Cohomology

Overview

Cohomology is the 'dual' theory to homology: instead of mapping simplices into a space, we assign algebraic values to them. The resulting cochain complex yields cohomology groups H^n(X; R) that carry additional structure — most notably a ring structure via the cup product — absent in homology. Cohomology is essential for characteristic classes, de Rham theory, Poincaré duality, and modern algebraic geometry.

Intuition

Where homology measures holes by assembling chains (sums of simplices), cohomology measures holes by assigning numbers (or elements of a coefficient group) to simplices, in a way consistent with the boundary maps. The cup product then lets you 'multiply' cohomology classes, giving cohomology the structure of a graded ring — a far richer invariant than homology groups alone.

Formal Definition

Definition

Given a chain complex C_*(X) and an abelian group R, the cochain complex is C^n(X; R) = Hom(C_n(X), R) with coboundary delta^n: C^n -> C^{n+1} defined by (delta^n phi)(sigma) = phi(∂_{n+1} sigma). The n-th cohomology group is H^n(X; R) = ker(delta^n) / im(delta^{n-1}). The cup product ∪: H^p ⊗ H^q -> H^{p+q} gives the cohomology ring H^*(X; R).

C^n(X;R) = \operatorname{Hom}(C_n(X),R)

Cochain group

(\delta^n \varphi)(\sigma) = \varphi(\partial_{n+1}\sigma)

Coboundary map

H^n(X;R) = \ker(\delta^n)/\operatorname{im}(\delta^{n-1})

n-th cohomology group

(\alpha \cup \beta)(\sigma) = \alpha(\sigma|_{[v_0,\ldots,v_p]})\cdot\beta(\sigma|_{[v_p,\ldots,v_{p+q}]})

Cup product formula

Theorems

Theorem 1: Universal Coefficient Theorem

There is a natural short exact sequence 0 \to \operatorname{Ext}^1(H_{n-1}(X),R) \to H^n(X;R) \to \operatorname{Hom}(H_n(X),R) \to 0, which splits (unnaturally).

Theorem 2: Poincaré Duality

For a closed orientable n-manifold M, there is an isomorphism H^k(M;\mathbb{Z}) \cong H_{n-k}(M;\mathbb{Z}) for all k.

Theorem 3: Künneth Formula for Cohomology

If H_*(Y;R) is torsion-free, then H^n(X\times Y;R) \cong \bigoplus_{p+q=n} H^p(X;R)\otimes_R H^q(Y;R).

Worked Examples

By the universal coefficient theorem and the known homology H_0(S^n) = Z, H_n(S^n) = Z, H_k(S^n) = 0 otherwise, we get:
$H^k(S^n;\mathbb{Z}) \cong \mathbb{Z} \text{ for } k=0,n; \quad 0 \text{ otherwise}.$
Let alpha be a generator of H^n(S^n; Z). The cup product alpha ∪ alpha lies in H^{2n}(S^n; Z) = 0 for n >= 1, so alpha^2 = 0.
The cohomology ring is the exterior algebra on one generator of degree n.
$H^*(S^n;\mathbb{Z}) \cong \mathbb{Z}[\alpha]/(\alpha^2), \quad |\alpha|=n.$

Answer: H^*(S^n; Z) is the exterior algebra Z[alpha]/(alpha^2) with |alpha| = n.

Practice Problems

Difficulty 7/10

Compute H^n(T^2; Z) for all n, where T^2 is the torus.

Difficulty 9/10

Prove that the cup product makes H^*(X; R) into a graded-commutative ring: alpha ∪ beta = (-1)^{pq} beta ∪ alpha for alpha in H^p and beta in H^q.

Difficulty 8/10

Use the universal coefficient theorem to compute H^2(Z/2Z; Z) given that H_1 = Z/2Z and H_2 = 0.

Common Mistakes

Common Mistake

Cohomology is just the dual of homology with no new information.

The cup product on cohomology provides strictly more information than homology alone. For example, CP^2 and S^2 ∨ S^4 have the same homology but different cohomology rings.

Common Mistake

The coboundary of a coboundary is nonzero.

Just as ∂∘∂ = 0 in the chain complex, we have delta∘delta = 0 in the cochain complex, since (delta^2 phi)(sigma) = phi(∂∘∂ sigma) = phi(0) = 0.

Quiz

Summary

Cohomology H^n(X; R) is defined via cochain complexes Hom(C_n(X), R) with coboundary maps dual to the boundary maps of homology.
The universal coefficient theorem relates cohomology to homology: H^n(X; R) fits in a sequence involving Hom(H_n, R) and Ext^1(H_{n-1}, R).
The cup product ∪ makes H^*(X; R) into a graded-commutative ring, providing finer invariants than homology.
Poincaré duality for closed orientable n-manifolds gives H^k(M) ≅ H_{n-k}(M).
Cohomology is the natural setting for characteristic classes, de Rham theory, and many structural results in geometry.