Moreover if U is any proper open subset of S1, then p−1(U ) is a disjoint union of open subsets {Ui} of R such that p : Ui→ U are homeomorphisms for all i

1. Fundamental Group of S¹.

Let p : R → S¹ be the map x 7→ e^2πix. For any open interval (a, b) with b − a < 2π, p : (a, b) → S¹ is a homeomorphism onto its image p(a, b). Moreover if U is any proper open subset of S¹, then p⁻¹(U ) is a disjoint union of open subsets {U_i} of R such that p : Ui→ U are homeomorphisms for all i. This is very easy to see from the map p. This motivates the definition of covering space.

Definition 1.1. We say that a continuous map π : E → X is a covering space of X if for each x ∈ X, there exists an open neighborhood U of x so that p⁻¹(U ) is a disjoint union of open subsets {U_i} of E with the property that p : U_i → U is homeomorphisms for all i.

This definition says that p : R → S¹ given by x 7→ e^2πix is a covering space of S¹. Proposition 1.1. (Lifting Lemma) Let σ : [0, 1] → S¹ be a path in S¹ with initial point 1. Then there exists a unique path σ⁰ in R with initial point 0 such that p ◦ σ⁰ = σ.

Proof. Let us prove the uniqueness first. Suppose σ⁰⁰ is another path with initial point 1 such that p ◦ σ⁰⁰ = σ. Then e^{2πi(σ0(t)−σ00(t))} = 1 for all t ∈ [0, 1]. Hence the image of σ⁰− σ⁰⁰ lies in Z. Since both σ⁰ and σ⁰⁰ are continuous map, so is σ⁰− σ⁰⁰ Since [0, 1] is connected, by the continuity of σ⁰− σ⁰⁰, the image of σ⁰− σ⁰⁰ is also connected. Hence σ⁰− σ⁰⁰ must be a constant function. Since σ⁰(0) = σ⁰⁰(0) = 0, we see that σ⁰(t) − σ⁰⁰(t) = 0 for all t ∈ [0, 1]

and thus σ⁰= σ⁰⁰ on [0, 1].

Since [0, 1] is compact and σ is continuous on [0, 1], σ is uniformly continuous. Hence there exists δ > 0 so that |f (t) − f (t⁰)| < 1/2 for |t − t⁰| < δ. We can choose a partition 0 = t0 < · · · < tn= 1 such that |ti− t_i+1| < δ/2 for 0 ≤ i ≤ n − 1.

Let B(a, ) be the open ball in C centered at a of radius . Consider the open subset Ui = B(ti, 1/2) ∩ S¹. Then for each i, p⁻¹(Ui) =S

αUi,α, where {Ui,α} is a family of disjoint open subsets of R so that p : Ui,α → U_i is a homeomorphism for all α. For i = 0, we choose U_0,α0 so that 0 ∈ U_0,α0. We see that σ|_[t0,t1] ⊂ U₀. We define σ₁⁰ : [t₀, t₁] → R by σ₁⁰ = p|⁻¹_U

0,α0 ◦ σ|_[t0,t1]. Then σ⁰₁ is a path in R with initial point 0 so that p ◦ σ1⁰ = σ|_[t0,t1]. Now we shall construct the path inductively. Assume that we have already construct a path σ_i⁰ : [0, ti] → R such that p ◦ σi⁰ = σ|_[0,ti]. We choose Ui,αi such that σ_i⁰(ti) ∈ Ui,αi. We define σ_i+1⁰ : [0, t_i+1] → R by

σ⁰_i+1= σ_i⁰? (p|⁻¹_U

i,αi ◦ σ|_[ti,ti+1]).

Then p ◦ σ_i+1⁰ = σ|_[0,ti]? σ|_[ti,ti+1] = σ|_[0,ti+1]. Here ? means the concatenation of curves.

Then we have already construct a curve σ⁰ : [0, 1] → R such that p ◦ σ⁰ = σ. In fact, the construction does not depend on the choice of partition of [0, 1] by the uniqueness proved above.

 We call σ⁰ the lift of σ.

Proposition 1.2. (Covering Homotopy Lemma) Let τ be another path in S¹ with initial point 1. Assume that

F : σ ' τ rel(0, 1).

Then there is a unique F⁰: I × I → R such that

F : σ⁰ ' τ⁰ rel(0, 1) and φ ◦ F⁰ = F.

1

2

Proof. The proof of the uniqueness of F⁰ is similar as above. We use the connectivity of I × I.

The interval I ×I is compact and F is continuous on I ×I. Hence F is uniformly continuous on I × I. Let U_i be as above for all i. For each point s ∈ I, we can find a neighborhood N_s of s and a partition {ti} of [0, 1] such that f (N × [t_i, ti+1]) ⊂ Ui for all i.

Let FNs,1 : Ns× [0, t₁] → R be the map defined by F_N⁰_s,1 = p|⁻¹_U

0,α0 ◦ F |_Ns×[0,t1]. Then p ◦ F_N⁰

s,1 = F |_Ns×[0,t1]. We can construct a map inductively F_N⁰

s : N × [0, 1] → R such that p ◦ F_N⁰_s = F |_Ns×[0,1].

