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First account for the unlabelled coefficient-sequence construction. The two short exact sequences of abelian groups areThe singular chain groups of $X$ are free abelian. Applying $\mathrm{Hom}(C_q(X),-)$ therefore preserves these exact sequences, degree by degree, giving short exact sequences of cochain complexes. The associated long exact sequence from a coefficient sequence gives the displayed maps in cohomology; the connecting maps are the integral Bockstein homomorphism $\hat{\beta }$ and the modulo-$n$ Bockstein homomorphism $\beta$. The first omitted map is multiplication by $n$, and the second is induced by $j$.

For the requested example, attach an $(i+1)$-cell to $S^i$ using a map of degree $n$. The resulting Moore space $X=M(\mathbb{Z}/n,i)$ has positive-degree cellular chain complexin degrees $i+1,i$. This construction also works for $i=1$, using the degree-$n$ map of the circle. In cellular cohomology with coefficients $\mathbb{Z}/n$, the differential is zero, so both $H^i$ and $H^{i+1}$ are $\mathbb{Z}/n$.

Lift the cochain taking value $1$ on the $i$-cell to a cochain with coefficients $\mathbb{Z}/n^2$. Its coboundary takes value $n$ on the $(i+1)$-cell, which is $j(1)$. The definition of the connecting homomorphism therefore sends the degree-$i$ generator to the degree-$(i+1)$ generator. HenceIt is nonzero for every $i\ge 1$ and $n\ge 2$, including composite $n$. This is the Bockstein on a cyclic Moore space.

An $R$-orientation of a rank-$d$ real vector bundle $E\to B$ is a locally coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. Equivalently, it is represented by a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to the chosen generator on each fibre pair. If $e(E)\in H^d(B;R)$ is the Euler class, the Gysin sequence of its unit sphere bundle contains

The diagonal quotient defining $L(p)$ is the three-dimensional lens space as a circle bundle $S^1\to L(p)\to S^2$ with Euler class $p$ times a generator. With integral coefficients, the only nontrivial Euler-class map is multiplication by $p:H^0(S^2;\mathbb{Z})\to H^2(S^2;\mathbb{Z})$. Exactness givesModulo $p$, the Euler class vanishes, so the same Gysin sequence giveswith all other groups zero.

For the coefficient sequence $0\to \mathbb{Z}\stackrel{p}{\to }\mathbb{Z}\to {\mathbb{F}}_p\to 0$, the long exact sequence from a coefficient sequence containsThe last map is zero and both middle groups have order $p$, so $\delta$ is an isomorphism. Reduction $H^2(L(p);\mathbb{Z})\to H^2(L(p);{\mathbb{F}}_p)$ is likewise an isomorphism. Their composite is the Bockstein isomorphism for a three-dimensional lens space

If $a^{\prime }=na$ is another generator, linearity of the Bockstein homomorphism and bilinearity of the cup product giveMoreover $t(a)\ne 0$: $\beta (a)$ is nonzero, and the Poincare duality pairing $H^1\times H^2\to H^3\cong {\mathbb{F}}_p$ is nondegenerate. If $h:L(p)\to L(p)$ is an orientation-reversing homotopy equivalence and $h^{\ast }a=na$, naturality givesCancelling the nonzero $t(a)$ yields $n^2\equiv -1(\mathrm{mod}p)$. Thus $-1$ must be a quadratic residue modulo $p$.

The short exact coefficient sequenceinduces the integral Bockstein homomorphismIf $\rho :H^{i+1}(X;\mathbb{Z})\to H^{i+1}(X;\mathbb{Z}/m)$ is reduction modulo $m$, thenEquivalently, if an integral cochain $a$ lifts a modulo-$m$ cocycle and $\delta a=mb$, then $\tilde{\beta }[a]=[b]$ and $\beta [a]=[b\mathrm{mod}m]$.

For integral lifts $a,c$ of classes $x,y$, the cup product coboundary formula isDividing by $m$ and reducing modulo $m$ proves the Bockstein derivation rule

Now assume that the stated closed five-manifold $M$ exists. Its top integral cohomology makes it connected and orientable. The long exact sequence from a coefficient sequence for multiplication by $p$ shows thatand that reduction $H^3(M;\mathbb{Z})\to H^3(M;\mathbb{Z}/p)$ is an isomorphism. HenceThe same coefficient sequence gives $H^4(M;\mathbb{Z}/p)=0$.

Choose $0\ne x\in H^2(M;\mathbb{Z}/p)$ and put $y=\beta x\ne 0$. By Poincare duality over $\mathbb{Z}/p$, the pairingis nondegenerate. Both factors are one-dimensional, so $x⌣y\ne 0$. But $x⌣x=0$ because $H^4(M;\mathbb{Z}/p)=0$, while the derivation rule and graded commutativity of the cup product giveThis contradicts $p>2$. Therefore no such manifold exists.

A degreewise short exact sequence of cochain complexes induces a long exact sequence from a coefficient sequence in the coefficient case, and a long exact cohomology sequence generally. Its degree-one connecting homomorphism lifts a quotient cocycle, takes its coboundary, and identifies that result in the subcomplex.