SO(3) is homeomorphic to the 3-sphere

Idea. Unit quaternions double-cover the rotation group.

Well defined. For $x\in\mathrm{Im}\,\mathbb{H}$ we have $\bar x=-x$. Then $\overline{qx\bar q}=q\bar x\bar q=-qx\bar q$, so $qx\bar q$ is again purely imaginary. The map $x\mapsto qx\bar q$ is $\mathbb{R}$-linear. The norm is multiplicative, $|qx\bar q|=|q|\,|x|\,|\bar q|=|x|$ for $|q|=1$, so $\Phi(q)\in O(3)$.

Homomorphism. $\Phi(q_1q_2)(x)=q_1q_2x\bar q_2\bar q_1=\Phi(q_1)\bigl(\Phi(q_2)(x)\bigr)$ since $\overline{q_1q_2}=\bar q_2\bar q_1$.

Determinant $1$. $q\mapsto\det\Phi(q)\in\{\pm1\}$ is continuous on the connected space $S^3$ and equals $1$ at $q=1$ ($\Phi(1)=\mathrm{id}$). So $\Phi(q)\in SO(3)$. Also $\Phi$ is continuous (polynomial in the coordinates of $q$).

Kernel. $\Phi(q)=\mathrm{id}$ iff $qx=xq$ for all imaginary $x$, iff $q$ is real (it commutes with $i,j,k$). Unit real quaternions: $q=\pm1$.

Surjectivity. Let $u$ be a unit imaginary quaternion, $\theta\in\mathbb{R}$ and $q=\cos\frac\theta2+u\sin\frac\theta2$ (so $\bar q=\cos\frac\theta2-u\sin\frac\theta2$ and $|q|=1$). Then:
- $q$ commutes with $u$, so $\Phi(q)u=qu\bar q=u$: $\Phi(q)$ fixes the axis $u$.
- Let $v$ be a unit imaginary quaternion orthogonal to $u$. Then $uv=-vu$ (orthogonal imaginary quaternions anticommute), hence $v\bar q=v\cos\frac\theta2-vu\sin\frac\theta2=qv$. Therefore $qv\bar q=q\cdot qv=q^2v=(\cos\theta+u\sin\theta)v=\cos\theta\,v+\sin\theta\,uv$.
Since $uv$ is a unit imaginary quaternion orthogonal to both $u$ and $v$, $\Phi(q)$ acts on the plane $u^\perp$ (basis $v,uv$) as the rotation by angle $\theta$. So $\Phi(q)$ is the rotation by $\theta$ about the axis $u$.
Every element of $SO(3)$ is a rotation about some axis: its eigenvalues are $e^{\pm i\alpha}$ and a real $\varepsilon$ of modulus $1$, with product $\varepsilon=1$ since $\det=1$, so $1$ is an eigenvalue, and in an orthonormal basis starting with a unit eigenvector it is $1\oplus R_\alpha$. So $\Phi$ is onto.

Homeomorphism. $\Phi$ induces a continuous bijection $\bar\Phi:S^3/\{\pm1\}\to SO(3)$. The source is compact, the target is Hausdorff, so $\bar\Phi$ is a homeomorphism. And $S^3/\{\pm1\}$ is $\mathbb{RP}^3$ (antipodal identification).

Check. $\Phi(i)$ is the rotation by $\pi$ about the $i$-axis: $i\,i\,(-i)=i$, $i\,j\,(-i)=-j$, $i\,k\,(-i)=-k$.

Remark. This is the origin of the use of quaternions for 3D rotations in computer graphics: two unit quaternions $q$ and $-q$ give the same rotation.