The map p: ℝ → S¹ defined by p(t) = e^{2πit} is a covering map. What does the structure of this covering space reveal about π₁(S¹)?
Aπ₁(S¹) = 0, since ℝ is simply connected and covers S¹
Bπ₁(S¹) ≅ ℤ, since the fiber over any point consists of countably many sheets corresponding to integer lifts
Cπ₁(S¹) ≅ ℝ, since the universal cover is ℝ
Dπ₁(S¹) is trivial because the covering map is surjective
ℝ is the universal cover of S¹ — it is simply connected and the deck transformations are integer translations t ↦ t + n. The group of deck transformations of the universal cover is isomorphic to π₁(X), so π₁(S¹) ≅ ℤ. Concretely: each integer n corresponds to winding once more around the circle, and the fundamental group of the circle is generated by a single loop. The fiber p⁻¹(1) = {0, ±1, ±2, ...} — one sheet per integer — reflects this ℤ structure directly.
Question 2 Multiple Choice
Two connected covering spaces p₁: X̃₁ → X and p₂: X̃₂ → X of a path-connected, locally path-connected, semi-locally simply connected space X correspond to subgroups H₁, H₂ ⊆ π₁(X). When is X̃₁ a covering space of X̃₂ (i.e., when does X̃₁ 'cover' X̃₂ in the hierarchy)?
AWhen H₁ and H₂ have the same index in π₁(X)
BWhen H₁ ⊆ H₂ — i.e., the subgroup corresponding to X̃₁ is contained in the subgroup corresponding to X̃₂
CWhen H₂ ⊆ H₁ — i.e., the subgroup corresponding to X̃₂ is smaller
DWhen H₁ and H₂ are conjugate subgroups of π₁(X)
The Galois correspondence for covering spaces is order-reversing: a larger subgroup corresponds to a 'smaller' (fewer sheets) covering space. If H₁ ⊆ H₂, then X̃₁ has 'more' of the fundamental group unwound — it is a more thorough unfolding — and X̃₁ covers X̃₂ (there is a covering map from X̃₁ to X̃₂). The universal cover corresponds to the trivial subgroup {e} ⊆ π₁(X), which is contained in every subgroup, consistent with the universal cover covering all other covering spaces.
Question 3 True / False
The group of deck transformations of the universal cover X̃ of a path-connected, locally path-connected space X is isomorphic to π₁(X).
TTrue
FFalse
Answer: True
This is one of the deepest facts about covering spaces. A deck transformation is a homeomorphism φ: X̃ → X̃ such that p ∘ φ = p — it permutes the sheets while remaining compatible with the projection. For the universal cover (simply connected), the deck transformations act freely and transitively on each fiber, and the group they form is isomorphic to π₁(X). This means X is homeomorphic to the quotient X̃ / π₁(X), so the topology of X is fully recovered from its universal cover and the action of its fundamental group — a beautiful algebraic-topological duality.
Question 4 True / False
A covering map p: X̃ → X is injective — distinct points in X̃ are mapped to distinct points in X.
TTrue
FFalse
Answer: False
A covering map is typically many-to-one, not injective. The defining property is that it is a local homeomorphism — every point has a neighborhood that maps homeomorphically — but globally, multiple distinct points in X̃ can map to the same point in X. For p: ℝ → S¹, every point on the circle has countably infinitely many preimages (all integers differing by 1). Covering maps are local homeomorphisms, not global homeomorphisms. A covering map is injective only in the trivial case where X̃ = X and p is the identity.
Question 5 Short Answer
Explain the correspondence between covering spaces of X and subgroups of π₁(X), using the example p: ℝ → S¹ to illustrate.
Think about your answer, then reveal below.
Model answer: Given a covering space p: X̃ → X, the induced map p*: π₁(X̃) → π₁(X) is injective, and its image H = p*(π₁(X̃)) is a subgroup of π₁(X). Different covering spaces correspond to different subgroups. For ℝ → S¹: ℝ is simply connected, so π₁(ℝ) = 0 and H = {e} — the trivial subgroup. This matches the universal cover, which corresponds to the trivial subgroup. The number of sheets equals the index [π₁(X) : H]; since H = {e} and π₁(S¹) ≅ ℤ, the index is infinite, matching the infinitely many preimages of each point on S¹.
The correspondence is a complete classification: (connected covering spaces of X up to isomorphism) ↔ (conjugacy classes of subgroups of π₁(X)). For normal subgroups, the corresponding covering space is a regular/normal cover, and the deck transformation group is isomorphic to π₁(X)/H. The universal cover sits at the bottom of this hierarchy (trivial subgroup, largest unfolding) and X itself corresponds to the full group π₁(X) (trivial cover, no unfolding).