Cohomology, Universal Coefficients & the Cup Product

Turning the arrows around: what cohomology is

By now you own the homology machine. From Guide 1 a space X gives a chain complex of free abelian groups C_k(X) with a boundary operator partial_k: C_k -> C_{k-1} satisfying partial^2 = 0, and homology H_k(X) = ker(partial_k) / im(partial_{k+1}) measures cycles modulo boundaries. Cohomology is built by one deceptively small move: apply the functor Hom(-, G) for an abelian group G, which reverses every arrow. The cochains are C^k(X; G) = Hom(C_k(X), G) — functions that eat a k-chain and spit out an element of G — and the coboundary delta: C^k -> C^{k+1} is the transpose of partial, defined by (delta phi)(c) = phi(partial c).

Because partial^2 = 0, automatically delta^2 = 0, so we have a cochain complex running the OTHER way — degrees go UP. The cohomology groups are H^k(X; G) = ker(delta^k) / im(delta^{k-1}): cocycles modulo coboundaries. A cocycle phi is a function on chains that vanishes on every boundary, i.e. phi(partial c) = 0 for all c; it is a coboundary if phi = psi composed with partial for some psi on lower chains. The pattern is exactly homology with the index raised and the arrows flipped, and you should hold onto that mental image: cochains are MEASUREMENTS made on chains, and a cocycle is a measurement that depends only on the homology class, not the representative.

The universal coefficient theorem: how much is genuinely new?

Here is the honest question a skeptic should ask: if cohomology is just homology dualized, does it carry any information that the Betti numbers and torsion of homology did not already give us? The universal coefficient theorem answers precisely. It says H^k(X; G) is built from H_k(X) and H_{k-1}(X) by a short exact sequence: 0 -> Ext(H_{k-1}(X), G) -> H^k(X; G) -> Hom(H_k(X), G) -> 0, and this sequence splits (non-naturally). In words: cohomology in degree k is the dual Hom(H_k, G) PLUS a correction term Ext(H_{k-1}, G) that detects torsion one dimension down.

Over a field — say G = R or G = Q — the Ext term vanishes (every module over a field is free), and Hom is just the vector-space dual. So over a field cohomology and homology have the SAME dimension: the cohomological Betti numbers equal the homological ones. This is exactly why de Rham cohomology, which is built over R, never sees torsion. The new content lives entirely with integer coefficients G = Z, where Ext(H_{k-1}, Z) resurrects the torsion of H_{k-1} and reincarnates it one degree UP in H^k. Cohomology is not 'more powerful than homology'; it repackages the same groups, but the repackaging is what makes the ring structure possible.

RP^2 worked through, integer coefficients (G = Z):

  Homology  :   H_0 = Z      H_1 = Z/2     H_2 = 0

  UCT in degree 2 :  0 -> Ext(H_1, Z) -> H^2 -> Hom(H_2, Z) -> 0
                     0 -> Ext(Z/2, Z) -> H^2 -> Hom(0, Z)  -> 0
                     Ext(Z/2, Z) = Z/2 ,   Hom(0,Z) = 0
                     ==>  H^2(RP^2; Z) = Z/2

  UCT in degree 1 :  0 -> Ext(H_0,Z) -> H^1 -> Hom(H_1,Z) -> 0
                     Ext(Z,Z)=0 , Hom(Z/2,Z)=0   ==>  H^1(RP^2; Z) = 0

  Cohomology  :  H^0 = Z      H^1 = 0       H^2 = Z/2

  Note the SHIFT: the torsion Z/2 sat in H_1 but reappears in H^2.

The universal coefficient theorem on the real projective plane: the Z/2 torsion in first homology re-emerges, shifted up one degree, in second cohomology — the cleanest small demonstration of the Ext correction term.

