Bloch Sphere and $SU(2) \to SO(3)$ map For any matrix $U \in SU(2)$ there is an associated map from $S^2$ (the surface of a 3-disk) to itself defined by $\pi \circ U$, where $\pi$ is the projection map from $\mathbb{C}^2$ to $CP(1)$, that is the map associating to a pure state of a qubit its representative on the Bloch sphere.
Is the map $\pi \circ U$ a rotation, and is the association $U \to \pi \circ U$ the usual projection map $SU(2)/\{\mathbb{1},-\mathbb{1}\} \sim SO(3)$?
 A: Yes it is. The map you consider acts as follows by definition. $SU(2) \ni U \mapsto R(U) \in SO(3)$ such that
$$U \frac{1}{2}\left(I + \vec{v}\cdot \vec{\sigma} \right)U^\dagger = \frac{1}{2}\left(I + R(U)\vec{v}\cdot \vec{\sigma} \right)\quad \forall \vec{v} \in \mathbb S^2$$
($R(U)$ is a rotation as it is linear and preserve the length of vectors)
which is easily seen to be  equivalent to write
$$U\vec{v}\cdot \vec{\sigma}U^\dagger =  R(U)\vec{v}\cdot \vec{\sigma} \quad \forall \vec{v} \in \mathbb R^3$$
This latter requirement is one of equivalent ways to define the natural surjective Lie Group homomorphism, locally isomorphism, map  $SU(2) \to SU(2)/\{-\mathbb{1},\mathbb{1}\} \sim SO(3)$
A: I might be lacking the mathematical depth you desire.
The map $\pi\circ U$ indeed is a rotation. $U$ can be decomposed in terms to the basis given by the Pauli matrices, $X,Y,Z$ and the identity $I$:
$U=\exp(-i\theta \hat n\cdot\vec\sigma/2) = \cos(\theta/2)I-i\sin(\theta/2)(n_xX+n_yY+n_zZ)$
where $\hat n$ is a unit vector.
Then, $\pi\circ U$ will be the $SO(3)$ rotation matrix by an angle $\theta$ about the $\hat n$ axis. Note, the factor 2 which physically means that a full rotation by $2\pi$ will lead to a factor of $-1$ in the spin/state space of the qubit, corresponding to the fact that $SO(3)$ is isomorphic only to the quotient group $SU(2)/\mathbb Z_2$. Also, I'm fairly sure the answer to your second question is "yes" but somebody more versed in Lie groups might want to elaborate on that.
