A loop on S¹ winds counterclockwise twice, then clockwise once. What is its homotopy class in π₁(S¹) ≅ ℤ?
A0, since it starts and ends at the same base point
B1, since the net winding number is 2 − 1 = 1
C3, since it made three total revolutions counting both directions
DIt is not a valid element of π₁(S¹) because it changes direction
The homotopy class is determined entirely by the winding number — how many net counterclockwise revolutions the loop completes. Going counterclockwise twice contributes +2 and clockwise once contributes −1, giving net winding number +1. The direction change is irrelevant; what matters is the total winding, which corresponds to the integer 1 in ℤ. Any two loops with the same winding number are homotopic, regardless of their specific paths.
Question 2 Multiple Choice
What makes the universal covering map p: ℝ → S¹ the key tool for computing π₁(S¹)?
AIt shows that ℝ and S¹ are homeomorphic, so they have the same fundamental group
BEvery loop γ in S¹ based at 1 lifts uniquely to a path γ̃ in ℝ starting at 0, and the endpoint γ̃(1) is always an integer equal to the winding number
CIt provides a path connecting any two points on S¹ without passing through the base point
DIt shows that S¹ is simply connected, so all loops are contractible
The covering map p(t) = e^{2πit} wraps ℝ around S¹. Its key property is the unique path lifting: every loop in S¹ lifts uniquely to a path in ℝ. Since the loop starts and ends at 1 ∈ S¹, the lifted path starts at 0 and must end at some integer n (because p(n) = 1 requires n ∈ ℤ). This integer n is exactly the winding number, and the map [γ] ↦ γ̃(1) is the isomorphism π₁(S¹) ≅ ℤ. Note that ℝ and S¹ are not homeomorphic — ℝ is simply connected (π₁(ℝ) = 0), which is precisely why lifting to ℝ is useful.
Question 3 True / False
Two loops on S¹ based at the same point are homotopic if and only if they have the same winding number.
TTrue
FFalse
Answer: True
This is the content of π₁(S¹) ≅ ℤ. Homotopy classes of loops are in bijection with integers via the winding number. Loops with the same winding number can be continuously deformed into each other; loops with different winding numbers cannot. The winding number is a complete invariant for loop homotopy on S¹.
Question 4 True / False
Since S¹ is a connected topological space, its fundamental group is expected to be trivial (i.e., nearly every loop is contractible to a point).
TTrue
FFalse
Answer: False
Connectivity and simple connectivity are different properties. A space is connected if it is in one piece; it is simply connected if every loop can be contracted to a point (i.e., π₁ = 0). The circle S¹ is connected but not simply connected: π₁(S¹) ≅ ℤ, meaning there are infinitely many distinct homotopy classes of loops. A loop that winds around the circle once cannot be continuously contracted to a point without leaving S¹. This is precisely what distinguishes S¹ from the disk D² (which is simply connected).
Question 5 Short Answer
Why must the endpoint γ̃(1) of a lifted loop γ̃: [0,1] → ℝ always be an integer, given that γ is a loop in S¹ based at 1?
Think about your answer, then reveal below.
Model answer: Because γ is a loop, its endpoint equals its starting point: γ(1) = γ(0) = 1 ∈ S¹. The lifted path γ̃ satisfies p(γ̃(t)) = γ(t) for all t, so at t = 1: p(γ̃(1)) = γ(1) = 1. Since p(t) = e^{2πit}, the condition p(γ̃(1)) = 1 means e^{2πi·γ̃(1)} = 1, which holds precisely when γ̃(1) is an integer. Thus the loop condition in S¹ forces the lift to end at an integer, and that integer is the winding number.
The integer constraint comes entirely from the loop condition combined with the definition of p. This is why the universal covering space ℝ is so useful: paths in ℝ are determined uniquely by their starting point, and the algebraic structure of ℤ ⊂ ℝ captures exactly which endpoints correspond to loops downstairs. The uniqueness of lifting guarantees that homotopic loops lift to paths with the same endpoint, making the winding number a well-defined homotopy invariant.