For \(X\) a finite CW complex and \(p: \tilde{X} \to X\) an \(n\) sheeted covering space, show that \(\chi(\tilde{X}) = n\chi(X)\).
Solution
Give \(X\) a finite CW structure. Pull this structure back along the \(n\)-sheeted covering \[ p:\widetilde X\to X. \]
1. Every open \(k\)-cell of \(X\) has exactly \(n\) lifts, each an open \(k\)-cell of \(\widetilde X\).
An open cell is contractible and hence simply connected. The restriction of the covering over the cell is therefore a disjoint union of homeomorphic copies of the cell. Since each point has exactly \(n\) preimages, there are exactly \(n\) such copies. The lifted characteristic maps provide a CW structure on \(\widetilde X\).
If \(c_k(X)\) denotes the number of \(k\)-cells, then \[ c_k(\widetilde X)=n\,c_k(X). \] Therefore \[ \begin{aligned} \chi(\widetilde X) &=\sum_k(-1)^k c_k(\widetilde X)\\ &=n\sum_k(-1)^k c_k(X)\\ &=\boxed{n\chi(X)}. \end{aligned} \]