That was an exam question asking for the homeomorphism between:

$\mathbb{S}^1 \times (a,b)$ and $\mathbb{R}$.

My guess: since $(a,b)$ is homeomorphic to $\mathbb{R}$, function $\mathbb{S}^1 \times (a,b) \rightarrow (a,b)$ is a homeomorphism and $(a,b)$ is homeomorphic to $\mathbb{R}$, so $\mathbb{S}^1 \times (a,b)$ is homeomorphic to $\mathbb{R}$? Is this correct?

  • $\begingroup$ What exactly is the function you allude to in "$\mathbb{S}^1\times (a,b)\rightarrow (a,b)$"? $\endgroup$ – Hayden Jul 21 '14 at 18:10
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    $\begingroup$ Was there an option to say they are not homeoomorphic? I haven't actually worked it out rigorously, but I think the open finite cylinder should have fundamental group isomorphic to the integers. Since R has trivial fundamental group and the fundamental groups are not isomorphic, the spaces can't be homeomorphic. $\endgroup$ – JohnnyMo1 Jul 21 '14 at 18:11

$S^1\times (a,b)$ is not homeomorphic to $\mathbb R$. An easy way to see this is that removing any point from $\mathbb R$ disconnects the space. However removing any point from $S^1\times (a,b)$ yields a path-connected space.

JohnnyM01's comments about the fundamental group are also true and give an alternate proof that these spaces are not homeomorphic.


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