Help about two-to-one homomorphism map $π:S^3 \to SO(3)$ Let the map $π:S^3 \to SO(3)$ . 
How can I define a function directly on this map show that $Im(π) \subseteq SO(3)$ ?
The sphere $S^3= \left\{ {a+bi +cj +dk : a^2+b^2+c^2+ d^2=1}\right\}\subseteq \mathbb{H}$ as the quaternions with length 1 is a commutative group? In which way i can see this this ?   
Thanks in advance!
 A: Let $L_q: H \to H$ denote left-multiplucation by the quaternion $q = a + bi + cj + dk$. Then in the basis $1, i, j, k$, the matrix $A_q$ for $L_q$ is something like
\begin{align}
A_q &= \begin{bmatrix}
a & -b & -c & -d \\
b &  a & -d &  c \\
c &  d &  a & -b\\
d & -c &  b &  a
\end{bmatrix}
\end{align}
where the first column contains the coefs of $q \cdot 1$; the second contains the coeffs of $q \cdot i$, and so on. 
You can similarly write out the matrix $B_q$of $R_q$, where $R_q w = w q$, i.e., right multiplication by $q$, again in the standard basis. 
Now for $q \in S^3$, consider the transformation 
$$
E: H \to H : w \mapsto L_q (R_{\bar{q}} (w))
$$
whose matrix is $M_q = A_q B_{\bar{q}}$.
Observe that both $L_q$ and $R_q$ are (when $\|q\| = 1$) isometries, so the matrices $A_q$ and $B_q$ are both orthogonal. The product is also orthogonal, and the vector $w = 1 + 0i + 0j + 0k$ is an eigenvector. So the matrix $M_q$ sends the pure-vector quaternion plane (i.e., all things of the form 
$w = bi + cj +dk$) to itself, and is an isometry. You can check by hand that the determinant is 1 for the case $q = 1+ 0i + 0j +0k$. Since that determinant varies continuously over $S^3$, and must be $\pm1$, it's alwasy 1. Hence the matrix $M_q$ represents a rotation of the $ijk$ plane, i.e., an element of $SO(3)$. Hence the map
$$
q \mapsto M_q
$$
is the map $\pi$ that you asked for. 
As @Jyrki points out, the group $S^3$, under quaternion multiplication, is not commutative, so there's no answer to part 2. 
