The long exact sequence for the pair (X, A) in cohomology runs: ... → H^n(X, A; G) → H^n(X; G) → H^n(A; G) → H^{n+1}(X, A; G) → ... How do the arrows compare to the homology long exact sequence?
AThe arrows point in the same direction as in homology
BThe arrows between spaces are reversed (maps go from the larger space to the subspace), and the connecting homomorphism increases dimension by 1 instead of decreasing it
COnly the connecting homomorphism changes direction
DThe cohomology sequence is not exact
In homology, the long exact sequence has i_*: H_n(A) → H_n(X) (inclusion pushes forward) and ∂: H_n(X,A) → H_{n-1}(A) (connecting homomorphism decreases dimension). In cohomology, the Hom functor reverses arrows: i*: H^n(X) → H^n(A) (restriction to subspace pulls back), and the connecting homomorphism δ: H^n(A) → H^{n+1}(X,A) increases dimension. Cohomology is contravariant — it pulls back along maps rather than pushing forward.
Question 2 True / False
The Kronecker pairing ⟨−, −⟩: H^n(X; Z) × H_n(X; Z) → Z is defined by evaluating a cocycle on a cycle. This pairing is always a perfect pairing (non-degenerate on both sides).
TTrue
FFalse
Answer: False
The Kronecker pairing is well-defined and natural, but it is NOT a perfect pairing in general. When H_n(X) has torsion, the pairing has a nontrivial kernel on the cohomology side: torsion elements in H^n pair trivially with all cycles. The universal coefficient theorem makes this precise: the free part of H^n pairs perfectly with the free part of H_n, but torsion in H^n comes from H_{n-1} and does not pair with H_n at all. Over a field, the pairing IS perfect.
Question 3 True / False
Singular cohomology is contravariant: a continuous map f: X → Y induces f*: H^n(Y) → H^n(X), reversing the direction.
TTrue
FFalse
Answer: True
A continuous map f: X → Y induces a chain map f_#: C_n(X) → C_n(Y) (by composition: σ ↦ f ∘ σ). Applying Hom(−, G) reverses the arrow: f^#: C^n(Y) → C^n(X) is defined by f^#(φ) = φ ∘ f_# (precomposition). This induces f*: H^n(Y) → H^n(X) on cohomology. Contravariance is essential: the cup product on H^*(X) is natural with respect to pullback (f*(α ∪ β) = f*(α) ∪ f*(β)), which would not work covariantly.
Question 4 Short Answer
Compute H^*(S^n; Z) and explain why the cohomology ring structure is trivial for spheres.
Think about your answer, then reveal below.
Model answer: H^k(S^n; Z) = Z for k = 0 and k = n, and 0 otherwise (same as homology, since the homology groups are free abelian and the universal coefficient theorem gives an isomorphism). The cohomology ring H^*(S^n; Z) ≅ Z[α]/(α^2) where α ∈ H^n is the generator: the only possible cup product of positive-degree classes would be α ∪ α ∈ H^{2n}(S^n), but H^{2n}(S^n) = 0 for dimensional reasons (2n > n when n > 0), so α^2 = 0. The ring structure is 'trivial' in the sense that all products of positive-degree classes vanish.
This triviality is specific to spheres and is a consequence of having cohomology concentrated in just two degrees. Compare with CP^n, which has H^*(CP^n; Z) ≅ Z[α]/(α^{n+1}) with α ∈ H^2 — here α^k ≠ 0 for k ≤ n, giving a rich ring structure. The cup product becomes a powerful distinguishing invariant precisely for spaces whose cohomology is spread across multiple degrees.