The cup product: a multiplication on cohomology

Now the payoff that justifies the whole detour through cochains. Homology groups are just groups; you can add classes but you cannot multiply them. Cohomology carries a product. Given cochains phi in C^p and psi in C^q, the cup product phi cup psi in C^{p+q} is defined on a singular simplex sigma: [v_0, ..., v_{p+q}] -> X by evaluating phi on the FRONT p-face and psi on the BACK q-face: (phi cup psi)(sigma) = phi(sigma restricted to [v_0,...,v_p]) times psi(sigma restricted to [v_p,...,v_{p+q}]). The Leibniz rule delta(phi cup psi) = delta phi cup psi plus (-1)^p phi cup delta psi shows the product of two cocycles is a cocycle and descends to cohomology.

The upshot is that the direct sum H*(X; R) = direct sum over k of H^k(X; R) becomes a graded ring, the cohomology ring, with the cup product as multiplication. It is graded-commutative: alpha cup beta = (-1)^{pq} beta cup alpha for alpha in degree p and beta in degree q. That sign is not optional bookkeeping — it is forced by the front-face/back-face asymmetry, and it means odd-degree classes anticommute and square to a 2-torsion element. If you took the forms route, this is literally the wedge product descending to de Rham cohomology: alpha cup beta corresponds to [alpha ^ beta], and the graded sign is the same sign that makes dx ^ dy = - dy ^ dx.

Why the ring sees more than the groups

Here is the concrete reason cohomology earns its keep. The torus T^2 and the wedge S^2 v S^1 v S^1 have the SAME homology and cohomology groups in every degree: a Z in degrees 0, two Z in the middle, a Z on top. As bare groups they are indistinguishable. But their RINGS differ. On T^2 the two degree-1 generators alpha, beta satisfy alpha cup beta = the top generator (a nonzero element of H^2) — the torus 'closes up' so the two circle directions multiply to the fundamental class. On the wedge, every cup product of positive-degree classes is ZERO, because positive-degree classes are supported on different pieces that meet only at the basepoint. The product structure tells the two spaces apart instantly.

A second example that every topologist carries in their head: complex projective space. The cohomology ring H*(CP^n; Z) is the truncated polynomial ring Z[h] / (h^{n+1}), where h is a single generator in degree 2. Everything is generated by one class and its powers, with the only relation being that h^{n+1} = 0 because there is nothing above degree 2n. Compare real projective space, whose ring is Z/2[w] / (w^{n+1}) with w in degree 1 and coefficients in Z/2 — the torsion the universal coefficient theorem warned you about now becomes the very coefficient ring of the multiplication. These compact descriptions are vastly more useful than listing groups degree by degree.

Computing a cohomology ring, step by step

Let us assemble everything on one example, the torus T^2, walking from chains to ring. This stitches together Guide 3's cellular homology, this guide's universal coefficient theorem, and the cup product, and it is the template you will run on every space you meet. The point to internalize is that the GROUPS come from the (co)chain complex, while the RING needs one extra geometric input — how the cells fit together — which here is encoded in the single nonzero product alpha cup beta.

Compute homology with the cellular complex from Guide 3. The torus has one 0-cell, two 1-cells a, b, one 2-cell, and all cellular boundary maps vanish, giving H_0 = Z, H_1 = Z (+) Z, H_2 = Z, with no torsion anywhere.
Apply the universal coefficient theorem. Since every homology group is free, all Ext terms vanish, so H^k(T^2; Z) = Hom(H_k, Z): cohomology equals H^0 = Z, H^1 = Z (+) Z, H^2 = Z. The torsion-free case means cohomology and homology look identical here.
Name generators: let alpha, beta in H^1 be dual to the two 1-cells a, b, and let mu in H^2 be the generator dual to the 2-cell (the fundamental cohomology class). So far this is pure linear algebra with no products.
Insert the cup product. The torus' 2-cell is glued along the commutator a b a^{-1} b^{-1}, which geometrically means the two circle directions span the top cell exactly once; so alpha cup beta = mu, while alpha cup alpha = 0 and beta cup beta = 0 by graded-commutativity. The ring is the exterior algebra Lambda[alpha, beta] on two degree-1 generators.
Read off the moral and look ahead. The single equation alpha cup beta = mu is what separates T^2 from S^2 v S^1 v S^1; it also says the pairing H^1 x H^1 -> H^2 is nondegenerate, which is exactly the Poincaré duality intersection form that Guide 5 will turn into the signature and the Lefschetz fixed-point count.