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If $X$ is a CW complex, show that there is a chain map $W_*(X) \to S_*(X)$ inducing isomorphisms in homology.

Here $W_p(X) = H_p(X^p,X^{p-1})$

Let $E$ be the CW decomposition of $X$ and let $M$ consist of one point chosen from each cell in $E$. In brief, we start with the triple $(X^k,X^k-M,X^{k-1})$, and take the long exact sequence, then use excision to get:

$W_p(X) = H_p(X^{(p)},X^{(p-1)}) \to H_p(X^{(p)},X^{(p)}-M) \to H_p(X^{(p)} - X^{(p-1)},X^{(p)}-X^{(p-1)}-M)$


$H_p(X^{(p)} - X^{(p-1)},X^{(p)}-X^{(p-1)}-M)\simeq \sum H_p(e_\lambda,e_\lambda-M)$

Let $r_p:W_p(X) \to \sum H_p(e_\lambda,e_\lambda-M)$ be this isomorphism.

The hint is to then define $s_p:\sum H_p(e_\lambda,e_\lambda -M) \to S_p(X)$ by sending a generator of $H_p(e_\lambda,e_\lambda - M)$ into $\Phi_{e_\lambda} \circ \alpha$, where $\alpha:(\Delta^p, \dot{\Delta}^p) \to (D^p,S^{p-1})$ is a homomorphism. Then the chain map is $\{s_p r_p \}$

So we are required to show that $d'_p S_p r_p \stackrel{?}{=} S_{p-1}r_{p-1}d_p$? (where the $d,d'$ are the boundary maps)

Any hints?

($S_*(X)$ must be singular homology)

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