CP^2 and S^2 ∨ S^4 have the same homology groups: H_0 = H_2 = H_4 = Z, all others zero. How does the cup product distinguish them?
AThe cup product on CP^2 has a generator α ∈ H^2 with α ∪ α ≠ 0 in H^4, while on S^2 ∨ S^4 all cup products of positive-degree classes are zero
BThe cup products are isomorphic but the spaces differ in homotopy groups
CCP^2 has torsion in cohomology that S^2 ∨ S^4 does not
DThe cup product only distinguishes orientable from non-orientable spaces
H^*(CP^2; Z) ≅ Z[α]/(α^3) where α ∈ H^2 is a generator and α^2 generates H^4. The ring has a nontrivial product. H^*(S^2 ∨ S^4; Z) has generators β ∈ H^2 and γ ∈ H^4, but β^2 = 0 (it maps to H^4 of S^2, which is trivial) and β ∪ γ = 0 for dimensional reasons. The ring is a square-zero extension. This algebraic difference reflects a genuine topological difference: CP^2 cannot be decomposed as a wedge sum, while S^2 ∨ S^4 is explicitly built as one.
Question 2 True / False
The cup product is graded commutative: α ∪ β = (-1)^{pq} β ∪ α for α ∈ H^p and β ∈ H^q.
TTrue
FFalse
Answer: True
Graded commutativity (also called supercommutativity) means that swapping two classes introduces a sign that depends on their degrees. For two even-degree classes, the product commutes. For two odd-degree classes, the product anti-commutes: α ∪ α = -α ∪ α, which implies 2(α ∪ α) = 0 (so α^2 is 2-torsion or zero). This sign convention arises naturally from the combinatorics of the front-face/back-face decomposition and is compatible with the Koszul sign rule that pervades graded algebra.
Question 3 Multiple Choice
At the cochain level, the cup product of f ∈ C^p and g ∈ C^q is defined by (f ∪ g)(σ) = f(σ|[v_0,...,v_p]) · g(σ|[v_p,...,v_{p+q}]). Why does this descend to a well-defined operation on cohomology?
ABecause the coboundary of a cup product satisfies the Leibniz rule: d(f ∪ g) = (df) ∪ g + (-1)^p f ∪ (dg)
BBecause all cochains are cocycles
CBecause the cup product commutes with all chain maps
DBecause Hom is an exact functor
The Leibniz rule d(f ∪ g) = (df) ∪ g + (-1)^p f ∪ (dg) is the key identity. It implies: if f and g are cocycles (df = 0, dg = 0), then d(f ∪ g) = 0, so the cup product of cocycles is a cocycle. Furthermore, if f is a coboundary (f = dh), then f ∪ g = (dh) ∪ g = d(h ∪ g) - (-1)^{p-1} h ∪ (dg) = d(h ∪ g) when g is a cocycle, so the cup product of a coboundary with a cocycle is a coboundary. Therefore the cup product descends to cohomology.
Question 4 Short Answer
Explain why the cup product makes cohomology a strictly finer invariant than homology for distinguishing topological spaces.
Think about your answer, then reveal below.
Model answer: Cohomology groups, as abelian groups, carry the same information as homology groups (related by the universal coefficient theorem). But the cup product introduces multiplicative relationships between classes in different degrees that have no homological analogue. Two spaces can have isomorphic cohomology groups in every degree but non-isomorphic cohomology rings, because the cup product structure detects how cohomology classes 'interact.' This multiplicative structure encodes topological features like the linking and intersection of submanifolds, the non-decomposability of spaces, and the complexity of fiber bundle structures.
The classic example — CP^2 vs S^2 ∨ S^4 — shows this concretely. Both have the same graded group Z ⊕ 0 ⊕ Z ⊕ 0 ⊕ Z, but the ring structures differ. There is no ring homomorphism between Z[α]/(α^3) and Z[β, γ]/(β^2, βγ, γ^2) that preserves degree, so the spaces are distinguishable. No amount of information about individual homology/cohomology groups can capture this difference.