# If $S\times\mathbb{R}$ is homeomorphic to $T\times\mathbb{R}$, and $S$ and $T$ are compact, can we conclude that $S$ and $T$ are homeomorphic?

If $S \times \mathbb{R}$ is homeomorphic to $T \times \mathbb{R}$ and $S$ and $T$ are compact, connected manifolds (according to an earlier question if one of them is compact the other one needs to be compact) can we conclude that $S$ and $T$ are homeomorphic?

I know this is not true for non compact manifolds.

I am mainly interested in the case where $S, T$ are 3-manifolds.

• Are $S$, $T$ metric spaces or otherwise constrained? You seem to be interested in manifolds? – Henno Brandsma May 7 '13 at 10:27
• yes i am interested in the result about manifolds. i edited the question and added this. thanks – user76556 May 7 '13 at 19:00
• Are manifolds with boundaries allowed? – Jacob H May 7 '13 at 19:26
• no just ordianry manifolds – user76556 May 8 '13 at 4:41
• $S$ and $T$ have the same homotopy type, only! – amine May 14 '13 at 21:19

## 1 Answer

For closed 3-manifolds, taking the product with $\mathbb{R}$ doesn't change the fundamental group, so if the two products are homemorphic, the original spaces have the same fundamental group, and closed 3-manifolds are uniquely determined by their fundamental group, if they are irreducible and non-spherical.

• Relatedly, the lens spaces $L(7,1)$ and $L(7,2)$ are known to be homotopy equivalent but not homeomorphic. One can prove that $TL(7,1)$ is diffeomorphic to $TL(7,2)$ and that these bundles are trivial. So, in this case, each of the spaces crossed with $\mathbb{R}^3$ provides an example (but I'm not sure what happens after crossing with just $\mathbb{R}$). – Jason DeVito Jun 3 '13 at 17:58
• it seems everything is ok. where can i read more details? and won't this irreducibility and non-sphericalness affect the argument? – user76556 Jun 3 '13 at 18:39
• I got it from arxiv.org/pdf/1205.0202.pdf. You're right, though, it doesn't apply everywhere; what case are you interested in? I usually do hyperbolic geometry, and it's true in that case. Jason DeVito's comment shows the question is interesting in the spherical case at least... – Brian Rushton Jun 3 '13 at 18:51
• i am looking for the general case. but it seems we are really close. i'll have a look at the article. tanx. – user76556 Jun 6 '13 at 20:07