Cellular homology of $X\times S^n$

Question

I want to compute the cellular homology for $X\times S^n$ where $X$ is a CW-complex. My current observation is that the $k-$th cellular group will be generated by cells $\{e^k_i\times e^0, e^{k-n}_j\times e^n\}$ where the upper index is the dimension of the cell and the lower index is just the index of cells in that dimensions. Well, intuitively, I think this structure will give the cellular homology group to be $H_k(X\times S^n)=H_k(X)\oplus H_{k-n}(X)$, but I'm not sure how to argue this rigorously. That being said, I want to know why we can just decompose those generators to two parts and use them generate the homology group on their own and then do direct sum to get the result. I'm appreciate if you can provide any insight on this, thanks!

Answer 1

You should be a little carefull when dealing with dimensions here. For instance, what is the dimension of $X$? Is it infinite dimensional? If $k$ is smaller than $n$, then $H_{k-n}(X)$ is the trivial group but it is not clear where $k$ ranges. In any case, for the sake of simplicity assume $dimX=n$ and use Mayer-Vietoris sequence, you should easly find an excisive couple: $X_{1}=X\times B_{1}$ and $X_{2}=X\times B_{2}$ where $B_1,B_2$ cells such that $S^{n}=B_1\cup B_2$ and $B_1\cap B_2$ is a deformation retract of $S^{n-1}$. So M-V and induction on $n$ should suit your problem. Just observe, in the case $n=0$, what is $X\times S^{0}$, I think your hunch is correct.