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By part (b), the moduli space is $M(T)=SU(2{)}^2/SU(2)$ under simultaneous conjugation. Regard $SU(2)$ as the unit quaternions and writewith $u,v\in {\mathbb{R}}^3$. Conjugation rotates both imaginary vectors by the same element of $SO(3)$. Their lengths and inner product determine an orbit, even when one vector vanishes or the vectors are collinear: an isometry between the two spanned planes can be extended to an orientation-preserving rotation of ${\mathbb{R}}^3$. Thus complete invariants areEvery allowed triple is realized by a pair of vectors with the prescribed lengths and dot product. The induced continuous bijection from the compact quotient to this closed subset of ${\mathbb{R}}^3$ is a homeomorphism.

Let $U$ be the subset where $A\ne \pm 1$, equivalently $-1<a<1$. It is open and dense because the excluded first holonomies can be approximated by noncentral ones. Rotate $u$ to the positive $i$ axis, so $A=a+\sqrt{1-a^2}i$. Its remaining stabilizer rotates the $j,k$ plane. If $B=b+ti+yj+zk$, the orbit is determined by $(b,t)$, andwith the remaining radius $\sqrt{1-b^2-t^2}$. These coordinates give the corrected resultContinuity of the inverse follows by taking $y=\sqrt{1-b^2-t^2}$ and $z=0$ as a representative. The complement consists of $A=1$ and $A=-1$, separately. In either case conjugation classes of $B$ are parametrized by its real part $b\in [-1,1]$. Therefore $M(T)\setminus U$ is two disjoint closed intervals, as requested.

The printed $S^2$ factor is incorrect: the factor must be a closed disk. The stabilizer acts by conjugation, fixing the circle of quaternions $b+ti$; it is not the free Hopf circle action on $S^3$. To rule out a different choice of dense open set with the printed topology, we can also identify the whole space. Replace $c$ by $d=ab+c$. The invariant region becomesThese are the Gram matrices of the three unit vectors $1,A,B$ in ${\mathbb{R}}^4$. Conversely, a positive-semidefinite matrix of this form can be realized by such vectors, hence by the original quaternion data. The set $\mathcal{C}$ is compact and convex and contains the identity matrix as an interior point in its three coordinates. Radial parametrization of a convex body therefore identifies it with a closed three-ball. ThusThis also agrees with Theorem 6.5 of the primary character-variety paper.

An open subset homeomorphic to the boundaryless three-manifold $S^2\times (0,1)$ cannot contain a boundary point of this closed ball. Otherwise a local chart, followed by inclusion into ${\mathbb{R}}^3$, would be a continuous injection from an open subset of ${\mathbb{R}}^3$ whose image contains a ball-boundary point while staying inside the closed ball, contradicting invariance of domain. Its complement would therefore contain the entire boundary sphere. But a sphere cannot embed in two disjoint intervals: its connected image would lie in one interval and be a nondegenerate compact interval, whose interior points disconnect it, unlike $S^2$. This proves that the literal printed pair of claims is impossible, and supplies the complete corrected SU2 character variety of the punctured torus description.