 Solution

Write $\rho :H^{\ast }(X;\mathbb{Z})\to H^{\ast }(X;\mathbb{Z}/n)$ for coefficient reduction. Compare the two coefficient sequences in part (a): the maps from the integral sequence to the finite sequence are reduction modulo $n$ on the left, reduction modulo $n^2$ in the middle, and the identity on the right. The square involving the injections commutes because $[na{]}_{n^2}=j([a{]}_n)$.

Naturality of the connecting homomorphism gives the Bockstein factorization through integral cohomologyOne can see this directly without a diagram: lift a modulo-$n$ cocycle to an integral cochain $a$. Its coboundary has the form $\delta a=nb$. Then $\hat{\beta }[a]=[b]$, whereas $\beta [a]=[b\mathrm{mod}n]$.

Exactness of the integral coefficient sequence gives $\hat{\beta }\rho =0$: a class obtained by reducing an integral cocycle has zero integral connecting class. ConsequentlyThis Bockstein square-zero identity holds without requiring $n$ to be prime. At the cochain level, ${\delta }^{2}a=0$ and the torsion-free integral cochain groups imply $\delta b=0$, which also makes the second connecting class vanish.