Homeomorphism of $SO(3)$? I am trying to get a better understanding of the homeomorphism of $SO(3)$ to the Real Projective Plane, so that ultimately I can show that $\pi_1(SO(3)) = \mathbb{Z}_2$.
From wikipedia and many other sources, $SO(3)$ is homeomorphic to $\mathbb{RP}^3$, but this section describes seemingly a topological space that is homeomorphic to $\mathbb{RP}^2$:

The Lie group SO(3) is diffeomorphic to the real projective space
  RP3.[3]
Consider the solid ball in R3 of radius π (that is, all points of R3
  of distance π or less from the origin). Given the above, for every
  point in this ball there is a rotation, with axis through the point
  and the origin, and rotation angle equal to the distance of the point
  from the origin. The identity rotation corresponds to the point at the
  center of the ball. Rotation through angles between 0 and −π
  correspond to the point on the same axis and distance from the origin
  but on the opposite side of the origin. The one remaining issue is
  that the two rotations through π and through −π are the same. So we
  identify (or "glue together") antipodal points on the surface of the
  ball. After this identification, we arrive at a topological space
  homeomorphic to the rotation group.
Indeed, the ball with antipodal surface points identified is a smooth
  manifold, and this manifold is diffeomorphic to the rotation group.

The description above makes sense to me, each point in the the solid ball in $\mathbb{R}^3$, represents a rotation with the 'amount of rotation' given by it's distance from the centre and the line from the point to the centre identifying the axis of rotation. Then identifying the anti-podal points on the surface - this seems to be pretty much the exact construction of $\mathbb{RP}^2$! (Except the ball here has radius $\pi$).
So I guess my confusion is: Is $SO(3)$ homeomorphic to $\mathbb{RP}^2$ or $\mathbb{RP}^3$ or both?
 A: To construct the projective plane, you take the sphere and identify antipodal points. That's not the same thing as what you're doing here, when you take the ball and identify antipodal points on the boundary sphere only. Then you get the interior of the the ball, which is homeomorphic to all of $\mathbb{R}^3$, together with a projective plane $\mathbb{RP}^2$ "at infinity", and this combined gives you projective space, $\mathbb{RP}^3$.
(By the way, $SO(3)$ is three-dimensional, so it would be strange if it were homeomorphic to a plane.)
A: Let's say there is a 3D rotation $(n, \omega)$ by an angle $\omega$ around an unit vector $n=n(\theta,\phi)=(n_x,n_y,n_z)$ in $\mathbb{R}^3$ with $n_x=\sin\theta\cos\phi,n_y=\sin\theta\sin\phi,n_z=\cos\theta,\theta\in[0,2\pi),\phi\in[0,\pi]$. This element of $SO(3)$ group could be represented as the rotation matrix given by Rodrigues' formula:
$\begin{equation}
R_n(\omega)=
\left[ \begin{array}{ccc}
\cos\omega+n_x^2(1-\cos\omega) & -n_z\sin\omega+n_x n_y(1-\cos\omega) & n_y\sin\omega+n_x n_z(1-\cos\omega)\\
n_z\sin\omega+n_y n_x(1-\cos\omega) & \cos\omega+n_y^2(1-\cos\omega) & -n_x\sin\omega+n_y n_z(1-\cos\omega)\\
-n_y\sin\omega+n_z n_x(1-\cos\omega) & n_x\sin\omega+n_z n_y(1-\cos\omega) & \cos\omega+n_z^2(1-\cos\omega)
\end{array}
\right ]\end{equation}$.
The corresponding elements of $SU(2)$ group are
$\begin{equation}
u(n,\omega)=
\left[ \begin{array}{cc}
\cos\frac{\omega}{2}-i\sin\frac{\omega}{2}\cos\theta & -\sin\frac{\omega}{2}\sin\theta\sin\phi-i\sin\frac{\omega}{2}\sin\theta\cos\phi \\
\sin\frac{\omega}{2}\sin\theta\sin\phi-i\sin\frac{\omega}{2}\sin\theta\cos\phi & \cos\frac{\omega}{2}+i\sin\frac{\omega}{2}\cos\theta
\end{array}
\right ]\end{equation}$
and $u(-n,2\pi-\omega)=-u(n,\omega)$.
