prove $RP^3\cong SO(3)$ [closed]

Suppose $RP^3$ is the real 3-dimensional projective space,prove the rotation group $SO(3)$ is homeophoric to $RP^3$.

Each rotation in $\Bbb R^3$ is characterized by an "oriented axis" $v\in S^2$ and an angle $\varphi\in [0,\pi]$ and the only relations are $(v,\pi)=(-v,\pi)$ and $(v,0)=(w,0)$ for each $v,w\in S^2$. If you represent $\Bbb RP^3$ as a 3-ball of diameter $\pi$ with identified antipodal points $v\cdot \pi=-v\cdot \pi$ for each $v\in S^2$, then the map $SO(3)\to \Bbb RP^3$ just maps $(v,\varphi)$ to $[v\cdot \varphi]$. The angle $\varphi\,\,\mathrm{mod}\,2\pi$ depends continuously on the rotation and the axis $v$ depends continuously on the rotation whenever $\varphi\neq 0$.

The following proposition is incredibly useful here:

Proposition The homomorphism $R: S^{3} \rightarrow SO(3)$ is surjective with $\ker{R}=\{\pm 1\}$ equal to the centre of $S^{3}$. In particular, every matrix of the form \begin{pmatrix} a^{2}+b^{2}-c^{2}-d^{2} & 2(bc-ad) & 2(bd + ac)\\ 2(bc+ad) & a^{2}+b^{2}-c^{2}-d^{2} & 2(cd-ad) \\ 2(bd-ac) & 2(cd+ad) & a^{2}-b^{2}-c^{2}+d^{2} \end{pmatrix} for some $\{a, b, c, d\} \in \mathbb{R}^{4}$ with $a^{2}+b^{2}+c^{2}+d^{2}=1$ The cosets of $\{\pm1\}$ in $S^{3}$ are simply pairs of antipodal points. Each pair determines a line in $\mathbb{R}^{4}$. the proposition above then determines that $SO(3)=\mathbb{RP}^{3}$ as topological spaces.

Aside

You can also regard $\mathbb{RP}^{3}$ as the quotient of a solid ball in $\mathbb{R}^{3}$ by identifying antipodal points on the boundary. Every element in $SO(3)$ is a rotation about some axis by some angle $\theta \in [-\pi, \pi]$