2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/119804We identify a universal group $U$ and show that $\Bbb H^3/G$ is $S^3$ when $G$ is a finite index subgroup of $U$ generated by elements of finite order.Typos corrected. Figures improvedGeometric TopologyMathematical PhysicsUniversal Cone Manifolds and the Poincaré Conjecture Itext