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?
Best Answer
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!