Proposition 1.1 Let p: X˜ → X be a covering map over a topological space X, let α: [0, 1] → X and β: [0, 1] → X be paths in X, where α(0) = β(0) and α(1) = β(1), and let                         α˜: [0, 1] → X˜ and β˜: [0, 1] → X˜ be paths in X˜ such that p ◦ α˜ = α and p ◦ β˜ = β. Suppose that α˜(0) = β˜(0) and that α ≃ β rel {0, 1}.

Then α˜(1) = β˜(1) and α˜ ≃ β˜ rel {0, 1}.

Proof Let x0 and x1 be the points of X given by

                x₀ = α(0) = β(0), x₁ = α(1) = β(1).

Now α ≅β rel {0, 1}, and therefore there exists a homotopy F: [0, 1]×[0, 1] → X such that F(t, 0) = α(t) and F(t, 1) = β(t) for all t ∈ [0, 1], F(0, τ ) = x₀ and F(1, τ ) = x₁ for all τ ∈         [0, 1].

It then follows from the Monodromy Theorem (The Monodromy Theorem)that there exists a continuous map G: [0, 1] × [0, 1] → X˜ such that p ◦ G = F and G(0, 0) = α˜(0). Then p(G(0, τ )) = x₀ and p(G(1, τ )) = x₁ for all τ ∈ [0, 1].

A straightforward application of (Let p: X~ → X be a covering map, let Z be a connected topological space, and let g:Z → X~ and h:Z → X~ be continuous maps.

Suppose that p ◦g = p ◦ h and that g(z) = h(z) for some z ∈ Z. Then g = h.) shows that any continuous lift of a constant path must itself be a constant path. Therefore G(0, τ ) = x˜₀

and G(1, τ ) = x˜₁ for all τ ∈ [0, 1], where x˜₀ = G(0, 0) = α˜ (0), x˜₁ = G(1, 0).

However

         G(0, 0) = G(0, 1) = ˜x0 = ˜α(0) = β˜(0),

         p(G(t, 0)) = F(t, 0) = α(t) = p(˜α(t))

and

           p(G(t, 1)) = F(t, 1) = β(t) = p(β˜(t))

for all t ∈ [0, 1]. Now Proposition (Let p: X~ → X be a covering map, let Z be a connected topological space, and let g:Z → X~ and h:Z → X~ be continuous maps.  Suppose that p ◦g = p ◦ h and that g(z) = h(z) for some z ∈ Z. Then g = h.)ensures that the lifts α˜and β˜ of the paths α and β are uniquely determined by their starting points. It follows that               G(t, 0) = α˜ (t) and G(t, 1) = β˜(t) for all t ∈ [0, 1]. In particular,  

    α˜(1) = G(0, 1) = x˜₁ = G(1, 1) = β˜(1).

Moreover the map G: [0, 1] × [0, 1] → X˜ is a homotopy between the paths α˜ and β˜ which satisfies G(0, τ ) = x˜₀ and G(1, τ ) = x˜₁ for all τ ∈ [0, 1]. It follows that α˜ ≃β˜                       rel {0, 1}, as required

Corollary 1.2 Let p: X˜ → X be a covering map over a topological space X, and let x˜0 be a point of X˜. Then the homomorphism

          p_#: π₁(X˜, x˜₀) → π₁(X, p(x˜₀))

of fundamental groups induced by the covering map p is injective

Proof Let σ₀ and σ₁ be loops in  X~ based at the point  x~₀, representing elements [σ₀] and [σ1] of π₁(X˜, x˜₀). Suppose that p_#[σ₀] = p_#[σ₁]. Then p◦σ₀ ≃ p ◦ σ₁ rel {0, 1}. Also σ₀(0) = x˜₀ = σ₁(0). Therefore σ₀ ≃σ₁ rel {0, 1}, by Proposition 1.1, and thus [σ₀] = [σ₁]. We conclude that the homomorphism  p_#: π₁(X˜, x˜₀) → π₁(X, p(x˜₀)) is injective

Corollary 1.3 Let p: X˜ → X be a covering map over a topological space X, let x˜₀ be a point of X˜, and let γ be a loop in X based at p(x˜₀). Then  [γ] ∈ p_#(π₁(X˜, x˜₀)) if and only if there exists a loop γ˜ in X˜, based at the point x˜₀, such that p ◦ γ˜ = γ.

Proof If γ = p ◦ γ˜ for some loop γ˜ in X˜ based at x˜₀ then [γ] = p_#[γ˜], and therefore [γ] ∈ p_#(π₁(X˜, x˜₀)).

Conversely suppose that [γ] ∈ p_#(π₁(X˜, x˜₀)). We must show that there exists some loop γ˜ in X˜ based at x˜₀ such that γ = p ◦ γ˜. Now there exists a loop σ in X˜ based at the point x ˜₀ such that [γ] = p_#([σ]) in π₁(X, p(x˜₀)).