Both $u(n,\omega)$ and $u(-n,2\pi-\omega)$ correspond to the same 3D rotation $R_n(\omega)$.
The $SO(3)$ group and $SU(2)$ group could be shown in 3D space as 3-ball as shown in the following picture.
SO(3) group and SU(2) group as 3-balls in 3D Each vector from the origin with the arrow along $n$ and with length $\omega$ represents $R_n(\omega)\in SO(3)$ when $\omega\le\pi$ and represents $u(n,\omega)\in SU(2)$ when $\omega\le2\pi$.
The $SO(3)$ group is a 3-ball with radius of $\pi$. The antipodal points on the surface of $\omega=\pi$ are identical 3D rotations. Within the 3-ball with radius of $\pi$, the $u(n,\omega)\in SU(2)$ is at the same point of the corresponding $R_n(\omega)\in SO(3)$. The $SO(3)$ group could not go beyond radius of $\pi$ in this picture, while the $SU(2)$ group could go up to radius of $2\pi$. The shell with $\omega\in[\pi, 2\pi]$ contains $u(n,\omega)=-u(-n,2\pi-\omega)$ for $\omega\le\pi$. Specially, $u(n,2\pi)=-u(-n,0)=-1$ for all $n$.
All above are with 3-ball in 3D space.
Now to answer the question: what about $SO(3)$ group with $\mathbb{RP}^3$?
For $u\in SU(2)$, suppose $u=\left[ \begin{array}{cc}
x_0-ix_3 & x_1+ix_2 \\
-x_1+ix_2 & x_0+ix_3
\end{array}
\right ]$. We have $x_0^2+x_1^2+x_2^2+x_3^2=1$. So the $SU(2)$ group is a 3-sphere $S^3$ which is in 4D space. And we know that both $u$ and $-u$ correspond to one $R\in SO(3)$, thus the $SO(3)$ group is homeomorphic to $\mathbb{RP}^3$ which is a projected 3-sphere by identifying antipodal points of $S^3$ in $\mathbb{R}^4$. So for $SO(3)$ group, $(-x_0, -x_1, -x_2, -x_3)$ is identical to $(x_0, x_1, x_2, x_3)$. We could not direct draw the picture of $\mathbb{RP}^3$ in 4D space. What we could do is to project 4D $\mathbb{RP}^3$ to 3D by scanning one variable, say $x_0\in[-1,1]$.
We have $x_1^2+x_2^2+x_3^2=1-x_0^2$. By scanning $x_0\in[-1,1]$, we will have a serial of 2D spheres $S^2$ with variant radius which assemble to a 3D ball as we discussed above.
For $x_0 = 1$, $x_1^2+x_2^2+x_3^2=0$, that's the origin point of the ball and $u=1$.
For $x_0 = 0$, $x_1^2+x_2^2+x_3^2=1$, that's the outer surface of $SO(3)$ group with radius $\omega=\pi$. And in this case, we could directly see the projective effect of $\mathbb{RP}^3$ in 3D space.
For $1\ge x_0 \ge 0$, those 2D spheres are together the ball for both $SO(3)$ group and $SU(2)$ group with $0\le \omega \le \pi$.
For $0\ge x_0 \ge -1$, those 2D spheres are together the shell for only $SU(2)$ group with $\pi\le \omega \le 2\pi$.
For $x_0 = -1$, $x_1^2+x_2^2+x_3^2=0$, that's the outer surface of $SU(2)$ group with radius $\omega=2\pi$ and $u=-1$.
So the projective effect of $\mathbb{RP}^3$ in 4D space for $SO(3)$ group could be viewed in 3D space by glue all pairs of corresponding points $(n,\omega)$ and $(-n,2\pi-\omega)$ across the surface $\omega=\pi$ within the ball with radius up to $\omega=2\pi$.
