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The antipodal map on the two-sphere reverses orientation, so its mapping torus is a nonorientable closed three-manifold. The Wang sequence gives $H_0=H_1=\mathbb{Z}$, $H_2=\mathbb{Z}/2$ and $H_3=0$. The universal coefficient theorem for cohomology therefore gives the cohomology ring $\mathbb{Z}[t,u]/(t^2,tu,u^2,2u)$ with $|t|=1$, $|u|=3$. The base-circle class satisfies $t^2=0$ by pullback; the other positive-degree products vanish for dimension reasons.

For $n\ge 2$, the Wang sequence and the mapping degree $(-1{)}^{n+1}$ of the antipodal map give $H_0=H_1=\mathbb{Z}$. If $n$ is odd, $H_n=H_{n+1}=\mathbb{Z}$; if $n$ is even, $H_n=\mathbb{Z}/2$ and $H_{n+1}=0$. All other groups vanish. At $n=1$ the mapping torus is a torus with $H_1={\mathbb{Z}}^2$. At $n=0$, the two-point swapping mapping torus is a circle. Every example is already a closed manifold; for $n\ge 1$ it is orientable exactly when $n$ is odd.

For $r=n-1$, the Wang sequence gives an additive splittingThe kernel is free, so the splitting exists as groups. The map in even degree is zero; the map in odd degree is multiplication by minus two.

This mapping torus fibers over the circle with fiber $T^{n-1}$ and inversion as monodromy. The Wang sequence computes its additive cohomology. Coordinate projections to the Klein bottle expose its mod-two cup products.

The mapping torus of $f:F\to F$ isIf $f$ is a homeomorphism, the projection to the circle $S^1$ is a fiber bundle with fiber $F$. For a general continuous map, the mapping torus need not be a fiber bundle, but its homology still has the Wang sequence.

For a continuous self-map $f:X\to X$, the mapping torus has the Wang sequence even when $f$ is not a homeomorphism. Cover the time circle by a seam neighborhood and an interval away from the seam. Their inverse images retract to $X$: the seam neighborhood is a mapping cylinder, and the other is a product. Their intersection retracts to two copies of $X$. The Mayer–Vietoris sequence map on homology is, after consistent choices of signs,Invertible changes of coordinates reduce it to the direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the isomorphism summands gives the Wang sequence. This proof does not assume a fiber bundle structure for a noninvertible $f$.

Modulo two the antipodal map acts trivially on fiber cohomology. The Wang sequence supplies a class $y$ restricting to the generator of $H^2(S^2;{\mathbb{F}}_2)$. The Leray-Hirsch theorem with classes $1,y$ gives a basis $1,t,y,ty$, so $ty\ne 0$. Pullback from the circle gives $t^2=0$, while dimension gives $y^2=0$. This cohomology ring agrees with that of $S^1\times S^2$, although the original manifold is nonorientable.

The space is the mapping torus of the antipodal map $A:S^2\to S^2$. Its mapping degree is $(-1{)}^3=-1$, so transporting a fiber orientation once around the base circle reverses it. $X$ is nonorientable.

On integral fiber homology, $A_{\ast }$ is the identity in degree zero and minus the identity in degree two. The Wang sequence, whose relevant map is $1-A_{\ast }$, therefore givesThe universal coefficient theorem for cohomology shifts the torsion contribution into degree three, because ${\mathrm{Ext}}_{\mathbb{Z}}(\mathbb{Z}/2,\mathbb{Z})=\mathbb{Z}/2$. HenceLet $t\in H^1(X;\mathbb{Z})$ be pulled back from the base circle, and $u$ the nonzero degree-three class. The square of the circle class vanishes, so $t^2=0$. All other products of positive-degree classes land above dimension three and vanish. This gives the cohomology ring of the antipodal two-sphere mapping torusModulo two, $A^{\ast }$ is the identity on fiber cohomology. The cohomological Wang sequence shows that restriction $H^2(X;{\mathbb{F}}_2)\to H^2(S^2;{\mathbb{F}}_2)$ is onto; choose $y$ mapping to its generator. The classes $1,y$ restrict to a fiber basis, so the Leray-Hirsch theorem makes the total cohomology free over $H^{\ast }(S^1;{\mathbb{F}}_2)$ on those two classes. Write $\bar{t}$ for the mod-two base class. The basis is $1,\bar{t},y,\bar{t}y$, and in particular $\bar{t}y\ne 0$. Since ${\bar{t}}^2=0$ by pullback and $y^2=0$ by dimension, the mod-two cohomology ring of the antipodal two-sphere mapping torus isThe argument of Question 3 works over any field, so its use here over ${\mathbb{F}}_2$ is justified. The ring agrees with that of $S^1\times S^2$; a cohomology ring alone does not determine orientability.

The space is the mapping torus of the antipodal map $A:S^n\to S^n$. For $n\ge 1$, its mapping degree is $\delta =(-1{)}^{n+1}$: $A$ is the restriction of $-I$ on ${\mathbb{R}}^{n+1}$, whose determinant is $(-1{)}^{n+1}$. The Wang sequence givesThe kernel terms are free abelian, so these short exact sequences split as groups. On $H_0(S^n)$ the map is the identity, and on $H_n(S^n)$ it is multiplication by $\delta$. For $n\ge 2$ this givesAt $n=1$, both contributions to degree one are present, giving $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^2$, with all other groups zero. At $n=0$, the antipodal map exchanges the two points of $S^0$, and the two cylinder intervals join into a single circle; thus $H_0=H_1=\mathbb{Z}$ and higher groups vanish. These are the homology of the antipodal sphere mapping torus cases.

The torus map acts on $H_1(T^2;\mathbb{Z})$ byIts determinant is one, so it is an orientation-preserving homeomorphism and acts as the identity on both $H_0(T^2)$ and $H_2(T^2)$.

To calculate the homology of its mapping torus, cover the time circle by two arcs. Each inverse image has the homotopy type of $T^2$, while their intersection has two components of that homotopy type. The map in the Mayer–Vietoris sequence can be writtenAdding the output coordinates and using the input coordinate $a+b$ separates an identity map from $f_{\ast }-I$. Canceling the identity summands gives the Wang sequence, and henceIn degree one,Its image and kernel are both generated by $(1,-1)$; its cokernel is $\mathbb{Z}$ since this image is primitive. Its Smith normal form is $\mathrm{diag}(1,0)$, so its torsion subgroup is zero. On the other two nonzero homology groups of the torus, $f_{\ast }-I=0$. The displayed short exact sequences split because their right-hand groups are free. ThereforeIn particular, the mapping torus contributes an additional degree-one class from the time circle, as well as the invariant degree-one class in the fiber.

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 $H^3$.

The quotient is the mapping torus ofA mapping torus of a cellular map has a finite CW structure. On integral homology, ${\tau }_{\ast }$ is $1$ on $H_0(T^2)$ and $H_2(T^2)$ and is $-I$ on $H_1(T^2)\cong {\mathbb{Z}}^2$. The Wang sequence therefore givesOver ${\mathbb{F}}_2$, the map ${\tau }_{\ast }$ is the identity. The Wang sequence splits as vector spaces into one copy of $H_q(T^2;{\mathbb{F}}_2)$ and one copy of $H_{q-1}(T^2;{\mathbb{F}}_2)$, yielding