Cohomology & Universal Coefficients

Dualize the chains and every arrow turns around: cochains, contravariance, and the theorem that moves torsion up a degree.

The idea

Cohomology assigns to a space a second sequence of groups, $H^{n}(X; G)$, built by dualizing the chain complex that computed homology. The groups themselves separate no spaces that homology could not already separate; the reason to introduce them is that they support a multiplication, which homology lacks.

Fix an abelian group $G$, the coefficients. Replace each chain group $C_{n}(X)$ by its dual $C^{n}(X; G) = \operatorname{Hom}(C_{n}(X), G),$ whose elements assign an element of $G$ to each cell, and replace the boundary map $\partial$ by the coboundary $\delta(\varphi) = \varphi \circ \partial$. Then $H^{n}(X; G) = \ker \delta^{n} \,/\, \operatorname{im} \delta^{n-1}.$

Dualizing reverses every arrow: a homomorphism defined on $B$ can only be precomposed with a map arriving at $B$, never followed by one. So $\delta$ raises degree where $\partial$ lowered it, and a map of spaces induces a map of cohomology in the opposite direction — cohomology is contravariant.

The universal coefficient theorem computes $H^{n}(X; G)$ from homology alone.

Theorem (Universal coefficient theorem).

For any space $X$, any coefficient group $G$ and any $n$, there is a short exact sequence, in the sense of Exact Sequences & the Long Exact Sequence, $0 \to \operatorname{Ext}(H_{n-1}(X), G) \to H^{n}(X; G) \to \operatorname{Hom}(H_{n}(X), G) \to 0,$ and it splits, so $H^{n}(X; G) \cong \operatorname{Ext}(H_{n-1}(X), G) \oplus \operatorname{Hom}(H_{n}(X), G).$

The $\operatorname{Hom}$ term reads the homology in the same degree; the correction term $\operatorname{Ext}(H_{n-1}(X), G)$ reads one degree lower and vanishes whenever $H_{n-1}(X)$ is free abelian. Over $\mathbb{Z}$ the effect is simple bookkeeping: free summands stay in their degree, and torsion moves up exactly one degree.

Ways to work on it

• Walkthrough. The cochain complex, why induced maps reverse, and the universal coefficient theorem run on familiar surfaces.
• Proof. Why the Ext of a cyclic group is a quotient of the coefficient group, computed from a free resolution.
• Practice. Compute cohomology from homology with the universal coefficient theorem, over several coefficient choices.
• Hardest. Recover a space's homology from its integral cohomology, then compute its cohomology in a new coefficient system.

Not sure where to start? Take the ten-question placement test.