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The Bockstein homomorphism for $0\to \mathbb{Z}/n\to \mathbb{Z}/n^2\to \mathbb{Z}/n\to 0$, with injection $[a]\mapsto [na]$, is coefficient reduction $\rho$ after the integral Bockstein homomorphism. The commuting maps of coefficient sequences prove this by naturality of the connecting homomorphism.

Reduction $\rho$ of the integral Bockstein homomorphism associated to $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to {\mathbb{F}}_2\to 0$ equals the cup product square on degree-one classes. One can see this on an ordered two-simplex: lift the values of a mod-two one-cocycle to zero or one. The half-coboundary is one precisely when both consecutive edge values are one, giving the cup-square cocycle.

Integral groups for $M_4$. Put $r=n-1$. The space $M_n$ is the inversion mapping torus of a torus, fibred over $S^1$ with fiber $T^r$. The cohomology ring of a torus is the exterior algebra on $r$ degree-one generators. Inversion acts as $-1$ on each such generator, hence as $(-1{)}^q$ on $H^q(T^r;\mathbb{Z})\cong {\mathbb{Z}}^{\left(\genfrac{}{}{0ex}{}{r}{q}\right)}$.

The Wang sequence therefore givesThe right-hand term is free, so this short exact sequence splits as a sequence of groups. In even fiber degree, $F^{\ast }-1=0$; in odd fiber degree it is multiplication by $-2$. For $r=3$ this computes the integral cohomology of an inversion mapping torus:The top torsion group is consistent with $M_4$ being nonorientable: inversion of its three-dimensional fiber reverses orientation.

Mod-two groups. Over ${\mathbb{F}}_2$, inversion acts as the identity on the fiber cohomology. The Wang sequence givesThuswith out-of-range binomial coefficients interpreted as zero. These are exactly the dimensions of the graded groups $H^{\ast }(T^n;{\mathbb{F}}_2)$.

The intersection pairing for $M_2$. The mapping torus of reflection of the circle is the Klein bottle. Let $a$ be the section loop through a fixed point of the reflection and $b$ a fiber circle. They form a basis of $H_1(M_2;{\mathbb{F}}_2)$. The section has a normal neighborhood homeomorphic to a Möbius band, so it is one-sided and a transverse displacement meets it once modulo two. The fiber is two-sided and can be displaced disjointly. The two loops meet once. The mod-two intersection pairing of the Klein bottle therefore has matrixTake the evaluation-dual basis $t,x\in H^1(M_2;{\mathbb{F}}_2)$, with $t(a)=1,t(b)=0$ and $x(a)=0,x(b)=1$. By Poincare duality, the cup product matrix in this dual basis is $Q^{-1}$, not $Q$:Writing $w$ for the nonzero top class gives $t^2=0$, $tx=w$ and $x^2=w$. Hence the cohomology ring isThese relations already force all degrees above two to vanish.

The ring for general $n$. For each of the $r$ fiber coordinates, projection induces $M_n\to M_2$. Pull back the classes above, calling the common base class $t$ and the fiber-coordinate classes $x_1,\dots ,x_r$. Naturality of the cup product gives $t^2=0$ and $x_i^2=t{x}_i$. The square-free products $x_I={\prod }_{i\in I}{x}_i$ restrict to the exterior algebra basis of the fiber cohomology. The Leray-Hirsch theorem then says that $x_I$ and $t{x}_I$ form an additive basis. This proves the full mod-two cohomology ring of an inversion mapping torus:For $n\ge 2$, $x_i^2=t{x}_i$ is nonzero by this basis description. In the cohomology ring of a torus every degree-one class squares to zero: the generators square to zero and the cross terms occur twice in characteristic two. The degree-one cup-square obstruction to ring isomorphism therefore proves that the rings are not isomorphic for any $n\ge 2$, despite their isomorphic graded groups. For $n=1$, both spaces are circles and the rings are isomorphic. Even if grading is forgotten, the rings differ for $n\ge 2$: every element of the torus ring has square either zero or one, whereas $x_i^2=t{x}_i$ is nonzero and is not the unit.

Integral cup products in $M_4$. To specify $H^{\ast }(M_4;\mathbb{Z})$ also as a ring, let $a\in H^1$ be the pullback of the positive generator of $H^1(S^1;\mathbb{Z})$. Let $\rho$ denote reduction modulo two and set ${\tau }_i=\hat{\beta }(x_i)$ using the integral Bockstein homomorphism. The degree-one Bockstein square identity gives $\rho {\tau }_i=x_i^2=t{x}_i$, so the ${\tau }_i$ are the three independent order-two classes in degree two. Choose free classes $b_{12},b_{13},b_{23}\in H^2$ which restrict to the corresponding two-fold fiber products and have reductions $\rho b_{ij}=x_i{x}_j$. Such choices exist: reduction in degree two is surjective because $H^3(M_4;\mathbb{Z})$ is free, and adding the ${\tau }_i$ removes any $t{x}_i$ terms from an initial lift.

The Wang sequence identifies $a{b}_{12},a{b}_{13},a{b}_{23}$ as a free basis of $H^3$. Let $\kappa$ generate $H^4\cong \mathbb{Z}/2$, with $\rho \kappa =t{x}_1{x}_2{x}_3$. Reduction in degree four is an isomorphism. The integral cup products in the four-dimensional inversion mapping torus are consequently determined byTogether with the unit, graded commutativity and vanishing above degree four, these give every product. For example, $\rho (b_{12}{b}_{13})=x_1^2{x}_2{x}_3=t{x}_1{x}_2{x}_3$, whereas $\rho (b_{12}^2)=t^2{x}_1{x}_2=0$. The products $a{\tau }_i$ vanish because they are torsion in the free group $H^3$.

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.