The operator-norm unit sphere of M_2(R) is a 3-sphere

Idea. Any two norms on a finite-dimensional space have homeomorphic unit spheres via radial projection. The Frobenius sphere is a round sphere, i.e. $S^3$.

Step 1: the operator norm in coordinates. For $|x|=1$, the triangle inequality gives $|zx+w\bar x|\le|z|+|w|$. For equality, choose $x=e^{i\varphi}$ so that $zx=ze^{i\varphi}$ and $w\bar x=we^{-i\varphi}$ have the same argument, i.e. $2\varphi\equiv\arg w-\arg z\pmod{2\pi}$ (if $z=0$ or $w=0$ any $\varphi$ works). Then $|zx+w\bar x|=|z|+|w|$. Hence $\|A\|=|z|+|w|$, and $S_{op}=\{|z|+|w|=1\}$.

Step 2: the Frobenius sphere is round. Since $\|A\|_F^2=\operatorname{tr}(A^TA)=a^2+b^2+c^2+d^2$ and $|z|^2+|w|^2=\frac14\bigl[(a+d)^2+(c-b)^2+(a-d)^2+(b+c)^2\bigr]=\frac12(a^2+b^2+c^2+d^2)$, we get $\|A\|_F^2=2(|z|^2+|w|^2)$. So
$$S_F=\{\|A\|_F=1\}=\{(z,w)\in\mathbb{C}^2:|z|^2+|w|^2=\tfrac12\}.$$
This is a Euclidean sphere of radius $1/\sqrt2$ in $\mathbb{C}^2=\mathbb{R}^4$, which is homeomorphic to $S^3$ by scaling by $\sqrt2$.

Step 3: radial projection. Define
$$\Phi:S_{op}\to S_F,\ A\mapsto\frac A{\|A\|_F},\qquad\Psi:S_F\to S_{op},\ B\mapsto\frac B{\|B\|}.$$
Both maps are well defined and continuous (norms are continuous and do not vanish on these spheres, because $A\neq0$ there). For $A\in S_{op}$, $\Psi(\Phi(A))=\dfrac{A/\|A\|_F}{\|A\|/\|A\|_F}=\dfrac A{\|A\|}=A$, and for $B\in S_F$, $\Phi(\Psi(B))=\dfrac{B/\|B\|}{\|B\|_F/\|B\|}=\dfrac B{\|B\|_F}=B$. So $\Phi$ is a homeomorphism.

Conclusion. $S_{op}\cong S_F\cong S^3$. Explicitly, the map $A\mapsto\dfrac{(z,w)}{\sqrt{|z|^2+|w|^2}}$ sends $S_{op}$ homeomorphically onto the unit sphere of $\mathbb{C}^2$, with inverse $(z,w)\mapsto\dfrac{(z,w)}{|z|+|w|}$.

Check. $I$ has $(z,w)=(1,0)$: $\|I\|=1$ and it goes to $(1,0)\in S^3$. $\operatorname{diag}(1,-1)$ has $(z,w)=(0,1)$, also in $S_{op}$ (an orthogonal reflection).

Correction of the source. The book writes the Frobenius sphere as $\{2(|z|^2+|w|^2)=2\}$, which is the sphere of Frobenius radius $\sqrt2$, not $1$; the sphere $\|A\|_F=1$ is $\{2(|z|^2+|w|^2)=1\}$ as above. The topology is unaffected.

Remark. The set $\{|z|+|w|=1\}$ is the join of two circles, $S^1*S^1\cong S^3$, and it is not a smooth round sphere: it has a flat (non-strictly-convex) structure along the circles $\{w=0\}$ and $\{z=0\}$ corresponding to the orthogonal matrices.

Lesson. Radial projection identifies the unit spheres of any two norms; to identify a sphere with $S^3$, compare it with a Euclidean sphere.