Show that $P\colon S^3\to SO(3)$ is a covering map.

Question

Please help, anyone?

This I have done so far:

identify $S^3$ with the quaternions of unit length and identify $R^3$ with the pure quaternions, that is, those of the form $\{b_i, c_j, d_k\}$, for $b,c,d \in R$. For every $x$ in $S^3$, let $P_x$ be conjugation of the pure quaternions by $x$, that is, $P_x(y)= xyx^{-1}$, where $y$ lies in $R^3$.

$P_x(y)$ is a pure quaternion. Therefore $P_x\colon R^3\to R^3$ is a linear transformation. With the standard $R^3$ basis, $P_x$ can be regarded as a $3\times 3$ matrix. 2. $P_x$ lies in $SO(3)$ and so a continuous function $P\colon S^3\to SO(3)$ is defined by $P(x)= P_x$.

How to go further?

Answer 1

To show that P is surjective, actually there is an explicit expression of the quaternion representing a rotation of angle $\theta$ against a unit axis $v$: $$x=\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\cdot v$$

Since the rotation is expressed as $xyx^{-1}$ so $(-x)y(-x)^{-1}=xyx^{-1}$, that is $-x$ maps to the same rotation.

Answer 2

As pointed out by Xipan Xiao the map $P:S^3\rightarrow SO(3)$ is surjective, moreover $P$ is a group homomorphism (of topological groups, it is easy to see that $P_{xy}=P_xP_y$) which has kernel $Ker\;P=\{\pm 1\}$ so you have that $SO(3)$ and $S^3/\{\pm 1\}$ are homeomorphic through the map $\overline{P}$ induced by $P$ on the quotient (on $S^3/\{\pm 1\}$ we put the quotient topology); the projection map $\pi:S^3\rightarrow S^3/\{\pm 1\}$ is a double covering ($S^3/\{\pm 1\}$ is the quotient of $S^3$ by the automorphism group $\{\pm 1\}$ whose action is free and properly discontinuous). In conclusion $P=\overline{P}\circ\pi$ is a double covering of $SO(3)$.