0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
6
$\begingroup$

$S^1\times (a,b)$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.