I do understand that the $k$ -cells of $X\times Y$ are $\{e^i_X\times e^{k-i}_Y\}$ where $e^i_X$ is an $i$-cell of $X$ and $e^{k-i}_Y$ is a $k-i$ cell of $Y$ . I have the attaching maps from the CW-structure of $X$ and $Y$, but I'm not being able to see how these give attaching maps from the $k$-cells of $X\times Y$ to its $k-1$-skeleton.
For starters , the boundary of $e^i_X\times e^{k-i}_Y$ consists of not just the product of the boundary of the individual cells but much more and I'm not being able to see how to map these extra things onto the $k-1$ -skeleton.
Can someone please help me understand this ? And also why is the product topology on $X \times Y$ , the same as the CW-topology induced by the their individual CW-structures(at least for finite CW-complexes)?
Best Answer
I guess that "cells" are supposed to be open cells. For $e^i_X$ we have an attaching map $\varphi : \partial D^i = S^{i-1} \to X^{(i-1)}$ and for $e^j_Y$ an attaching map $\psi : \partial D^j = S^{j-1} \to Y^{(j-1)}$. The attaching maps give us characteristic maps $\bar \varphi : D^i \to X^{(i)}$ and $\bar \psi : D^j \to Y^{(j)}$.
Moreover $ D^{i+j} \approx D^i \times D^j$, thus $S^{i+j} = \partial D^{i+j} \approx \partial D^i \times D^j \cup D^i \times \partial D^j$. The attaching map for $e^i_X \times e^j_Y$ is given by $$\chi : \partial D^i \times D^j \cup D^i \times \partial D^j \to X^{(i-1)} \times Y^{(j)} \cup X^{(i)} \times Y^{(j-1)} \subset (X\times Y)^{(i+j-1)}, \\ \chi(a,b) = \begin{cases} (\varphi(a), \bar \psi(b)) & (a,b) \in \partial D^i \times D^j \\ (\bar \varphi(a), \psi(b)) & (a,b) \in D^i \times \partial D^j \end{cases}$$