Topological rigidity of compact manifolds in dimension three - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:17:45Z http://mathoverflow.net/feeds/question/77400 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77400/topological-rigidity-of-compact-manifolds-in-dimension-three Topological rigidity of compact manifolds in dimension three Shinpei 2011-10-06T21:32:32Z 2011-10-06T22:16:11Z <p>The Borel Conjecture asserts that homotopy equivalent aspherical closed manifolds are homeomorphic, which is still open in general. But, for three-dimensional manifolds, this conjecture holds (I read this in <a href="http://www-fourier.ujf-grenoble.fr/~lbessier/english_principal.pdf" rel="nofollow">Bessieres-Besson-Boileau</a>), whose proof depends on the geometrization theorem (Perelman). </p> <p>Question: Does the relative version of the Borel conjecture also hold for compact 3-manifolds with boundary (by the geometrization)? The relative version: If there is a homotopy equivalence between two compact aspherical manifolds that is a homeomorphism between their boundaries, are those manifolds homeomorphic?</p> http://mathoverflow.net/questions/77400/topological-rigidity-of-compact-manifolds-in-dimension-three/77404#77404 Answer by Richard Kent for Topological rigidity of compact manifolds in dimension three Richard Kent 2011-10-06T22:16:11Z 2011-10-06T22:16:11Z <p>Yes.</p> <p>When the manifolds are Haken this is a theorem of Waldhausen. See Ian Agol's answer <a href="http://mathoverflow.net/questions/35680/complete-knot-invariant/35687#35687" rel="nofollow">here</a>.</p> <p>Since your manifolds are aspherical, they are irreducible by the Poincar&eacute; conjecture. Since they have boundary and are irreducible, they are Haken.</p>