Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 16 / 1 / 2 / Solution

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.