1. Covering Space
A continuous map p : E → X is said to be a covering space of X if for each x ∈ X has an open neighborhood U and a family of disjoint open subsets {Ui} of E such that
(1) p−1(U ) =`
iUi
(2) each map p|Ui : Ui→ U is a homeomorphism.
We say that a morphism p : (E, e0) → (X, x0) between pointed spaces is a covering space if p : E → X is a covering space and p(e0) = x0.
Proposition 1.1. (Unique Lifting Theorem) Let p : (E, e0) → (X, x0) be a covering space.
Let f : (Y, y0) → (X, x0) be any morphism. Assume that Y is connected. If there exists a morphism f0 : (Y, y0) → (E, e0) such that p ◦ f0 = f, it is unique.
Proof. Let f00: (Y, y0) → (E, e0) be a morphism such that p ◦ f00= f. We want to show that f00= f0. Let A = {y ∈ Y : f0(y) = f00(y)}. Since f00(y0) = f0(y0) = e0, y0 ∈ A. Therefore A is nonempty. Let y be a point in A and z = f0(y) = f00(y). Then pf0(y) = pf00(y) = f (y).
Denote x = f (y). Since p is a covering space of X, we choose an open neighborhood U of x so that p−1(U ) is a disjoint union of open sets {Ui} in E so that each p : Ui → U is a homeomorphism. Choose i so that z ∈ Ui. Let us consider V = (f0)−1(Ui) ∩ (f00)−1(Ui).
Then V is an open neighborhood of y. Moreover, for all v ∈ U, f0(v) and f00(v) both lie in Ui. Since pf0(v) = pf00(v) = f (v), by p : Ui → U is a homeomorphism, f0(v) = f00(v).
Hence V ⊂ A. This implies that y is an interior point of A. Since y is arbitrary in A, A is open. To show that A is closed, we need to show that D = Y \ A is open, i.e.
D = {y ∈ Y : f0(y) 6= f00(y)} is open.
Let y ∈ D. Then f0(y) 6= f00(y). Since p(f0(y)) = p(f00(y)) = f (y) in X, we know f0(y) ∈ Ui and f00(y) ∈ Uj for some i, j so that Ui∩ Uj = ∅. Consider W = (f0)−1(Ui) ∩ (f00)(Uj).
Then W is open in Y. Moreover, for all v ∈ W, f0(v) 6= f00(v) (since Ui∩ Uj 6= ∅.) Therefore W ⊂ D, i.e. y is an interior point of D. Since y is arbitrary in D, D is open in Y . We see that A = Y \ D is closed.
Since Y is connected, A is a nonempty closed and open subset of Y, Y = A. This implies
that f0 = f00. We complete the proof.
Proposition 1.2. (Path Lifting Theorem) Let p : (E, e0) → (X, x0) be a covering space. if σ is a path with initial point x0, then there exists a unique path σ0 in E with initial point e0 such that p ◦ σ0 = σ.
Proof. Let Y = [0, 1]. Using the previous lemma, if σ0 : (Y, 0) → (E, e0) exists, then it is unique. We have already construct σ0 in the case E = R and X = S1. The construction to
the general case is similar.
Proposition 1.3. Let p : (E, e0) → (X, x0) be a covering space. Let f : (Y, y0) → (X, x0) be a morphism of pointed spaces. Suppose that f0 : (Y, y0) → (E, e0) is a lift of f. Then any homotopy F : Y × I → X with F (y, 0) = f (y) can be lifted to a homotopy F0: Y × I → E with F0(y, 0) = f0(y).
Proof. The construction of F0 is similar to that in the case when E = R and X = S1. Corollary 1.1. If σ and τ are paths in X with initial point x0 and σ ' τ rel(0, 1). Then
σ0 ' τ0 rel(0, 1).
In particular, σ0 and τ0 the the same end point.
Corollary 1.2. p∗ : π1(E, e0) → π1(X, x0) is a monomorphism.
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Proof. Let [σ0] ∈ ker p∗. Then [p ◦ σ0] = [ex0]. In other words, p ◦ σ0 ' ex0 rel(0, 1). Then we know that the constant loop e0 is the lift of ex0. By the previous corollary, σ0' e0 rel(0, 1).
In other words, [σ0] = [e0] in π1(E, e0).
Notice that the lift σ0 of a loop σ at x0 in X need not be to a loop in E but will be a point in p−1(x0). For each [σ] ∈ π1(X, x0), we can define a point σ0e(1) ∈ p−1(x0), where σ is a representative of [σ] and σ0 is its lift in E with initial point e. This is well-defined by Corollary 1.1. Hence we obtain an action of π1(X, x0) on p−1(x0) by
e · [σ] = σ0e(1).
The stabilizer of e0 is the subgroup of π1(X, x0) consisting of [σ] so that e0· [σ] = e0.
In other words, σe0
0(1) = e0. Hence σe00 is a loop in E at e0. Hence p ◦ σe00 = σ implies that p∗[σe00] = [σ]. In other words, the stabilizer of e0 is the image of p∗.
When E is path connected, the action of π1(X, x0) on p−1(x0) is transitive. Hence p−1(x0) can be identified with the coset space G/Ge0. In other words, we have:
Corollary 1.3. Let E be a path connected space. Then the map [σ] 7→ e0· [σ] induces a bijection between the set of all cosets of Im p∗ onto the fiber.
Let p : E → X be any covering space of X. An automorphism of p : E → X is a homeomorphisms f : E → E so that p ◦ f = p. The group of automorphisms of p is denoted by Aut(p).
Theorem 1.1. Let p : (E, e0) → (X, x0) be a covering space. Suppose that E is simply connected. Then
Aut(p) ∼= π1(X, x0).
