Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 114 / 1

For a CW complex $X$, the cellular chain complex isthe free abelian group generated by the oriented $n$-cells. Its differential is the compositeNaturality of the long exact sequences of the triples $X^{n-2}\subset X^{n-1}\subset X^n$ makes two successive connecting maps compose to zero, so $d_{n-1}{d}_n=0$.

The cellular boundary formula says that the coefficient of an $(n-1)$-cell $e_{\alpha }^{n-1}$ in the boundary of $e_{\beta }^n$ is the degree of

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The quotient of the equator is $S^1/(z\sim -z)\cong S^1$, giving one zero-cell and one one-cell. The interiors of the upper and lower hemispheres give two two-cells. Each boundary circle maps to the quotient equator by the degree-two covering, so, after choosing orientations,is the cellular chain complex; changing one orientation only changes one sign. Therefore

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The point $x$ on the quotient equator has two preimages. Small discs around them become four half-discs glued along their common diameter, so a neighborhood of $x$ is the cone on a graph with two vertices joined by four edges. The Excision theorem and the local homology from a link identifywhere $L$ is this four-edge graph. It is connected and has first Betti number $4-2+1=3$. Hence

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No. If a finite CW complex $Y$ had exactly one two-cell, then $C_2^{\mathrm{cell}}(Y)\cong \mathbb{Z}$. The isomorphism $H_2(Y)\cong H_2(X)\cong \mathbb{Z}$ forces $d_2=0$ and $\mathrm{im}{d}_3=0$: a subquotient of $\mathbb{Z}$ can remain infinite cyclic only in this way. Consequentlyis a subgroup of the free abelian group $C_1(Y)$ and is therefore torsion-free. This contradicts the homotopy-invariant calculation $H_1(Y)\cong H_1(X)\cong \mathbb{Z}/2$.

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