I have heard that $SO(3)$ is "isomorphic" to a sphere. Is this true? if so, how do we prove it? i am not claiming that $SO(3)$ and a sphere are isomorphic. instead, i am asking if they are, and if so, for an intuitive proof. 
 A: No, the 3D rotation group $SO(3) \cong SU(2)/ \mathbb{Z}_2  \cong S^3/ \mathbb{Z}_2\cong RP^3$ is isomorphic to the real 3-dimensional projective space.
On the other hand, the double cover
$$SU(2)~\cong~\left\{\left. \begin{pmatrix} \alpha & \beta \cr -\bar{\beta} & \bar{\alpha} \end{pmatrix} \in {\rm Mat}_{2\times 2}(\mathbb{C}) \right| \alpha, \beta\in\mathbb{C}, |\alpha|^2+|\beta|^2=1\right\}~\cong~S^3  $$ is a 3-sphere.
A: It's not. It's isomorphic to a sphere and a circle; i.e.
$$\mathrm{SO}(3)\cong S^2\times S^1. $$
To see why, look into Euler angles or Tait-Bryan angles. Those angles can be grouped as parameterizing a unit vector (element of $S^2$, a sphere) and a single rotation angle (element of $S^1$, a circle).
In the more general case, you can think of an element of $\mathrm{SU}(N)$ as defining a new orthonormal basis in terms of an old one, in a way that preserves handedness. Now just go through the process of picking an orthornormal basis: the first unit vector you pick is confined to be on a unit sphere in $N$-dimensions, so it's in $S^{N-1}$. The second unit vector has to be perpendicular to the first, so it's limited to a sphere of one lower dimension $S^{N-2}$. Repeat the process until there is only one vector left to pick in the basis, and the "special" requirement, the requirement that the matrix have determinant $1$, picks which of the two possible vectors it could be. Otherwise, for $O(N)$ you'd have another factor of $\mathbb{Z}_2$ in your space.
