The set of rotations in $\mathbb{R}^{3}$ with the operation of composition is a group known as $SO(3)$. $S^{3}$ with the operation of quaternion multiplication is also a group. Verify that $S^{3}/\{-1,1\} \approxeq SO(3)$
What is the meaning of the quotient topology on $\{-1,1\}$? I know what is $S^{3}$ but not $S^{3}/\{-1,1\}$.
How do verify? I am supposed to find a homeomorphism right?
The set $S^3/\{\pm1\}$ is the set $\{\pm v\,|\,v\in S^3\}$. Let $\pi\colon S^3\longrightarrow S^3/\{\pm1\}$ be the map defined by $\pi(v)=\pm v$. Then a set $A\subset S^3/\{\pm1\}$ is open if and only if $\pi^{-1}(A)$ is an open subset of $S^3$. In other words, the topology of $S^3/\{\pm1\}$ is the final topology with respect to $\pi$.
If $q\in S^3$ and if $v\in\mathbb H$ is of the type $ai+bj+ck$ (that is, $v$ is purely quaternionic), then $q^{-1}.v.q$ is also of the same type. Besides $\|v\|=\bigl\|q^{-1}.v.q\bigr\|$. This defines a map from $S^3$ into $SO(3)$, which happens to be a group homomorphim.