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. Express the homotopy grops of $Z$ in terms of the homotopy groups of the spheres.
What I know: There's a result that says that since $\Bbb Z_2$ is a finite group and $S^m \times S^n$ is Hausdorff, then 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, $G(p) \cong \pi_1(Z)$.
So, since $m,n>1$, $\pi_1(S^m \times S^n)=\pi_1(S^m \times S^n) = \pi_1(S^m)\times \pi_1(S^n) =0\times 0 = 0$, and we have $\pi_1(Z)=\Bbb Z_2$, but how can this be expressed in terms of $ \pi_1(S^n)=0$ and $ \pi_1(S^m)=0$? And this is only for $\pi_1(Z)$, what about for $\pi_q(Z)$ for $q>1$? I can't find any result that can be used in that case.
Well for $\pi_1$ you have an "absolute" expression, so you can consider that to be "in terms of $\pi_1(S^n), \pi_1(S^m)$" (since it's in terms of nothing, just as $f: x\mapsto 2$ expresses $2$ as a function of $x$)
As for $\pi_n,n\geq 2$, you probably have a result that relates $\pi_n(X)$ to $\pi_n(Y)$ when $Y\to X$ is a covering map, don't you ?
If you don't, a hint is that $S^n,n\geq 2$ is simply-connected, so the lifting theorem allows you to lift maps $S^n\to X$ to maps $S^n\to Y$.