Product of CW complexes question I am having trouble understanding the product of CW complexes. I know how to actually do the computations and all, I just don't understand how exactly it works.
So here's my questions specifically: If $X,Y$ are CW-complexes then say $e,f$ are $p,q$ cells on $X,Y$ respectively, then we know that $e \times f$ will be a $p+q$ cell in $X \times Y$. But this cell we have to think of as $D^{p+q}$ with some identification on the boundary sphere. But what we have here is $(D^p \times D^q)$. So I guess we need to know that we have homeomorphism of pairs $(D^{p+q},S^{p+q-1}) \cong (D^p \times D^q, S^{p-1} \times D^q \cup D^p \times S^{q-1})$. But that's what I do not get. How exactly do this homeomorphism work?
I kinda see it in the case $p=q=1$ (it's just that a square and a disc are homeomorphic with canonical identifications in the boundaries). But I am having trouble defining it or actually visualizing it in higher dimensions. Any help?
Thanks!

I hope the following picture of a solid torus with the cells labelled will help: 

It is taken from Nonabelian Algebraic Topology (pdf available there), which also compares algebraic modelling of products of cells as in your question and in terms of cubes; these are  much easier to understand since the $m$-cube $I^m$ is the topological product of $m$ copies of the unit interval $I=[0,1]$ and so $I^{m+n} \cong I^m \times I^n$. The relation between cells and cubes is used a lot in algebraic topology. The natural cell structure of $I^m$ is more complicated than that of the $m$-cell $D^m= e^0 \cup e^{m-1} \cup e^m$, but you gain in other ways, so one needs both models. 
