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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 $S^n$?

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You said "diffeomorphic". Does this mean you know $X$ is a manifold? –  Todd Trimble Dec 13 '11 at 14:40

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If $X$ is a smooth manifold (and this is the only case when you can speak of a diffeomorphism 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.

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Great thanks for answer. Of course $X\times R$ is homeomorphic (not diffeomorphic) to $S^n\times R$. –  Olga Dec 14 '11 at 7:01

This question (or a close relative) is discussed and answered in

When factors may be cancelled in homeomorphic products?

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