TITLE: prove $RP^3\cong SO(3)$
QUESTION [8 upvotes]: Suppose $RP^3$ is  the real 3-dimensional projective space,prove  the rotation group $SO(3)$ is homeophoric to $RP^3$.

REPLY [10 votes]: 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 $\theta \in [-\pi, \pi]$