ) → (X, x)
is a based covering space and X̃1 is simplyconnected, it satisfies the following universal
property:
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Universal covering space property. For every based covering p2 : (X̃2 , x̃2 ) →
(X, x), there exists a unique morphism of based covering spaces f : (X̃1 , x̃1 ) →
(X̃2 , x̃2 ).
(X̃2 , x̃2 )
f
(X̃1 , x̃1 )
p1
p2
(X, x)
These remarks justify the following
Definition 3. A universal covering space is defined to be a covering space that is simply
connected. One can refer to a universal cover as the universal cover since it is unique
up to isomorphism.
Problem 2. Prove that if f : (X̃1 , x̃1 ) → (X̃2 , x̃2 ) is a morphism from a based coverp1
p2
ing (X̃1 , x̃1 ) → (X, x0 ) to a based covering (X̃2 , x̃2 ) → (X, x0 ), then f : (X̃1 , x̃1 ) →
(X̃2 , x̃2 ) itself is a based covering.
1.4
The Galois group of a covering
By the uniqueness of lifts, based covering spaces have no automorphisms. Now consider unbased coverings.
p1
p2
Definition 4. A morphism from a covering X̃1 → X to a covering X̃2 , → X is a
continuous function f : X̃1 → X̃2 satisfying p2 = f p1 .
f
X̃1
X̃2
p01
p2
X
Definition 5. An automorphism of a cover p : X̃ → X is called a covering transformation or a deck transformation. Let G(X̃) := Aut(X̃) denote the group of automorphisms of a cover p : X̃ → X. The group G(X̃) is called the Galois group or the deck
group of the cover. If necessary, one can write G(X̃, X) to indicate the dependence on
the base space X.
Now, we have two groups are associated to a covering p : (X̃, x̃) → (X, x):
namely, the fundamental group H = π1 (X̃, x̃) ⊂ G := π1 (X, x) and the Galois group
G(X̃). For spaces X satisfying a necessary local condition, we have a “Galois theory”
of covering spaces, where based covers p : (X̃, x̃) → (X, x) correspond bijectively to
subgroups H and
• Morphisms between covering spaces X̃1 → X̃2 correspond to inclusions of subgroups H1 → H2 .
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• There is a universal cover Ỹ corresponding to H = {e} and G(Ỹ ) ' G. All
other covers are obtained as quotients of the universal cover.
• A covering is called regular if G(x̃) acts transitively on the fibers. Regular covers
correspond to normal subgroups G, and for a regular cover X̃, we have G(X̃) ∼
=
G/H.
• If X̃1 → X̃2 is a morphism between regular covers that is itself a regular cover,
then G(X̃2 , X) ∼
= G(X̃1 , X)/G(X̃1 , X̃2 ).
One very good way to get insight into the automorphism group of a covering space is
by looking at an action π1 (X, x) on the fibers p−1 (x). Before describing this action,
let me state what I called “a necessary local conditon” on the space X in order to have
a Galois theory of coverings of X. Suppose that p : X̃ → X is a covering and that X̃ is
simply connected. Choose any point x ∈ X and let U be an elementary neighborhood
of x. Choose a point x̃ in p−1 (x) and let V be a component of p−1 (U ) containing x̃.
Then we have a diagram of pointed spaces
(V, x̃)
j
p
p|V
(U, x)
(X̃, x̃)
i
(X, x)
and by applying π1 we obtain the diagram of groups
π1 (V, x̃)
j∗
p∗
(p|V )∗
π1 (U, x)
π1 (X̃, x̃)
i∗
π1 (X, x)
Since p|V is a homeomorphism, (p|V )∗ is an isomorphism. The fact that (p|V )∗ is
an isomorphism and π1 (X̃, x̃) is trivial by assumption implies that the image of i∗ is
trivial. You can see that since i∗ must factor as the composition:
π1 (U, x)
(p|V )−1
∗
π1 (V, x̃)
j∗
π1 (X̃, x̃)
p∗
π1 (X, x)
i∗
and the map p∗ must be trivial.
Conclusion. If X has a universal cover, then every point x ∈ X has a neighborhood
U so that the image of i∗ : π1 (U, x) → π1 (X, x) is trivial.
