What are the homology groups of S^3 (the 3-sphere)?
AH_0 = Z, H_1 = Z, H_2 = Z, H_3 = Z
BH_0 = Z, H_3 = Z, all others zero
CH_0 = Z, H_1 = Z^3, all others zero
DH_k = Z for all k ≥ 0
The homology of S^n is concentrated in dimensions 0 and n: H_0(S^n) ≅ Z (connected), H_n(S^n) ≅ Z (fundamental class detecting the n-dimensional 'hole'), and H_k(S^n) = 0 for all other k. For S^3: H_0 = Z, H_3 = Z, and H_1 = H_2 = 0. The vanishing of H_1 means S^3 is simply connected (which also follows from the fundamental group being trivial), and the vanishing of H_2 means there are no 2-dimensional holes.
Question 2 True / False
The homology of S^n can be computed inductively using the Mayer-Vietoris sequence by decomposing S^n into two hemispheres whose intersection is S^{n-1}.
TTrue
FFalse
Answer: True
Decompose S^n as U ∪ V where U and V are open hemispheres (slightly overlapping at the equator). Both U and V are contractible (each deformation retracts to a point), so H_k(U) = H_k(V) = 0 for k > 0. The intersection U ∩ V deformation retracts to the equatorial S^{n-1}. The Mayer-Vietoris sequence then gives: ... → H_k(U) ⊕ H_k(V) → H_k(S^n) → H_{k-1}(S^{n-1}) → H_{k-1}(U) ⊕ H_{k-1}(V) → ... Since H_k(U) = H_k(V) = 0 for k > 0, this yields isomorphisms H_k(S^n) ≅ H_{k-1}(S^{n-1}) for k ≥ 2, which inductively computes the homology starting from H_*(S^0) = Z ⊕ Z (two points).
Question 3 True / False
S^2 has trivial fundamental group (π_1 = 0) but nontrivial H_2 ≅ Z. This shows that homology detects topological features invisible to the fundamental group.
TTrue
FFalse
Answer: True
The fundamental group detects 1-dimensional holes (non-contractible loops). The 2-sphere has no non-contractible loops — every loop on S^2 can be shrunk to a point — so π_1(S^2) = 0. But S^2 encloses a 2-dimensional cavity, detected by H_2(S^2) ≅ Z. This is precisely the kind of higher-dimensional information that homology captures and the fundamental group misses. More generally, the n-sphere (n ≥ 2) has trivial fundamental group but nontrivial H_n, showing that higher homology groups are genuinely new invariants.
Question 4 Short Answer
Explain the geometric meaning of the generator of H_n(S^n) ≅ Z.
Think about your answer, then reveal below.
Model answer: The generator of H_n(S^n), called the fundamental class [S^n], is represented by a singular n-cycle that 'wraps once around' the sphere. For any triangulation, this is the sum of all n-simplices with consistent orientations. The integer k ∈ Z = H_n(S^n) represents a cycle that wraps around S^n exactly k times (with sign indicating orientation). The fact that this group is Z (not Z/mZ) means the fundamental cycle has infinite order: wrapping around the sphere k times is never homologous to zero for any k ≠ 0.
The fundamental class is the homological version of the orientation of the sphere. For an oriented sphere, there is a canonical choice of generator (the fundamental class compatible with the orientation). This fundamental class is essential for degree theory: a map f : S^n → S^n sends [S^n] to some multiple d·[S^n], and this integer d is the degree of f, which determines f up to homotopy.
Question 5 Short Answer
Using the Mayer-Vietoris induction, what is H_1(S^2)?
Think about your answer, then reveal below.
Model answer: H_1(S^2) = 0. The Mayer-Vietoris sequence with contractible hemispheres gives H_1(S^2) ≅ H_0(S^1)/im(H_0(U) ⊕ H_0(V) → H_0(S^1)). Since S^1 is connected and both hemispheres are connected, the map Z ⊕ Z → Z is surjective (both hemispheres include into the connected equator), so the connecting homomorphism H_1(S^2) → H_0(S^1) lands in the kernel of Z → Z ⊕ Z, giving H_1(S^2) ≅ H_0(S^1)/image ≅ 0 after working through the exact sequence carefully.
More directly: S^2 is simply connected, so H_1(S^2) = π_1(S^2)^{ab} = 0 by the Hurewicz theorem. The Mayer-Vietoris computation confirms this by the algebraic machinery. The vanishing of H_1 for S^n with n ≥ 2 is a general pattern reflecting the simple-connectivity of higher-dimensional spheres.