 Solution

For $v=(v_1,v_2,v_3)$ define the skew-symmetric matrixThen $\hat{v}x=v\times x$. This is a linear isomorphism ${\mathbb{R}}^3\to \mathfrak{so}(3)$, with inverse $\varphi (\xi )=({\xi }_{32},{\xi }_{13},{\xi }_{21})$. Since rotations preserve the cross product,Thus $\varphi$ intertwines the Adjoint representation with the standard representation. The vector triple-product identity gives, for every $x$,Consequently $[\hat{v},\hat{w}]=\widehat{v\times w}$ and the chosen normalization satisfiesNo undetermined scale remains in this explicit matrix convention.

Identify $SU(2)$ with the unit quaternions usingThe 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 $q^{-1}=\overline{q}$, soThus conjugation preserves $\mathrm{Im}\mathbb{H}\cong {\mathbb{R}}^3$. 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 $\{\pm 1\}$. 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 $qv{q}^{-1}$ therefore givesThis 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, andThis is the quaternionic form of the Adjoint double cover from SU(2) to SO(3). Its differential sends an imaginary quaternion $u$ to $2\hat{u}$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.