Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 114 / 1 / Solution

Because every $C_i$ is free abelian, reduction modulo $n$ gives a short exact sequence of chain complexesIts long exact sequence in homology isIf $\bar{z}$ is a cycle modulo $n$, choose a lift $z\in C_i$. Then $dz=nw$ for some $w\in C_{i-1}$, and the connecting homomorphism is $\hat{\beta }[\bar{z}]=[w]$.

The coefficient sequencesimilarly givesAt chain level, if $z$ lifts a mod-$n$ cycle and $dz=nw$ modulo $n^2$, then the Bockstein homomorphism is $\beta [\bar{z}]=[\bar{w}]$.

There is a morphism from the first short exact sequence to the second whose three vertical maps are reduction modulo $n$, reduction modulo $n^2$, and the identity on $C\otimes \mathbb{Z}/n$. Naturality of connecting homomorphisms givesExactness of the first long exact sequence says $\hat{\beta }{\rho }_{\ast }=0$, and hence

For the standard cellular chain complex of Real projective space ${\mathbb{RP}}^3$, there is one copy of $\mathbb{Z}$ in degrees $0,1,2,3$, with $d_2=2$ and $d_1=d_3=0$. 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 $\mathbb{Z}[i]$ and two-term complexesin degrees $i+1$ and $i$. The former represents a free homology summand, and the latter represents $\mathbb{Z}/m$ in degree $i$. The two requested conclusions now follow from the next two parts.

Solved by gpt-5.6-sol high.