It turns out that the necessary condition described in the conclusion above is sufficient. Be aware of the following terminology: a space X with the property that every
point x ∈ X has a neighborhood U so that the image of i∗ : π1 (U, x) → π1 (X, x) is
trivial is called semi-locally simply conencted.
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2
The structure of the Galois group of a covering
The Galois group of a covering p : X̃ → X acts on the fiber p−1 (x) over a point
x ∈ X. The fundamental group π1 (X, x) also acts on the fiber of p−1 (x). The Galois
group action is compatible with the action of the fundamental group and the structure
of the Galois group of a covering can be understood completely by studying the action
of π1 (X, x) on p−1 (x). First, we review the basic ideas of group actions.
2.1
Group actions
Here’s a quick review of group actions. The kind that are relevant for covering spaces
are right actions.
Definition 6. A group G acts on a set S on the right if there is a function
S×G→S
(s, g) 7→ sg
satisfying se = s for all s ∈ S and s(gg 0 ) = (sg)g 0 for all g, g 0 ∈ S and for all s ∈ S.
If G acts on a set S then every element g ∈ G determines a bijection σg : S → S
by s 7→ sg, the inverse being σg−1 . If σg = idS only for g = e ∈ G then the action is
called faithful. The action of G defines an equivalence relation on S by s ∼ s0 ⇔ ∃g ∈
G with sg = s0 . The equivalence classes of this action are called orbits. Specifically,
given an element s, the orbit of s is defined to be Os = {sg : g ∈ G}. The action
is called transitive if there is only one orbit. An element s ∈ S is called a fixed point
of the action if Os = {s}. Giving a subgroup of permutations of S is another way to
describe group actions.
For every s ∈ S, there is a subgroup Gs ⊂ G called the stabilizer or the isotropy
subgroup of s defined by Gs = {g ∈ G : sg = s}. If s ∼ s0 then the stabilizers Gs
and Gs0 are conjugate subgroups. Specifically, if sh = s0 , then Gs0 = h−1 Gs h. The
action is called free if Gs = {e} for all s ∈ S. Note an element s ∈ S is a fixed point
for the action if Gs = G.
The Orbit-Stabilizer Theorem. There is a bijection between the orbit Os and Gs \G,
the set of right cosets of Gs .
Proof. The bijection is given by Gs g 7→ sg.
If G acts on two sets S and T , a function f : S → T is G equivariant if f (sg) =
f (s)g for all s ∈ S and all g ∈ G. Sets with right G actions together with G-equivariant
maps form a category called the category of “right G sets.” If G acts transitively on S,
then S is called a homogeneous G set. Note that an automorphism of homogeneous G
sets is determined by its value at one point—any two automorphisms of homogeneous
G sets that agree at one point must be equal.
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Now, an orbit Os in a G-set S is itself a homogeneous G-set. Also, the set of right
cosets Gs \G is a G-set with the action defined by (Gs g)h = Gs gh. The bijection
Os ↔ Gs \G in the orbit-stabilizer theorem defines an isomorphism of G sets.
Theorem 2. Let S be a homogeneous G set. There exists an automorphism of S
sending s → s0 if and only if Gs = Gs0 .
Proof. Suppose S is a homogeneous G set and Gs = Gs0 . Then there exists g ∈ G
with sg = s0 . For any t ∈ G, t = sh for some h ∈ G. Define a G-set automorphism
f : S → S so that f (t) = s0 h. By contruction, if f is well defined it is an equivariant
bijection satisfying f (s) = s0 . The condition that Gs = Gs0 gaurantees that f is well
defined.
Theorem 3. Let S be a homogeneous G set and let s ∈ S. Let Gs be the stabilizer of
s and let N (Gs ) be the normalizer of Gs . Then the group of automorphisms of S is
isomorphic to N (Gs )/Gs .
