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Another one that has been bugging me:

Say $X$ is a finite CW complex. What is the simplest CW structure on $S^n \times X$

So I assume that $E$ is the family of cells in $X$ and $\Phi = \{ \Phi_e:e \in E \}$ is the family of attaching maps (technically I guess $\Phi_e | S^{k-1}$ is the attaching map of a $k$-cell).

I think that if I take the usual CW structure on the $n$-sphere (1 0-cell, $e^0_s$ and 1 $n$-cell $e^n_s$) that the family of cells $E'$ of $S^n \times X$ is just $E'=\{ e^0_s \times e, e^n_s \times e: e \in E \}$

But I am unsure how to attach it? I guess we only really need to worry in the instances we are attaching a 0-cell and an $n$-cell, else we can just use the usual maps.

Writing down the cellular chain complex is not too bad - it will just have an extra copy of $\mathbb{Z}$ in the $0$-th and $n$-th position.

Can we then calculate $H_k(S^n \times X)$ in terms of $H_k(X)$? (Again, the boundary formulas will only change in the $0$-th and $n$-th case

Edit: And can we do it without the Kunneth formula?

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@user8268 - thanks - I will edit the question. I thought it can be done without the Kunneth formula –  Juan S Apr 6 '11 at 8:22
    
for homology you only need to know "what cells of dimension $k-1$ are on the boundary of a dim-$k$-cell". In your example the chain complex is just the direct sum of the chain complex of $X$ and of its copy, where everything is multiplied by the dim-$n$ cell of $S^n$. Hence $H_k(S^n\times X)=H_k(X)\oplus H_{k-n}(X)$. –  user8268 Apr 6 '11 at 9:00
    
(what I said is true for $n>1$. Anyway, the chain complex for $X\times Y$ is simply the tensor product of the chain complexes of $X$ and of $Y$) –  user8268 Apr 6 '11 at 9:51
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1 Answer 1

up vote 5 down vote accepted

Here is a nice solution I was just shown:

Use the retraction $r:X \times S^n \to x \times \{ x_0 \}$ given by $(x,s) \mapsto (x,x_0)$ along with the LES of the pair $(X \times S^n, X \times \{ x_0 \})$ to show $$H_k(X \times S^n) \simeq H_k(X \times \{ x_0 \} ) \oplus H_k(X \times S^n, x \times \{ x_0 \} )$$

Then take $S^n$ as the upper and lower hemisphere, $D^n_+$ and $D^n_-$ with $D^n_+ \cap D^n_- \sim S^{n-1}$ such that $x_0 \in D^n_+ \cap D^n_-$. Then use a a relative Mayer-Vietrois sequence with the sets

$A = X \times D^n_+$

$B = X \times D^n_-$

$C = D = X \times \{ x_0 \}$

to get $$H_K(X \times S^n,X \times \{ x_0 \} ) \simeq H_{k-1}(X \times S^{n-1}, x \times \{ x_0 \} )$$

Iterate this and use excision to get

$$H_K(X \times S^n,X \times \{ x_0 \} ) \simeq H_{k-n}(X)$$

Combining the above we get

$$H_k(X \times S^n) \simeq H_k(X) \oplus H_{k-n}(X)$$ exactly as @user8268 says

Nice!

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