I am working on the following problem:
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 $X_1/G_1 \times X_2/G_2$.
The following is the definition that I have for a covering space action:
- (1) If $G$ is a group and $X$ is a topological space, then an action of $G$ on $X$ is a homomorphism $\phi:G \rightarrow \text{Homeo}(X)$, where $\text{Homeo}(X)$ is the set of all homeomorphisms $X \rightarrow X$. (Each $g \in G$ is associated to a homeomorphism $\phi(g):X \rightarrow X$, but we will write $g:X \rightarrow X$ in place of $\phi(g):X \rightarrow X$.)
- (2) If, in addition, each $x \in X$ has a neighborhood $U$ such that $g_1(U) \cap g_2(U) \neq \emptyset$ implies $g_1 = g_2$ for all $g_1,g_2 \in G$, then the action of $G$ on $X$ is said to be a covering space action.
It is easy for me to check that the action of $G_1 \times G_2$ on $X_1 \times X_2$ defined in the problem is indeed an action according to (1), but I'm struggling with showing that it is a covering space action according to (2).
Suppose $(x_1,x_2) \in X_1 \times X_2$, $U_1$ is a neighborhood of $x_1 \in X_1$ that yields the covering space action of $G_1$ on $X_1$ via (2), and $U_2$ is a neighborhood of $x_2 \in X_2$ that yields the covering space action of $G_2$ on $X_2$ via (2). My idea was to consider the neighborhood $U_1 \times U_2$ of $(x_1,x_2)$. I would have to show that, for all $(g_1,g_2),(g_3,g_4) \in G_1 \times G_2$, we have that $(g_1,g_2)(U_1 \times U_2) \cap (g_3,g_4)(U_1 \times U_2) \neq \emptyset$ implies that $(g_1,g_2) = (g_3,g_4)$. How can I go about showing this? From there, how can I go about showing the desired homeomorphism $(X_1 \times X_2)/(G_1 \times G_2) \cong X_1/G_1 \times X_2/G_2$?
Thanks!