most recent 30 from http://mathoverflow.net 2013-06-20T07:56:26Z http://mathoverflow.net/feeds/question/83337 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83337/on-product-of-some-spaces Olga 2011-12-13T14:36:31Z 2011-12-13T16:13:24Z

<p>I know that $X$ is a topological space, $R$ is the real line, $S^n$ is n-sphere and $X\times R$ is diffeomorphic to $S^n\times R$. Is it true that $X$ is homeomorphic to $S^n$? </p>

http://mathoverflow.net/questions/83337/on-product-of-some-spaces/83340#83340 Igor Rivin 2011-12-13T14:53:36Z 2011-12-13T14:53:36Z

<p>This question (or a close relative) is discussed and answered in</p> <p><a href="http://mathoverflow.net/questions/26385/when-factors-may-be-cancelled-in-homeomorphic-products" rel="nofollow">http://mathoverflow.net/questions/26385/when-factors-may-be-cancelled-in-homeomorphic-products</a></p>

http://mathoverflow.net/questions/83337/on-product-of-some-spaces/83341#83341 Vitali Kapovitch 2011-12-13T14:54:48Z 2011-12-13T16:13:24Z

<p>If $X$ is a smooth manifold (and this is the only case when you can speak of a <strong>diffeomorphism</strong> between $X\times \mathbb R$ and $\mathbb S^n\times\mathbb R$) then this is true by Poincare. If $X$ is not assumed to be a manifold then this is false. For example, there is a theorem of Edwards that if $Y$ is a closed $(n-1)$-dimensional manifold and a homology sphere then $X$ equal to suspension of $Y$ satisfies that $X\times \mathbb R$ is homeomorphic to $\mathbb S^n\times\mathbb R$. There are many examples of homology spheres already in dimension $3$ which are not spheres so any of them will work.</p>