I am trying to understand (simplify) the following space
$Z = S^2 \times S^2 / E$
Let $(a_1 , a_2 , a_3) \in S^2$ and $(n_1 , n_2 , n_3) \in S^2$. Here ${a_1}^2 + {a_2}^2 + {a_3}^2 = 1$ and ${n_1}^2 + {n_2}^2 + {n_3}^2 = 1$.
The equivalence relations $E$ are the following
1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$
2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)]$
where $(\pi; a_1,a_2,a_3)$ is a rotation of angle $\pi$ along the axis $(a_1, a_2, a_3)$. So the point $(\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)$ is the mirror image of $(n_1, n_2, n_3)$ in the plane whose normal is $(a_1, a_2, a_3)$.
The objective is to simplify the equivalence relation (2) by a homeomorphism of $S^2 \times S^2$. For example, the approach I am trying is to rotate the second $S^2$ such that the mirror plane becomes horizontal (XY plane). Since the normal of the mirror plane is $(a_1, a_2, a_3)$, the amount by which the second $S^2$ has to be rotated depends on the $(a_1, a_2, a_3)$ it corresponds to. And for this transformation to be a homeomorphism the rotation operations should vary continuously. The transformation should not complicate the first equivalence relation; one way this condition could be imposed is by ensuring that the rotation operation corresponding to $(a_1, a_2, a_3)$ and $(-a_1, -a_2, -a_3)$ are be the same. I am not able to come up with such a transformation. All my attempts to simplify (2) end up complicating (1).
To be more specific, I want to get a homeomorphism between the space $Z$ and the following space:
$Z^' = S^2 \times S^2 / E^' $
The equivalence relations $E^'$ are the following
1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$
2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (n_1, n_2, -n_3)]$
Any ideas are appreciated. Please let me know if you have any questions.