Show that if \(p_1: \tilde{X}_1 \to X_1\) and \(p_2: \tilde{X}_2 \to X_2\) are covering spaces, so is their product \(p_1 \times p_2: \tilde{X}_1 \times \tilde{X}_2 \to X_1 \times X_2\).
Solution
1. Fix \((x_1,x_2)\in X_1\times X_2\) and choose evenly covered neighborhoods\[ x_i\in U_i\subseteq X_i \qquad(i=1,2) \]for the coverings \(p_i\).
For each \(i\), the covering-space property gives a decomposition \[ p_i^{-1}(U_i)=\coprod_{\alpha_i\in A_i}V^i_{\alpha_i}, \] where each \[ p_i|_{V^i_{\alpha_i}}:V^i_{\alpha_i}\to U_i \] is a homeomorphism.
2. The product \(U_1\times U_2\) is an open neighborhood of \((x_1,x_2)\) in \(X_1\times X_2\).
Products of open sets form a basis for the product topology.
3. The inverse image of this product neighborhood is\[ (p_1\times p_2)^{-1}(U_1\times U_2) = \coprod_{(\alpha_1,\alpha_2)\in A_1\times A_2} \left(V^1_{\alpha_1}\times V^2_{\alpha_2}\right). \]
One has \[ (p_1\times p_2)^{-1}(U_1\times U_2) =p_1^{-1}(U_1)\times p_2^{-1}(U_2). \] Substituting the disjoint sheet decompositions from 1 and distributing the Cartesian product gives the displayed disjoint union. Each product sheet is open in \(\widetilde X_1\times\widetilde X_2\).
4. On every product sheet,\[ p_1\times p_2: V^1_{\alpha_1}\times V^2_{\alpha_2} \longrightarrow U_1\times U_2 \]is a homeomorphism.
It is the product of the two homeomorphisms \[ p_1|_{V^1_{\alpha_1}} \quad\text{and}\quad p_2|_{V^2_{\alpha_2}}. \] Its inverse is the product of their continuous inverses.
5. Hence \(p_1\times p_2\) is a covering map.
The point \((x_1,x_2)\) was arbitrary, and <1>2–<1>4 show that it has an evenly covered neighborhood.