Cochains and the universal coefficient theorem
Dualize a chain complex correctly and retain the Ext term and noncanonical splitting in cohomology calculations.
On this page
The question for today
What you’ll understand
- Calculate a coboundary by precomposing with the boundary map.
- State the cohomological universal coefficient sequence over a PID.
- Compute torsion cohomology that ordinary evaluation on cycles misses.
Chains assemble geometric pieces; cochains assign coefficients to those pieces in a way compatible with linear combinations. It is tempting to say that cohomology is simply the dual of homology, but integer torsion obstructs that shortcut. The universal coefficient theorem identifies the extra term precisely. We will distinguish the cochain complex from its cohomology, keep the coefficient module explicit, and track why a natural exact sequence can admit a splitting without providing a natural direct-sum decomposition.
Cochains reverse arrows and raise degree
Let R be a principal ideal domain, let be a chain complex of free R-modules, and let G be an R-module. Define and . Thus coboundary raises degree, and follows from boundary squared zero. Cocycles vanish on boundaries; coboundaries evaluate to zero on cycles. Consequently evaluation gives a homomorphism from cohomology to R-linear functionals on homology. The chosen convention has no extra degree sign in the coboundary definition.
If the chain groups have finite chosen bases, this coboundary is represented by the transpose boundary matrix with coefficients acting on G. Infinite bases require care: Hom of a direct sum is a product, so singular cochains can assign nonzero values to infinitely many generating simplices. They are not restricted to finite-support assignments. A continuous map induces a chain map forward and a cochain map backward by precomposition, making cohomology contravariant in spaces, in contrast to homology's covariance.
Evaluation can have a nontrivial kernel
The universal coefficient theorem gives a short exact sequence with on the left and on the right. The evaluation map is surjective, but the Ext term can represent cohomology classes invisible to evaluation on n-dimensional cycles. It measures the failure of the Hom functor to preserve exactness when applied to the relevant chain relations. The degree n−1 is essential: torsion in one homology degree can contribute to cohomology one degree higher.
The sequence is natural in the appropriate maps of complexes and coefficient modules. Under these PID and free-chain hypotheses it splits, but generally not naturally. One may use a chosen splitting to identify abstract module types, yet cannot silently transport that decomposition through every map of spaces. Over a field the Ext term vanishes and evaluation canonically identifies cohomology with the vector-space dual of homology. Even then an infinite-dimensional dual is not automatically identifiable with the original vector space by a preferred basis.
Compute Ext using a small free resolution
For integer coefficients, use the free resolution with positive m. Applying Hom into G shows that is the subgroup killed by m, while . These two constructions are a kernel and a cokernel of the same multiplication map, not interchangeable definitions. For G equal to Z, the Hom term is zero and the Ext term is Z/m.
For spaces, singular chains satisfy the free-chain hypotheses, and cellular cochains can be used for computations on CW complexes. Degree zero also illustrates the distinction between sums and products: H₀ is the direct sum of one copy of the coefficient ring per path component, while H⁰ with module coefficients gives the corresponding product of assignments. In a finite connected example both reduce to one copy, making it easy to miss this difference until an infinite-component space or an infinite chain basis is considered.
Work it through
Integer cohomology of the projective plane
Use RP2's cellular chain complex Z in degrees two, one, and zero, with d2=2 and d1=0, to compute its integer cohomology. Relate the degree-two answer to the universal coefficient theorem.
Dualizing into Z gives cochains Z in the same three degrees, with delta0=0 and delta1 multiplication by two. Thus H⁰=Z and H¹=ker(×2)/0=0; the superscripts denote cohomology.
There are no degree-three cochains, so H²=Z/2Z from the cokernel of delta1. This is cohomology in degree two despite integral homology H2(RP2;Z) being zero.
The universal coefficient sequence in degree two has Ext1(Z/2,Z)=Z/2 on the left and Hom(H2,Z)=0 on the right. Hence the degree-two class comes entirely from Ext and has zero evaluation on every two-dimensional integral homology class.
The cohomology groups are H⁰=Z, H¹=0, and H²=Z/2Z. Simply taking Hom of each integer homology group would miss H².
The living notebook
Explore the model
Interactive lab · modular
A kernel and a quotient are different constructions
Pause and Reset, then set both controls to modulus m=6 and additive step a=2. Predict the subgroup 2G={0,2,4}, leaving two cosets in G/2G. Reset and set m=6 and a=3; predict orbit {0,3}, which is the subgroup killed by multiplication by two. Both abstract groups have two elements here, but one arose as a quotient and the other as a kernel. The instrument displays additive orbits, not Ext resolutions or cochains.
- Residue r0
- 0
- Orbit period · m/gcd(m,a)
- 12
- Distinct residues visited
- 1 of 12
Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forEqual group orders do not make the Hom kernel and Ext cokernel the same map or the same construction.
Pause & reason
Check your understanding
Your turn
Try it yourself
Repeat the RP2 cochain calculation with G=Z/6Z. Find H¹ and H², distinguishing the subgroup killed by two from the quotient by twice G.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The coboundary delta1 is still multiplication by two, now within Z/6Z. List its kernel and image separately.
Compare with the solution
The kernel of multiplication by two is {0,3}, so H¹ is Z/2Z. The image is {0,2,4}, so H² is G/2G, also Z/2Z. H⁰ is G because delta0 is zero. Here Hom(Z/2,G) supplies H¹ and Ext1(Z/2,G) supplies H². Their equal abstract isomorphism type does not justify identifying the actual kernel inclusion with the quotient map or assuming a canonical universal-coefficient splitting in other examples.
Keep the idea
Take it with you
- Coboundary reverses boundary arrows and raises degree.
- The Ext term detects cohomology invisible to ordinary evaluation on cycles.
- A noncanonical splitting computes module types without supplying natural coordinates.
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