Given covering space actions of groups \(G_1\) on \(X_1\) and \(G_2\) on \(X_2\), show that the action of \(G_1 \times G_2\) on \(X_1 \times X_2\) defined by \((g_1, g_2)(x_1, x_2) = (g_1(x_1), g_2(x_2))\) is a covering space action, and that \((X_1 \times X_2)/(G_1 \times G_2)\) is homeomorphic to \(X_1/G_1 \times X_2/G_2\).
Solution
1. The product action of \(G_1\times G_2\) on \(X_1\times X_2\) is free.
Suppose \[ (g_1,g_2)(x_1,x_2)=(x_1,x_2). \] Then \[ g_1x_1=x_1,\qquad g_2x_2=x_2. \] Since each \(G_i\) acts as a covering space action, in particular freely, we obtain \[ g_1=e_1,\qquad g_2=e_2. \]
2. The product action is a covering space action.
Fix \((x_1,x_2)\in X_1\times X_2\). Since the action of \(G_i\) is a covering space action, choose an open neighborhood \(U_i\) of \(x_i\) such that \[ g_iU_i\cap U_i=\varnothing \qquad(g_i\ne e_i). \] Set \[ U=U_1\times U_2. \] If \[ (g_1,g_2)U\cap U\ne\varnothing, \] then necessarily \[ g_1U_1\cap U_1\ne\varnothing, \qquad g_2U_2\cap U_2\ne\varnothing. \] Hence \(g_1=e_1\) and \(g_2=e_2\). Thus distinct translates of \(U\) are disjoint, which is exactly the local condition for a covering space action.
3. Let\[ q:X_1\times X_2\to (X_1\times X_2)/(G_1\times G_2) \]and\[ q_i:X_i\to X_i/G_i \]be the quotient maps.The map\[ q_1\times q_2:X_1\times X_2\to (X_1/G_1)\times(X_2/G_2) \]is constant on \((G_1\times G_2)\)-orbits.
For every \((g_1,g_2)\), \[ (q_1\times q_2)(g_1x_1,g_2x_2) =(q_1x_1,q_2x_2), \] since each \(q_i\) is constant on \(G_i\)-orbits.
4. Hence there is a unique continuous map\[ \Phi:(X_1\times X_2)/(G_1\times G_2) \to (X_1/G_1)\times(X_2/G_2) \]with\[ \Phi([(x_1,x_2)])=([x_1],[x_2]). \]It is bijective.
3.Surjectivity is immediate.For injectivity, suppose\[ ([x_1],[x_2])=([y_1],[y_2]). \]Then there exist \(g_i\in G_i\) with\[ y_i=g_ix_i. \]Therefore\[ (y_1,y_2)=(g_1,g_2)(x_1,x_2), \]so the two points lie in the same \((G_1\times G_2)\)-orbit.
5. The map \(\Phi\) is a homeomorphism.
Each quotient map \(q_i\) is open, since for an open set \(V_i\subseteq X_i\), \[ q_i^{-1}(q_i(V_i))=\bigcup_{g_i\in G_i}g_iV_i \] is open. Hence \(q_1\times q_2\) is open on the basis of product-open sets and therefore open. Since \[ q_1\times q_2=\Phi\circ q \] and \(q\) is a quotient map, this implies that \(\Phi\) is open. A continuous bijective open map is a homeomorphism. Thus \[ \boxed{(X_1\times X_2)/(G_1\times G_2)\cong X_1/G_1\times X_2/G_2.} \]