Let N_s⁰ be another open neighborhood of s. Then we can construct a map F_N⁰ 0

s such that p◦F_N⁰ 0

s = F |_N⁰

s×[0,1]. We know that N_s⁰∩N_sis an open neighborhood of s. By the uniqueness, we know that F_N⁰

s|_Ns∩N⁰

s = F_N⁰

s|_Ns∩N⁰

s. We can use this to construct a continuous map F_N⁰_s∪N0

s : (Ns∪ N_s⁰) × [0, 1] → R so that p ◦ F_N⁰_s∪N_s⁰ = F |_(Ns∪N_s⁰)×[0,1].

Since {N_s : s ∈ [0, 1]} forms an open cover of [0, 1] and [0, 1] is compact, we can find s1, · · · , sn such that {Nsi : 1 ≤ i ≤ n} forms a finite open subcover of [0, 1]. Let Ns1 be the open neighborhood of s₁ containing 0 and denote F₁⁰ = F_Ns1. Let N_s2 be an open neighborhood so that N_s2∩ N_s1 6= ∅. Let F₂⁰ = F_Ns1∪N_s2. Inductively assume that N_sj is an open set such that N_sj∩ (Sj−1

i=1N_si) 6= ∅ with F_j−1⁰ is constructed. We can define F_j⁰ to be F_(N⁰

s1∪···N_sj−1)∪N_sj. Then we set F⁰ = F_n⁰. It is easy to see that p ◦ F⁰ = F. Moreover if F is a homotopy between σ and τ, then F⁰ is a homotopy between their lifts σ⁰ and τ⁰. This

completes the proof of our assertion. 

This proposition immediately implies that:

Corollary 1.1. The end point of σ⁰ depends only on the homotopy class of σ.

Theorem 1.1. The fundamental group π₁(S¹, 1) is isomorphic to Z.

Proof. Let σ be a loop of S¹ at 1 and σ⁰ be its lift in R. Since σ is a loop, σ(1) = e^2πiσ0(1) = 1.

Hence σ⁰(1) ∈ Z. For each [σ] ∈ π1(S¹, 1), we choose a representative of σ and define χ : π1(S¹, 1) → Z

by χ([σ]) = σ⁰(1). Then χ is a well-defined map by Corollary 1.1.

Let [σ] and [τ ] in π₁(S¹, 1). Choose representatives σ and τ of [σ] and [τ ] respectively.

Denote σ⁰(1) = m and τ⁰(1) = n. Then σ⁰(1) + τ⁰(1) = m + n. Let τ⁰⁰ be the path from m to m + n defined by τ⁰⁰= τ⁰+ m. We know that σ⁰? τ⁰⁰ is the lift of σ ? τ with the initial point 0 by the uniqueness of the lift. Hence

χ([σ][τ ]) = χ([σ ? τ ]) = (σ⁰? τ⁰⁰)(1) = m + n = χ([σ]) + χ([τ ]).

Therefore χ : π₁(S¹, 1) → Z is a group homomorphism.

Given n ∈ Z, define σ⁰(t) = nt for t ∈ [0, 1] and σ(t) = e^2πint. Then χ([σ]) = n. Hence χ is surjective. If [σ] ∈ ker χ, then σ⁰(1) = 0, i.e. σ⁰ is a loop at 0. Then we know that σ⁰ is homotopic to the constant loop e0 at 0. Hence σ is homotopy to the constant loop e₁= e^2πie0 at 1. Hence ker χ is trivial. We complete the proof.  Proposition 1.3. Let (X₁, x₁) and (X₂, x₂) be two pointed spaces. Then there is a natural identification:

π₁(X₁× X₂, (x₁, x₂)) ∼= π1(X₁, x₁) × π₁(X₂, x₂).

3

Proof. Let p1 and p2 be the projections of X1 × X₂ to X1 and X2 respectively. Then p1

and p₂ induce group homomorphisms:

(pi)∗ : π1(X1× X₂, (x1, x2)) → π1(Xi, xi), i = 1, 2.

Hence we obtain a group homomorphism

((p1)∗, (p2)∗) : π1(X1× X₂, (x1, x2)) → π1(X1, x1) × π1(X2, x2).

This group homomorphism is in fact an isomorphism. Let us construct its inverse.

Let σ1 and σ2 be loops at x1 and x2 respectively. Then we define a loop σ1× σ₂ at (x₁, x₂) by setting

(σ₁× σ₂)(t) = (σ₁(t), σ₂(t)).

For any given [σ_i] ∈ π₁(X_i, x_i), i = 1, 2, we choose a representative σ_i of [σ_i] for each i.

Then we consider the map

π1(X1, x1) × π1(X2, x2) → π1(X1× X₂, (x1, x2)) given by ([σ₁], [σ₂]) 7→ [σ₁× σ₂]. Then this is the required inverse map.

 If we take T = S¹× S¹, then we obtain immediately that

π₁(T, (1, 1)) ∼= Z².