Proof. Fix s ∈ S. The stabilizer of sg is g −1 Gs g, so h ∈ Gsg ⇔ sgh = sg ⇔ h ∈
g −1 Gs g. So Ssg = Gs ⇔ g ∈ N (Gs ). Therefore N (Gs )/Gs corresponds bijectively
to the set of elements in S whose stabilizer is Gs , which by the previous theorem shows
that N (Gs )/Gs corresponds bijectively to the set of G-set automorphisms of S. One
should check that the bijective corresponence is a group isomorphism.
2.2
The action of π1 (X, x) on the fibers p−1 (x) of a cover
Let G = π1 (X, x) and let S = p−1 (x). For any [γ] ∈ G and x̃ ∈ S define
x̃[γ] := the endpoint of the lift of γ to a path beginning at x̃.
Note that this definition makes S into a (right) homogeneous G set. The stabilizer
Gx̃ ' π1 (X̃, x̃). The orbit-stabilizer theorem implies that S ' G/Gs .
Corollary 2. The cardinality of S equals the index of Gs in G and is independent of
the point s. The cardinality of S is called the index of the cover.
It is straightforward to check that if f : X̃ → X̃ is a deck transformation, then f
induces an automorphism of G-sets on S = p−1 (x). Since two deck transformations
that agree at a point must be equal, the assignment of a G-set automorphism of S to a
deck transformation is injective. In fact, it is bijective.
Theorem 4. G(X̃) ∼
= Aut(S)
Proof. As a solution to the lifting problem, there exists a deck transformation mapping
x̃1 → x̃2 if and only if the subgroups π1 (X̃, x̃1 ) and π1 (X̃, x̃2 ) are equal. In terms of
G-sets, this means the stabilizer subgroups are the same Gx̃1 ' Gx̃2 , which is precisely
the condition for there to exist a G-set automorphism of S.
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Corollary 3. G(X̃) ∼
= N (π1 (X̃, x̃))/π1 (X̃, x̃).
Proof. In general, the automorphism group of a homogeneous G-set S is isomorphic
N (Gs )/Gs for any s ∈ S.
Corollary 4. If X̃ is a regular cover, then G(X̃) ∼
= π1 (X, x)/π1 (X̃, x̃).
Proof. If X̃ is regular, then G(X̃) acts transitively on the fiber, which implies all the
stabilizer subgroups of Gs are equal and normal in G, hence N (π1 (X̃, x̃)) = π1 (X, x).
Corollary 5. X̃ is a universal cover if and only if G(X̃) ∼
= π1 (X, x).
Proof. If X is simply connected, then π1 (X̃, x̃) is trivial and its normalizer is the whole
group G := π1 (X, x), and conversely.
3
The existence of a universal cover
In order to complete the Galois picture of covering spaces, we establish the existence
of covering spaces.
Theorem 5. If X is semi-locally simply connected then there exists a universal cover.
Proof. Here’s how to construct the universal cover. Choose a basepoint point x ∈ X
and consider the path space P(X, x) ' hom((I, 0), (X, x)). Note that the path space
is a pointed space (with basepoint given by the constant path) and is contractible. Let
X̃ be the quotient of P(X, x) obtained by identifying paths that are path-homotopic
and let x̃ = [cx ]. There is a natural map
p : (X̃, x̃) → (X, x)
[γ] 7→ [γ](1)
To finish the proof, check that if X is semi-locally simply connected then X̃ is a cover
of X. Then, check that G(X̃) = π1 (X, x) to conclude that X̃ is simply connected,
hence a universal cover of X.
Theorem 6. If X is semi-locally simply connected then for every subgroup H of the
fundamental group π1 (X, x) there exists a covering map p : (X̃, x̃) → (X, x) unique
up to isomorphism such that H = p∗ (π1 (X̃, x̃)).
Proof. Let Ỹ be the universal cover of X and identify H with a subgroup K ⊂ G(Ỹ ).
Then define X̃ = Y / ∼ where y ∼ y 0 if and only there exists a k ∈ K with y = ky 0 .
Choose an appropriate p. The details are left as an exercise.
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References
[1] A. Hatcher. Algebraic Topology. Cambridge Univ. Press, 2002.
[2] W. S. Massey. A basic course in Algebraic Topology. Springer Verlag, 1991.
[3] J. P. May. A Concise Course in Algebraic Topology. Univ. of Chicago Press, 1999.
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