For finite CW complexes \(X\) and \(Y\), show that \(\chi(X \times Y) = \chi(X)\chi(Y)\).
Solution
Give \(X\) and \(Y\) finite CW structures. If \(c_p(X)\) and \(c_q(Y)\) denote the numbers of cells in dimensions \(p\) and \(q\), then the product CW structure on \(X\times Y\) has one cell \[ e^p\times e^q \] of dimension \(p+q\) for every ordered pair of cells \((e^p,e^q)\). Hence the number of \(r\)-cells of \(X\times Y\) is \[ c_r(X\times Y)=\sum_{p+q=r}c_p(X)c_q(Y). \] Therefore \[ \begin{aligned} \chi(X\times Y) &=\sum_r(-1)^r c_r(X\times Y)\\ &=\sum_{p,q}(-1)^{p+q}c_p(X)c_q(Y)\\ &=\left(\sum_p(-1)^p c_p(X)\right) \left(\sum_q(-1)^q c_q(Y)\right)\\ &=\boxed{\chi(X)\chi(Y)}. \end{aligned} \]