Then γ ≃ p ◦ σ rel {0, 1}. It follows from the Path Lifting Theorem for covering maps that there exists a unique path γ˜: [0, 1] → X˜ in X˜ for which γ˜ (0) = x˜₀ and p ◦ γ˜ = γ. It then follows from Proposition 1.1 that γ˜ (1) = σ(1) and γ˜ ≃ σ rel {0, 1}. But σ(1) = x˜₀. Therefore the path γ˜ is the required loop in X˜ based the point x˜₀ which satisfies                  p ◦ γ˜ = γ.

Corollary 1.4 Let p: X˜ → X be a covering map over a topological space X, let w₀ and w₁ be points of X˜ satisfying p(w₀) = p(w₁), and let α: [0, 1] → X˜ be a path in X˜ from w0 to w₁. Suppose that [p ◦ α] ∈ p_#(π₁(X, w˜₀)). Then the path α is a loop in X˜, and thus w₀ = w₁.

Proof It follows from Corollary 1.3 that there exists a loop β based at w0 satisfying p ◦ β = p ◦ α. Then α(0) = β(0). Now Proposition (Let p: X˜ → X be a covering map, let Z be a connected topological space, and let g:Z → X˜ and h:Z → X˜ be continuous maps. Suppose that p ◦ g = p ◦ h and that g(z) = h(z) for some z ∈ Z. Then g = h.) ensures that the lift to X˜ of any path in X is uniquely determined by its starting point. It follows that α = β. But then the path α must be a loop in X˜, and therefore w₀ = w₁, as required.

Corollary 1.5 Let p: X˜ → X be a covering map over a topological space X.  Let α: [0, 1] → X and β: [0, 1] → X be paths in X such that α(0) = β(0) and α(1) = β(1), and let α.β⁻¹ be the loop in X defined such that

Let α˜: [0, 1] → X˜ and β˜: [0, 1] → X˜ be the unique paths in X˜ such that p ◦ α˜ = α, and p ◦ β˜ = β. Suppose that α˜(0) = β˜(0). Then α˜(1) = β˜(1) if and only if                                [α.β⁻¹] ∈ p_#(π₁(X˜, x˜₀)), where x˜₀ = α˜(0) = β˜(0).

Proof Suppose that α˜(1) = β˜(1). Then the concatenation α˜.β˜⁻¹ is a loop in X˜ based at x˜₀, and [α.β⁻¹] = p_#([α˜.β˜⁻¹]), and therefore [α.β⁻¹] ∈ p_#(π₁(X ˜, x˜₀)).  Conversely suppose that α˜ and β˜ are paths in X˜ satisfying p ◦ α˜ = α,  p ◦ β˜ = β and α˜(0) = β˜(0) = x˜₀, and that [α.β⁻¹] ∈ p_#(π₁(X˜, x˜₀)). We must show that α˜(1) = β˜(1). Let γ: [0, 1] → X be the loop based at p(x˜₀) given by γ = α.β⁻¹. Thus

Then [γ] ∈ p_#(π₁(X˜, x˜₀)). It follows from Corollary 1.3 that there exists a loop γ˜ in X˜ based at x˜₀ such that p ◦ γ˜ = γ. Let αˆ: [0, 1] → X˜ and Let βˆ: [0, 1] → X˜ be the paths in X˜ defined such that αˆ (t) = γ˜ (1/2t)    and    βˆ(t) = γ˜ (1 −1/2t) for all t ∈ [0, 1]. Then

           α˜(0) = αˆ (0) = β˜(0) = βˆ(0) = x˜₀,  p ◦ αˆ = α = p ◦ α˜ and p ◦ βˆ = β = p ◦ β˜.

But Proposition (Let p: X˜ → X be a covering map, let Z be a connected topological space, and let g:Z → X˜ and h:Z → X˜ be continuous maps. Suppose that p ◦ g = p ◦ h and that g(z) = h(z) for some z ∈ Z. Then g = h.) ensures that the lift to X˜ of any path in X is uniquely determined by its starting point.

Therefore α˜ = α ˆand β˜ = βˆ. It follows that

α˜(1) = α(1ˆ) = γ˜ (1/2) = βˆ(1) = β˜(1),

as required.

Theorem 1.6 Let p: X˜ → X be a covering map over a topological space X.

Suppose that X˜ is path-connected and that X is simply-connected. Then the covering map p: X˜ → X is a homeomorphism.

Proof We show that the map p: X˜ → X is a bijection. This map is surjective  (since covering maps are by definition surjective). We must show that it is injective. Let w₀ and w₁ be points of X˜ with the property that p(w₀) = p(w₁). Then there exists a path α: [0, 1] → X˜ with α(0) = w0 and α(1) = w₁, since X˜ is path-connected. Then p ◦ α is a loop in X based at the point x₀, where x₀ = p(w₀). However π₁ (X,p(w₀)) is the trivial group, since X is simply- connected. It follows from Corollary 1.4 that the path α is a loop in X˜ based at w₀, and therefore w₀ = w₁. This shows that the the covering map p: X˜ → X is injective. Thus the map p: X˜ → X is a bijection, and thus has a well-defined inverse p⁻¹: X → X˜. It now follows from Lemma (Let p: X^~ → X be a covering map. Then p(V ) is open in X  for every open set V in  X^~. In particular, a covering map p: X^~ → X is a homeomorphism if and only if it is a bijection.) that p: X˜ → X is a homeomorphism, as required.

