Cellular homology of suspension

Question

As part of an assignment I would like to show directly from the definition of cellular homology that if $X$ is a CW-complex then: $$H^{\mathrm{Cell}}_{n+1}(S(X);\mathbb{Z})\cong H^{\mathrm{Cell}}_n(X;\mathbb{Z})$$ That is I'm supposed to find a CW-structure on the unreduced suspension of X and show it's cellular homology is the same as the cellular homology of X without applying theorems like Mayer-Vietoris or using in advance that it's true for singular homology.

However, I am aware of the proof that $H^{\mathrm{Cell}}_n(X;\mathbb{Z})\cong H_n(X;\mathbb{Z}) $ which can be shown directly from the definition so this at least can be used.

For us, the chain complex of a CW-complex $Y$ is denoted $C_{n}^{CW}(Y)=H_n(Y^n,Y^{n-1})$ with chain maps coming from compositions in long exact sequences of pairs as in the definition in p.139 in Hatcher's book.

The CW-structure that I guessed we should use is induced by the suspension. for any $n$-cell in $X$ we would have two $n$-cells in $S(X)$.

Will appreciate feedback, stuck on this for quite some time without progress.

Answer 1

There is a cell structure on a product (see appendix of Hatcher) of cell complexes, with cells given by pairs of cells of the factors. This way in particular you get a cell structure on $X\times I$. If you use the standard cell structure on $I$ then the map $X\times I \to SX$ is the quotient by a subcomplex, which also has a natural cell structure.

The differentials behave as you expect, so you should be able to work out the cellular chain complex and its homology. The idea is something like “$d(e\times f) = de\times f \pm e\times df$ But $df=0$ because of the quotient by a subcomplex step.”