I have the following Lie group homomorphism:
$\phi: Sp(1) \rightarrow SO(3)$ given by $\phi(q)x=qxq^{-1}$, where $x \in \mathbb{R}^3$ and $Sp(1)$ is the set of quaternions with module $1$.
I want to prove it is surjective, i.e., every matrix in $SO(3)$ comes by $\phi$ from an element $q \in Sp(1)$.
The way to do this is to see that your map $\phi(q)x=qxq^{-1}$ is actually a linear isometry and orientation preserving. The quaternion $qxq^{-1}$ has real part $0$ for any quaternion $q$ with modulus $1$. Now the point is the map $\phi$ sends $q$ and $-q$ to the same rotation of $\mathbb{R^3}$. The following document has the formula for a rotation matrix associated to a unit quaternion, which I will not attempt to derive here....Rotation Matrix (incidentally that document says the expression for this matrix in $SO(3)$ is due to Euler...)