Let p: X˜ → X be a covering map over some topological space X, and let x₀ be some chosen basepoint of X. We shall investigate the dependence of the subgroup p_#(π₁(X˜, x˜)) of π₁(X, x₀) on the choice of the point x˜ in X˜,  where x˜ is chosen such that p(x˜) = x₀. We first introduce some concepts from group theory.

Let G be a group, and let H be a subgroup of G. Given any g ∈ G, let gHg⁻¹ denote the subset of G defined by

gHg⁻¹ = {g₀ ∈ G : g₀ = ghg⁻¹ for some h ∈ H}.

It is easy to verify that gHg⁻¹ is a subgroup of G.

Definition Let G be a group, and let H and H0 be subgroups of G. We say that H and H₀ are conjugate if and only if there exists some g ∈ G for which H₀ = gHg⁻¹ Note that if                    H₀ = gHg⁻¹      then  H = g⁻¹H₀g. The relation of conjugacy is an equivalence relation on the set of all subgroups of the group G. Moreover conjugate subgroups of G are isomorphic, since the homomorphism sending h ∈ H to ghg⁻¹ is an isomorphism from H to gHg⁻¹ whose inverse is the homorphism sending h 0 ∈ gHg⁻¹ to g⁻¹h₀g.

A subgroup H of a group G is said to be a normal subgroup of G if ghg⁻¹ ∈ H for all h ∈ H and g ∈ G. If H is a normal subgroup of G then gHg⁻¹ ⊂ H for all g ∈ G. But then g⁻¹Hg ⊂ H and H = g(g⁻¹Hg)g⁻¹ for all g ∈ G, and therefore H ⊂ gHg⁻¹ for all g ∈ G. It follows from this that a subgroup H of G is a normal subgroup if and only if gHg⁻¹ = H for all g ∈ G. Thus a subgroup H of G is a normal subgroup if and only if there is no other subgroup of G conjugate to H.

Lemma 1.7 Let p: X˜ → X be a covering map over a topological space X.

Let x₀ be a point of X, and let w₀ and w₁ be points of X˜ for which p(w₀) = x₀ = p(w₁). Let H₀ and H₁ be the subgroups of π₁(X, x₀) defined by

H₀ = p_#(π₁(X, w˜₀)), H₁ = p_#(π₁(X, w˜1)).

Suppose that the covering space X˜ is path-connected. Then the subgroups H₀ and H₁ of π₁(X, x₀) are conjugate. Moreover if H is any subgroup of π₁(X, x₀) which is conjugate to H0 then there exists an element w of X˜ for which p(w) = x and p_#(π₁(X, w˜ )) = H.

Proof Let α: [0, 1] → X˜ be a path in X˜ for which α(0) = w₀ and α(1) = w₁. (Such a path exists since X˜ is path-connected.) Then each loop σ in X˜ based at w₁ determines a corresponding loop α.σ.α⁻¹ in X˜ based at w0, where

(This loop traverses the path α from w0 to w1, then continues round the loop σ, and traverses the path α in the reverse direction in order to return from w₁ to w₀.) Let                  η: [0, 1] → X be the loop in X based at the point x₀ given by η = p ◦ α, and let ϕ: π₁(X, x₀) → π₁(X, x₀) be the automorphism of the group π₁(X, x₀) defined such that ϕ([γ]) = [η][γ][η] ⁻¹ for all loops γ in X based at the point x₀. Then p ◦ (α.σ.α⁻¹) = η.(p ◦ σ).η⁻¹, and therefore p_#([α.σ.α⁻¹]) = [η]p_#([σ])[η]⁻¹ = ϕ(p_#([σ])) in π₁(X, x₀). It follows that ϕ(H₁) ⊂ H₀. Similarly ϕ⁻¹ (H₀) ⊂ H₁, where ϕ⁻¹ ([γ]) = [η]⁻¹ [γ][η] for all loops γ in X based at the point x₀. It follows that ϕ(H₁) = H₀, and thus the subgroups H₀ and H₁ are conjugate Now let H be a subgroup of π₁(X, x₀) which is conjugate to H₀. Then H₀ = [η]H[η]⁻¹ for some loop η in X based at the point x₀. It follows from the Path Lifting Theorem for covering maps (Theorem (Path Lifting Theorem)  that there exists a path α: [0, 1] → X˜ in X˜ for which α(0) = w₀ and p ◦ α = η. Let w = α(1).

Then p_#(π₁(X, w˜ )) = [η]⁻¹H₀[η] = H,  as required.