Wang sequence

For the mapping torus of $f:F\to F$, the Wang exact sequence contains

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.

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.

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.

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 $f$.