Consider the product of the spheres $S^m \times S^n$, with $m,n>1$. Define a continuous action $\Bbb Z_2 \times S^m \times S^n \to S^m \times S^n$ with $g*(x,y)=(-x,-y)$ and let $Z$ be the orbit space of this action. Find a representant of each equivalence class of covering spaces of $Z$.
What I know: There's a result that says that since $\Bbb Z_2$ is a finite group, $S^m \times S^n$ is Hausdorff and the action is free, we have that the quotient map to the orbit space $p: S^m \times S^n \to Z$ is a covering space. There's another result that says that $\Bbb Z_2 \cong G(p)$, where $G(p)$ is the set of deck tranformations, and also, there's a third result that says that if $S^m \times S^n$ is simply connected, $G(p) \cong \pi_1(Z)$, and then we have that $\pi_1(Z)\cong \Bbb Z_2$.
Consider the following theorem:
Using this theorem, the subgroups of $\pi_1(Z)\cong \Bbb Z_2$ are only the trivial subgroup $0$ and $\Bbb Z_2$. The covering space $p: S^m \times S^n \to Z$ is universal, so we need to compute $(S^m \times S^n) / 0 = S^m \times S^n$ and $(S^m \times S^n) / \Bbb Z_2=Z$ (Right?). The representant of the first one is $p$, and a representant of the second one is the identity id$:Z \to Z$. Am i right ?