Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford / 2009 / B / 5 / a / Solution

A CW complex is built from a discrete set of zero-cells by attaching $k$-discs along maps of their boundary spheres into the previous skeleton. Its topology is the weak topology determined by the characteristic maps, and each cell closure meets only finitely many cells. The k-skeleton $X^k$ is the union of cells of dimension at most $k$, with $X^{-1}=\varnothing$.

Set $C_k^{CW}(X)=H_k(X^k,X^{k-1};\mathbb{Z})$. This is the free abelian group on the $k$-cells. The cellular boundary map is ${\partial }_k=q_{k-1}{\delta }_k$ with the maps in the question. In the long exact sequence of the pair $(X^{k-1},X^{k-2})$, the consecutive maps areExactness gives ${\delta }_{k-1}{q}_{k-1}=0$. HenceAt the lowest indices put negative chain groups equal to zero.

The cellular homology is $H_n^{CW}(X)=\ker {\partial }_n/\mathrm{im}{\partial }_{n+1}$. For a field $\mathbb{F}$, use $C_k^{CW}(X;\mathbb{F})=H_k(X^k,X^{k-1};\mathbb{F})$, the vector space with basis the $k$-cells, with the same connecting-and-quotient construction. Its homology is $H_n^{CW}(X;\mathbb{F})$. Equivalently tensor the integral free cellular chain complex with $\mathbb{F}$. The cellular homology theorem identifies these with singular homology using the same coefficients.