Questions: Homotopy Exact Sequence of a Fibration

Question 1

The Hopf fibration S^1 → S^3 → S^2 has fiber S^1, total space S^3, and base S^2. Using the long exact sequence, what is π_2(S^2)?

• A. 0
• B. Z
• C. Z/2Z
• D. Z ⊕ Z

Answer: (B)

The long exact sequence gives: ... → π_2(S^3) → π_2(S^2) →^∂ π_1(S^1) → π_1(S^3) → ... Now π_2(S^3) = 0 and π_1(S^3) = 0 (S^3 is 2-connected). By exactness, the connecting homomorphism ∂: π_2(S^2) → π_1(S^1) is an isomorphism. Since π_1(S^1) ≅ Z, we get π_2(S^2) ≅ Z. This recovers the known result via a completely different method from Hurewicz or Mayer-Vietoris.

Question 2

For a covering space p: E → B with fiber F (a discrete set of n points), the long exact sequence of the fibration gives π_k(E) ≅ π_k(B) for all k ≥ 2.

• T. True
• F. False

Answer: True

If F is discrete, then π_k(F) = 0 for all k ≥ 1 (discrete spaces have trivial higher homotopy groups). The long exact sequence gives: ... → π_k(F) → π_k(E) → π_k(B) → π_{k-1}(F) → ... For k ≥ 2, both π_k(F) = 0 and π_{k-1}(F) = 0, so by exactness, π_k(E) → π_k(B) is an isomorphism. This is a well-known property of covering spaces: they share all higher homotopy groups with the base. Only π_1 can differ, and the exact sequence relates π_1(E) to π_1(B) via the fundamental group of the fiber.

Question 3

A fibration p: E → B with contractible total space E has π_n(B) ≅ π_{n-1}(F) for all n ≥ 1.

• T. True
• F. False

Answer: True

If E is contractible, then π_n(E) = 0 for all n ≥ 1. The long exact sequence gives: ... → 0 → π_n(B) →^∂ π_{n-1}(F) → 0 → ... By exactness, the connecting homomorphism ∂ is an isomorphism for all n ≥ 1. This is the key computation for path spaces and loop spaces: the path space PB → B has contractible total space and fiber ΩB (the loop space), giving π_n(B) ≅ π_{n-1}(ΩB). This relationship between a space and its loop space is fundamental to homotopy theory.

Question 4

Explain the difference between a fibration and a fiber bundle, and why the long exact sequence applies to fibrations.

Model answer: A fiber bundle has local trivializations: every point of the base has a neighborhood U such that p^{-1}(U) ≅ U × F. A fibration is more general: it only requires the homotopy lifting property (any homotopy in the base can be lifted to the total space). Every fiber bundle over a paracompact base is a fibration, but not every fibration is a fiber bundle. The long exact sequence requires only the homotopy lifting property, not local triviality, so it applies to all fibrations. This generality is important because many natural constructions (path spaces, mapping spaces) produce fibrations that are not fiber bundles.

The homotopy lifting property is the key: it allows any map from a sphere into the base to be lifted to the total space (up to homotopy), which is exactly what's needed to relate the homotopy groups. The proof of the long exact sequence constructs the connecting homomorphism ∂ using the lifting property: a map S^n → B lifts to a map D^{n+1} → E whose boundary (a map S^n → F) gives the image under ∂.