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Assume future null geodesic completeness. The compactness of the future horismos of a trapped surface makes $E^{+}(S)$ a nonempty compact achronal boundary without edge. Projection along a complete timelike vector field into the Cauchy hypersurface is injective, open by invariance of domain, and closed by compactness. Connectedness forces surjectivity, contradicting the noncompact Cauchy hypersurface. This global step completes the contradiction.^[ref]

Invariance of domain says that a continuous map that is an injective function between topological manifolds without boundary of the same dimension is an open map. It is useful in the Penrose singularity theorem: projecting a compact achronal boundary along timelike curves onto a Cauchy hypersurface gives a nonempty subset that is both open and closed, forcing that connected Cauchy hypersurface to be compact.

First show that the boundary map is onto and has degree $\pm 1$, and that the fundamental-group injection is actually an isomorphism. An injective map between closed surfaces is a homeomorphism onto an open-and-closed subset by invariance of domain. Thus each component of $\partial M$ maps homeomorphically to a distinct component of $\partial N$.

Orient $M$ and $N$, and writeApply the boundary map in the relative homology sequence. The boundary of the right side has coefficient $d$ on every boundary component of $N$. On the left, an image boundary component has coefficient $+1$ or $-1$, and an omitted component has coefficient zero. Since at least one component is present, $d=\pm 1$; then no component can be omitted and all signs agree. Therefore

Let $H=f_{\ast }{\pi }_1(M)$ and lift $f$ to the connected cover $p:N_H\to N$ corresponding to $H$. If the cover were noncompact, then $H_3(N_H,\partial N_H;\mathbb{Z})=0$. The lifted relative fundamental class would have zero boundary, but its boundary is a nonzero sum of distinct boundary fundamental classes, since the original boundary map is injective and the boundary components are compact. This is impossible. The cover therefore has a finite number $r$ of sheets. Degree multiplicativity givesHence $r=1$. The image of $f_{\ast }$ is the whole target fundamental group, and the assumed injection makes $f_{\ast }$ an isomorphism.

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.

Take an open neighbourhood $\tilde{U}$ on which $p$ is one-to-one and a homeomorphism onto its image. By invariance of domain for topological surfaces, that image is open. Shrink $\tilde{U}$ so the image lies in a complex chart $(V,z)$ of $R$, and use $z\circ p$ as a chart of $\tilde{R}$. These charts cover $\tilde{R}$. On any overlap their transition is the transition between the original charts of $R$, restricted to an open set, so is holomorphic with holomorphic inverse. They define the desired complex structure, and in these charts $p$ is locally the identity. Thus $p$ is holomorphic. The openness supplied by invariance of domain is important when the hypothesis is phrased only as a homeomorphism onto the image.

Let $f:\mathbb{C}\to Y$ be holomorphic and let $\pi :\mathbb{D}\to Y$ be the stated covering. Choose a lift of $f(0)$. The covering-space lifting theorem says that a map from a simply connected, locally path-connected domain has a unique lift after fixing this initial point: lift paths from zero, and homotopy lifting makes their endpoint independent of path. Since $\mathbb{C}$ is simply connected, this constructs $\tilde{f}:\mathbb{C}\to \mathbb{D}$ with $\pi \circ \tilde{f}=f$. On a small neighbourhood, the lift is a biholomorphic local inverse of $\pi$ composed with $f$, so it is holomorphic. Its modulus is less than one, and Liouville theorem therefore makes it constant. Hence every holomorphic map $\mathbb{C}\to Y$ is constant.

The relevant Penrose singularity theorem states: a connected time-oriented four-dimensional globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a nonempty compact orientable boundaryless trapped surface, and the null convergence condition is future null-geodesically incomplete. Under the Einstein field equations, the null energy condition supplies the curvature hypothesis. A trapped surface here is spacelike and has both future normal null expansions strictly negative. The conclusion is an incomplete null geodesic, not necessarily a divergent curvature invariant at an identifiable point.

Suppose, for contradiction, that every future null geodesic is complete. Write $S$ for the trapped surface andThe equality follows because the causal future of a compact set is closed in a globally hyperbolic spacetime. This future horismos is a closed achronal boundary. It is nonempty: the restriction of a Cauchy time function to compact $S$ has a minimum, and a point at that minimum cannot be chronologically preceded by another point of $S$.

Normalize the two future null normal directions along $S$ against a smooth future timelike field, fixing the affine scale continuously. Their initial null expansions are continuous and strictly negative. Compactness of the normalized normal bundle gives a uniform $c>0$ with ${\theta }_{\pm }\le -c$ everywhere on $S$. The preceding null focusing theorem forces a conjugate point to a spacelike surface along each normal generator within affine length $L=2/c$.

By the supplied boundary-generator result, every point of $E^{+}(S)$ is reached by an orthogonal future null geodesic which has no earlier conjugate point. No such boundary generator can remain on the boundary beyond its first focal point. Every point of $E^{+}(S)$ therefore lies inThe parameter set is compact. Future null completeness makes its geodesic flow defined throughout this common finite interval, and smooth dependence on initial data makes its image $K$ compact. Since $E^{+}(S)$ is closed and contained in $K$, it is compact. This is the compactness of the future horismos of a trapped surface step; a merely pointwise finite bound would not suffice without compactness and uniform normalization.

Choose a smooth complete timelike vector field, obtained if necessary by positive rescaling against a complete auxiliary Riemannian metric. Its inextendible integral curves of a vector field meet a chosen smooth Cauchy hypersurface $C$ exactly once. Projection along those curves defines a continuous map ${\pi }_C:E^{+}(S)\to C$. The achronal set property makes this map an injective function: two boundary points on the same timelike integral curve of a vector field would be timelike related.

An achronal boundary is a topological hypersurface without boundary, even at nonsmooth generator junctions; the submanifold property allowed in the question supplies this fact. Thus domain and codomain are both three-dimensional topological manifolds without boundary. Invariance of domain makes ${\pi }_C(E^{+}(S))$ open in $C$. Compactness makes it closed in the Hausdorff space $C$, and it is nonempty. Connectedness of the spacetime gives connectedness of $C$, so this image is all of $C$. That would make $C$ compact, contradicting the noncompact Cauchy hypersurface hypothesis. The completeness assumption is false, proving the theorem.

The noncompactness hypothesis and strict trapping are essential to this version. Global hyperbolicity cannot simply be omitted in the compactness and projection steps, and nonpositive initial expansion with zeros does not supply the uniform focusing bound used here.