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Suppose $S$ and $T$ are connected manifolds such that: 1) $S \times \mathbb{R}$ is homeomorphic to $T \times \mathbb{R}$, 2) $S$ is compact. can we conclude that $T$ is compact?

user76556
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Since $S$ and $T$ are locally connected, we may with out loss of generality assume, that $S$ and $T$ are connected. Now $S \times \mathbb R$ and hence $T \times \mathbb R$ have exactly two ends.

A connected and locally connected space is compact iff it has no ends. Suppose $T$ is not compact, then it would have at least one end. Now the product $T \times \mathbb R$ would have exactly one end, which is the desired contradiction.

Alexander Thumm
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    what is an end? – user76556 May 07 '13 at 07:35
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    @user76556: See Wikipedia. – joriki May 07 '13 at 07:35
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    @user76556: An end is basically a connected component at infinity. $\mathbb R$ has two of them (often denoted by $+\infty$ and $-\infty$). A space is compact iff it has no ends. The product of two connected spaces has one end if both factors have at least one end (draw a picture) and $n$ ends if one factor has zero and the other factor has $n$ ends. – Alexander Thumm May 07 '13 at 07:39
  • I need this result for manifolds. and they are connected too. – user76556 May 07 '13 at 09:51
  • I unaccepted it so that others don't get a wrong answer but i have the answer i was looking for. thank you. – user76556 May 07 '13 at 09:53
  • @user76556: You should accept an answer not for the benefit of others, but to say that you are satisfied with it. Maybe you could edit your question, specifying that you were interested to the case of connected manifolds? – A.P. May 07 '13 at 21:59
  • ok thanks for explaining – user76556 May 08 '13 at 04:45