Topological structure of a quotient of ${\rm{SU}}(2)\times{\rm{SU}}(2)$ I'm trying to understand the topology of the product of two three dimensional spheres $\mathbb{S}^3\times \mathbb{S}^3$ quotiented by the action of $\pm 1$ sending a pair of points $(x,y)$ to the corresponding pair of antipodal points $(-x,-y)$. My hunch is that this may be better understood by identifying $\mathbb{S}^3$ with the special unitary group $SU(2)$ via the well known homeomorphism and understanding the quotient of the topological group $SU(2)\times SU(2)$ by the group $H=\{I_2\times I_2,-I_2\times -I_2\}$ as a topological space but I haven't yet been successful with that either. I'd like to know if this approach is worthwhile or if I should be looking in another direction to recognize the space $\mathbb{S}^3\times \mathbb{S}^3/\pm1$.
 A: The quotient $S^3\times S^3/\{\pm (1,1)\}$ is diffeomorphic (but not Lie isomorphic) to $S^3\times \mathbb{R}P^3$  (Here, $\mathbb{R}P^3$ is naturally diffeomorphic to $SO(3)$, so we give it the Lie group structure $SO(3)$ has).
An explicit map is given as follows.  Identifying $S^3$ as the unit quaternions, define $\tilde{f}:S^3\times S^3\rightarrow S^3\times \mathbb{R}P^3$ by $\tilde{f}(p,q) = (pq, [q])$ where $[q]$ denotes the class of $q$ in $\mathbb{R}P^3$.
Notice that $$\tilde{f}(-p,-q) = \left((-p)(-q), [-q]\right) = (pq,[q]) = \tilde{f}(p,q)$$ so $\tilde{f}$ descends to a map on $S^3\times S^3/\pm(1,1)$, which I'l call $f$.
The map $\tilde{f}$ is clearly smooth, so $f$ is as well.  Further, $f$ has an inverse given by $f^{-1}(p,[q]) = [pq^{-1},q]$.  This is also clearly smooth, so we have that $f$ is a diffeomorphism.
Why aren't they isomorphic as Lie groups?  I claim that $S^3\times \mathbb{R}P^3$ has a normal subgroup isomorphic to $SO(3)$ while $S^3\times S^3/\pm(1,1)$ doesn't.
First, since $\mathbb{R}P^3$ is isomorphic to $SO(3)$, the subgroup $\{e\}\times \mathbb{R}P^3$ is isomorphic to $SO(3)$ and is normal in $S^3\times\mathbb{R}P^3$.
Second, to see there are no normal subgroups isomorphic to $SO(3)$ in $S^3\times S^3/\pm(1,1)$ note that the Lie algebra of $S^3\times S^3/\pm(1,1)$ is $\mathfrak{so}(3)\oplus\mathfrak{so}(3)$ and $\mathfrak{so}(3)$ is simple, so the only nontrivial ideals are $\mathfrak{so}(3)\oplus 0$ and $0\oplus \mathfrak{so}(3)$.  In $S^3\times S^3$, these exponentiate to the two different $S^3$ factors and only can easily check that under the projection $\pi:S^3\times S^3\rightarrow S^3\times S^3/\pm(1,1)$, $\pi$ is injective on each factor.
It follows that the two ideals in $\mathfrak{so}(3)\oplus\mathfrak{so}(3)$ exponentitate to $S^3$s in $S^3\times S^3/\pm(1,1)$, so, in particular, they are not isomorphic to $SO(3)$.
