If a finite CW complex \(X\) is the union of subcomplexes \(A\) and \(B\), show that \(\chi(X) = \chi(A) + \chi(B) - \chi(A \cap B)\).
Solution
Because \(A\) and \(B\) are subcomplexes of \(X\), every open cell of \(X=A\cup B\) lies in \(A\) or in \(B\), and it lies in both exactly when it is a cell of \(A\cap B\).
Let \(c_i(Z)\) denote the number of \(i\)-cells of a finite CW complex \(Z\). For each \(i\), ordinary inclusion–exclusion for the finite sets of \(i\)-cells gives \[ c_i(X)=c_i(A)+c_i(B)-c_i(A\cap B). \] Taking alternating sums, \[ \begin{aligned} \chi(X) &=\sum_i(-1)^i c_i(X)\\ &=\sum_i(-1)^i c_i(A) +\sum_i(-1)^i c_i(B) -\sum_i(-1)^i c_i(A\cap B)\\ &=\boxed{\chi(A)+\chi(B)-\chi(A\cap B)}. \end{aligned} \]