Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 16 / 1

Use integral cellular homology. All three vertices become a single vertex, and all three oriented edges become one oriented circle, with generator $a$. The face goes round that circle three times in the same direction. Thus the quotient has a CW complex structure with one cell in each of dimensions zero, one and two, and cellular chain complexThe degree-two boundary is injective and its cokernel is $\mathbb{Z}/3$. ThereforeThe attaching word is $a^3$, so the Seifert-van Kampen theorem also gives ${\pi }_1(X)\cong \mathbb{Z}/3$. This is the degree-three circle-attachment model of a Moore space.

Let $a$ be the projective-line loop and $b$ the core loop of the Möbius band. The boundary of the Möbius band goes twice around its core. Its boundary inclusion is a cofibration, so replacing the gluing map by a homotopic cellular degree-$n$ map preserves the homotopy type. More precisely, the band is a mapping cylinder of the boundary double covering onto the core circle. After attaching its other end and collapsing the connecting tree, the resulting space has a homotopy-equivalent CW complex with one vertex, edges $a,b$, and two two-cells with attaching wordsThe first is the projective-plane face; the second records equality of the attached boundary loop $a^n$ and the band boundary loop $b^2$. Taking exponent sums gives the cellular chain complexSince $\det D=4$, its kernel is zero. The first invariant factor in the Smith normal form is the greatest common divisor of its entries, $d=\gcd (2,n)$, and the second is $4/d$. HenceThis includes $n=0$ and negative degrees. The fundamental group after attaching a Möbius band is $⟨a,b\mid a^2=1,b^2=a^n⟩$, consistent with the computed abelianization. The calculation is an example of homology after attaching a Möbius band to the real projective plane.

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.

For the triangle quotient, the universal coefficient theorem for homology gives $H_j(X;{\mathbb{F}}_2)=0$ for every $j>0$, because its only positive integral homology is $\mathbb{Z}/3$. A connected positive-dimensional closed manifold has a nonzero top mod-two fundamental class, independently of orientability. Thus this quotient cannot be homotopy equivalent to such a manifold. A connected zero-dimensional closed manifold is a point, which its nonzero $H_1(X;\mathbb{Z})$ also excludes.

For the attached-band space, the same universal coefficient theorem for homology gives nonzero $H_2(X;{\mathbb{F}}_2)$ and zero homology above degree two. A connected closed manifold homotopy equivalent to it would consequently have dimension two. Its Euler characteristic would be one. The classification theorem for surfaces says that a closed connected surface of Euler characteristic one is the real projective plane. But its integral $H_1$ is $\mathbb{Z}/2$, differing from both possibilities computed in part 2. Thus this space is never homotopy equivalent to a closed manifold.

The antipodal mapping torus is already a closed manifold: gluing the two boundary components of $S^n\times [0,1]$ by a diffeomorphism gives a smooth $(n+1)$-manifold without boundary. For $n\ge 1$, it is orientable exactly when $(-1{)}^{n+1}=1$, namely when $n$ is odd. In that case $-I$ on ${\mathbb{R}}^{n+1}$ is isotopic to the identity through rotations, so the mapping torus is diffeomorphic to $S^n\times S^1$. The $n=0$ case is a circle as above. Only the third family can be homotopy equivalent to a closed manifold.