Proof. Let φ : E → E be an automorphism of p : (E, e0) → (X, x0). Since p ◦ φ = p, we find p ◦ φ(e) = x0 for all e ∈ p−1(x0). Hence φ induces a map on the fiber p−1(x0), i.e.
φ : p−1(x0) → p−1(x0).
Now, we want to construct a class in π1(X, x0) from φ. Since e0 ∈ p−1(x0), and E is path connected, we can choose a path σ0 connecting e0 and φ(e0). Then p ◦ σ0 is a loop at x0 in X. It is natural to consider the class [p ◦ σ0] in π1(X, x0). By π1(E, e0) = {0}, we can see that [p ◦ σ0] is independent of the choice of σ0. Then we set
χ : Aut(p) → π1(X, x0)
by χ(φ) = [p ◦ γ0]. It is routine to check that χ is a group homomorphism.
Let us check that χ is a monomorphism. Assume that χ(φ) = [ex0]. In other words, p◦σ0 ' ex0 rel (0, 1). Here σ0is the path connecting e0and φ(e0). Hence σ0 ' e0 rel (0, 1).We know that φ(e0) = e0. Since idE : (E, e0) → (E, e0) is another automorphism of p so that idE(e0) = e0, by the unique lifting theorem, we know φ = idE. Hence χ is a monomorphism (since E is connected.).
Let us prove χ is surjective. Let [σ] ∈ π1(X, x0). We want to construct an automorphism of p : (E, e0) → (X, x0). Let e ∈ E be any point. Choose a path τ0 connecting e0 and e and denote τ = p ◦ τ0. Set x = p(e). Then τ−1στ is a loop at x. Hence e · [τ−1στ ] is an element in p−1(e). We define
φ(e) = e · [τ−1στ ].
Then we want to check that φ : E → E is well-defined. Since π1(E, e0) = 0, any two paths connecting e0 and e are homotopic. Hence φ is well-defined (it depends only on [σ].). We
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check that χ(φ) = [σ] if we can prove that φ is a homeomorphism. To show that φ and φ−1 are continuous, we use the fact that E is locally path connected. Definition 1.1. If p : (E, e0) → (X, x0) is a covering space of X such that E is simply connected, we call p the universal covering space. (One can show that is unique up to isomorphisms in the category of pointed spaces if it exists).
A space X is said to be locally contractible if for each x ∈ X there exists an open neighborhood U of x so that U is contractible.
Theorem 1.2. Let (X, x0) be a pointed space. If X is connected and locally contractible, then X has a universal covering space E.
Proof. The construction of E depends on the based point x0 of X. Let X be the set of all paths in X with initial point x0. We write α ∼ β if α ' β rel (0, 1). Then ∼ is an equivalence relation on X . Let E = X / ∼ be the quotient space. Define p : E → X by p[α] = α(1).
Let us equip E a topology as follows. Let (α, V ) be the subset of E consisting of all [αβ], where V is an open neighborhood of p[α] and β is a path in V with initial point α(1).
Then the family {(α, V )} forms a subbase for a topology on E such that p is continuous and open.
Since X is locally constractible, one can show that p is a covering space. By the con-
struction of E, E is simply connected.
Corollary 1.4. Under the same hypothesis, for every subgroup H of π1(X, x0), there exists a covering space p : (E, e0) → (X, x0) unique up to equivalence so that H = p∗π1(E, e0).
Proof. Let eX be the universal covering space for X and G be its automorphism group.
Identify G with π1(X, x0). Then the subgroup H of G acts on eX. Hence we take E = eX/H.
This is the required covering space.
Definition 1.2. A covering space p : E → X is called normal if for each x ∈ X, and each pair of lifts e, e0 of x, there is an automorphism of p sending e to e0.
Proposition 1.4. (Lifting Criterion) Let p : (E, e0) → (X, x0) be a covering space and f : (Y, y0) → (X, x0) be a morphism with Y path connected and locally path connected.
Then a lift f0 : (Y, y0) → (E, e0) of f exists if and only if f∗π1(Y, y0) ⊂ p∗π1(E, e0).
Proof. The only if part is obvious.
Let y ∈ Y and γ be a path connecting y0 and y. Then f ◦ γ is a path starting at x0 to f (y). Then there is a unique lift (f ◦ γ)0 starting at e0 in E. We define
f0(y) = (f ◦ γ)0(1).
Then we can check that f0 : Y → E is a well-defined map if f∗π1(Y, y0) ⊂ p∗π1(E, e0). The continuity of f0 follows from the fact that Y is locally path connected (and p is a covering space).
Proposition 1.5. A covering p : E → X is a normal covering iff H = p∗π1(E, e0) is a normal subgroup of π1(X, x0).
Proof. We see that [γ] ∈ π1(X, x0) is in N (H) if and only if p∗π1(E, e) = p∗π1(E, e0). By the lifting criterion, it is equivalent to the existence of a deck transformation taking e0 to e. Hence it is equivalent to say that H is a normal subgroup of π1(X, x0).
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Theorem 1.3. Let p : (E, e0) → (X, x0) be a covering space. Then Aut(p) ∼= N (H)/H, where N (H) is the normalizer of H in π1(X, x0). In particular, if p is a normal covering, then Aut(p) is isomorphic to π1(X, x0)/H.
Proof. Let [γ] ∈ N (H). Then construct an automorphism φ from γ sending e0 to e for some e ∈ E. Then define χ : N (H) → Aut(p) by [γ] → φ. (The construction is similar as above).
Then χ is a surjective homomorphism. Its kernel consists of classes [γ] lifting to loops in E. These are exactly p∗π1(X, x0) = H. Hence we proved that
N (H)/H ∼= Aut(p).
The rest of the proof is given as follows. If p is a normal covering, N (H) = π1(X, x0).
