Questions: Cochain Complexes and Cohomology

Question 1

If C_n is a chain complex with boundary d_n, the coboundary map d^n: C^n → C^{n+1} is defined by d^n(f) = f ∘ d_{n+1} (precomposition with the boundary). Why does d^{n+1} ∘ d^n = 0?

• A. Because cohomology is always trivial
• B. Because d^{n+1}(d^n(f)) = f ∘ d_{n+1} ∘ d_{n+2} = f ∘ 0 = 0, since d ∘ d = 0 in the chain complex
• C. Because Hom(−, G) is an exact functor
• D. Because f is a homomorphism

Answer: (B)

The coboundary of d^n(f) is d^{n+1}(f ∘ d_{n+1}) = (f ∘ d_{n+1}) ∘ d_{n+2} = f ∘ (d_{n+1} ∘ d_{n+2}) = f ∘ 0 = 0. The vanishing of the composed coboundary follows directly from d ∘ d = 0 in the original chain complex. This is the dual of the fundamental property of chain complexes, and it guarantees that cohomology is well-defined. Note that Hom(−, G) is NOT exact in general (it is only left exact), which is why cohomology and homology can differ.

Question 2

A cochain f ∈ C^n(X; Z) = Hom(C_n(X), Z) is a cocycle if d^n(f) = 0, meaning f ∘ d_{n+1} = 0. What does this mean geometrically?

• A. f assigns zero to every n-simplex
• B. f assigns the same integer to homologous n-cycles
• C. f vanishes on all boundaries — f(d_{n+1}(c)) = 0 for every (n+1)-chain c
• D. f is a constant function on simplices

Answer: (C)

The cocycle condition d^n(f) = 0 means f ∘ d_{n+1} = 0, i.e., f(d_{n+1}(c)) = 0 for all (n+1)-chains c. In other words, f vanishes on all n-boundaries. This means f 'sees' only the homology: its value on an n-cycle depends only on the homology class, not the specific representative. Cocycles are the linear functionals on chains that respect the boundary structure, and their cohomology classes are well-defined pairings with homology classes.

Question 3

Cohomology with coefficients in a field k (e.g., Q or Z/pZ) is isomorphic to the linear dual of homology: H^n(X; k) ≅ Hom_k(H_n(X; k), k).

• T. True
• F. False

Answer: True

When the coefficient group is a field k, the Hom functor is exact, and the universal coefficient theorem simplifies to H^n(X; k) ≅ Hom(H_n(X; k), k). This is the vector space dual. Over a field, cohomology and homology are dual vector spaces of the same dimension, so they carry equivalent information. Over the integers, the situation is more subtle: H^n(X; Z) ≅ Free(H_n(X; Z)) ⊕ Torsion(H_{n-1}(X; Z)), mixing information from adjacent homology groups.

Question 4

Why does algebraic topology study both homology and cohomology, rather than just one?

Model answer: Cohomology has a natural ring structure via the cup product, which homology lacks. The cup product H^p(X) × H^q(X) → H^{p+q}(X) encodes how cohomology classes interact multiplicatively, and this ring structure is a strictly finer invariant than the additive group structure alone — spaces with isomorphic homology groups can have non-isomorphic cohomology rings. Additionally, cohomology is the natural setting for duality theorems (Poincare duality), characteristic classes, and obstruction theory.

The multiplicative structure is the key advantage. For example, CP^2 and S^2 ∨ S^4 have isomorphic homology groups in every dimension, but their cohomology rings differ: CP^2 has a generator α ∈ H^2 with α^2 ≠ 0 in H^4, while in S^2 ∨ S^4 all cup products of positive-degree classes vanish. Cohomology detects this difference; homology cannot.