Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 127 5 Solution 2026-09-28
Let . A pairis transgressive pair when, in the long exact sequence of the pair ,where is identified with reduced cohomology. In the Serre spectral sequence, this says that survives to the transgression andunder the edge identifications, modulo the usual earlier-differential indeterminacy.
The Kudo transgression theorem says that if is transgressive and , thenis transgressive. To prove it, use relative Steenrod squares. Naturality givesand stability, equivalently compatibility with the suspension isomorphism, makes squares commute with the connecting map:Applying to proves the theorem. The properties used are naturality, stability, additivity, and the instability conditions for and .
Let be the generator. Instability givesThe Cartan formula says the total square is multiplicative, soComparing components yields the complete formulawith the binomial coefficient reduced modulo two.
Finally, is the Bockstein homomorphism associated withIf a mod-two cocycle representing is lifted to an integral cochain , write . Then modulo two represents . But , because integral cochains are torsion-free and . Thus itself is a cocycle lift, so its Bockstein vanishes. Thereforefor every space and every .
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 165 4 iii Solution Created 2026-09-24 Updated 2026-09-25
The group is one-dimensional, so it suffices to consider its nonzero class . This class is represented by the quotient homomorphism , which lifts to the identity homomorphism with coefficients in . Its Bockstein homomorphism fortherefore vanishes. Since the first Steenrod square is this Bockstein and for every degree-one class,The zero degree-one class plainly has square zero as well.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 114 1 i Solution Created 2026-09-24 Updated 2026-09-25
A one-term summand becomes a one-term complex after reduction modulo . Its Bockstein homomorphism vanishes, so it contributes one summand to Bockstein homology in degree .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 114 1 Solution Created 2026-09-24 Updated 2026-09-25
Because every is free abelian, reduction modulo gives a short exact sequence of chain complexesIts long exact sequence in homology isIf is a cycle modulo , choose a lift . Then for some , and the connecting homomorphism is .
The coefficient sequencesimilarly givesAt chain level, if lifts a mod- cycle and modulo , then the Bockstein homomorphism is .
There is a morphism from the first short exact sequence to the second whose three vertical maps are reduction modulo , reduction modulo , and the identity on . Naturality of connecting homomorphisms givesExactness of the first long exact sequence says , and hence
For the standard cellular chain complex of Real projective space , there is one copy of in degrees , with and . Modulo two all cellular differentials vanish, while the Bockstein is the identity from degree two to degree one and zero elsewhere. Therefore
Finally, Smith normal form decomposes a bounded chain complex of finitely generated free abelian groups, up to chain isomorphism and contractible summands, into one-term complexes and two-term complexesin degrees and . The former represents a free homology summand, and the latter represents in degree . The two requested conclusions now follow from the next two parts.
Steenrod algebra 2026-10-06
The Steenrod algebra organizes stable mod- cohomology operations. At it is generated by the Steenrod squares; at an odd prime it is generated by the Steenrod reduced powers and Bockstein homomorphism. Cohomology is a module over this algebra, and every continuous map induces a homomorphism preserving its action.