= Solution
For $v=(v_1,v_2,v_3)$ define the <skew-symmetric matrix>
$$
\widehat v=\begin{pmatrix}0&-v_3&v_2\\v_3&0&-v_1\\-v_2&v_1&0\end{pmatrix}.
$$
Then $\widehat v x=v\times x$. This is a <linear isomorphism> $\mathbb R^3\to\mathfrak{so}(3)$, with inverse $\phi(\xi)=(\xi_{32},\xi_{13},\xi_{21})$. Since rotations preserve the <cross product>,
$$
R\widehat vR^{-1}=\widehat{Rv}\qquad(R\in SO(3)).
$$
Thus $\phi$ intertwines the <Adjoint representation> with the standard <representation>. The vector triple-product identity gives, for every $x$,
$$
[\widehat v,\widehat w]x=v\times(w\times x)-w\times(v\times x)=(v\times w)\times x.
$$
Consequently $[\widehat v,\widehat w]=\widehat{v\times w}$ and the chosen normalization satisfies
$$
\boxed{\phi([\xi,\eta])=\phi(\xi)\times\phi(\eta).}
$$
No undetermined scale remains in this explicit <matrix> convention.
Identify $SU(2)$ with the <unit quaternions> using
$$
q=z+w j\longmapsto\begin{pmatrix}z&w\\-\overline w&\overline z\end{pmatrix},\qquad |z|^2+|w|^2=1.
$$
The relations $jz=\overline z j$ and $j^2=-1$ verify multiplication, and every <matrix> in $SU(2)$ has this form. If $v$ is imaginary, then $\overline v=-v$. Quaternionic <conjugation> reverses <product order> and $q^{-1}=\overline q$, so
$$
\overline{qvq^{-1}}=q\overline v q^{-1}=-qvq^{-1}.
$$
Thus <conjugation> preserves $\operatorname{Im}\mathbb H\cong\mathbb R^3$. Multiplicativity of the <quaternion> norm shows that this action preserves the <inner product>. The <unit quaternions> form the <connected> three-sphere, so its <determinant>, equal to one at the identity, is always one. We obtain a homomorphism $C:SU(2)\to SO(3)$.
Its <group kernel> consists of <unit quaternions> commuting with every imaginary <quaternion>. Commuting with both $i$ and $j$ forces a <quaternion> to be real, so the <group kernel> is exactly $\{\pm1\}$. To prove <surjectivity> explicitly, let $u$ be a unit imaginary <quaternion> and put $q=\cos(\theta/2)+u\sin(\theta/2)$. For imaginary $u,v$, multiplication satisfies $uv=-u\cdot v+u\times v$. Expanding $qvq^{-1}$ therefore gives
$$
qvq^{-1}=v\cos\theta+(u\times v)\sin\theta+u(u\cdot v)(1-\cos\theta).
$$
This is the <Rodrigues rotation formula> about axis $u$. Every element of $SO(3)$ has such an axis-angle description: an odd-dimensional <orthogonal matrix> of <determinant> one has an eigenvector of <eigenvalue> one, and its action on the perpendicular plane is a plane rotation. Hence $C$ is onto, and
$$
\boxed{SU(2)/\{\pm I_2\}\cong SO(3).}
$$
This is the quaternionic form of the <Adjoint double cover from SU(2) to SO(3)>. Its differential sends an imaginary <quaternion> $u$ to $2\widehat u$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.
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