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Let $S^1 = \{ \omega \in \mathbb{C}: \ |\omega| =1 \}$. I need to figure out the relation $\sim$ between two elements of a set $S^1 \times \mathbb{R}$ such that quotient group $S^1 \times \mathbb{R} / _{\sim}$ will be izomorphic with torus.

Please help. Thanks Tommy.

tommy
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    $S^1\times \mathbb{R}$ describes a cylinder. How can you manipulate a cylinder such that it becomes a torus? – Soby Jun 04 '19 at 11:37
  • Infinite cylinder I can cut into equal pieces (finite length cylinders) and roll them such that each of them becomes a torus. Each piece is izomorphic with torus. Quotient group is a set of cosets so each coset is one torus. How to describe it mathematically? – tommy Jun 04 '19 at 11:54
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    If I cut infinite cylinder $S^1 \times \mathbb{R}$ into pieces of lenght say $l$ then: $(\omega_1 , u) \sim (\omega_2 , t) \leftrightarrow \omega_1 = \omega_2$ and there exist $k \in \mathbb{Z}$ such that $|u-t| = k \cdot l$ Is this correct? – tommy Jun 04 '19 at 11:56
  • Nicely done, Tommy! – Cameron Buie Jun 04 '19 at 12:01

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You've already deduced a scheme for equivalence relations that will do the job, but I would suggest a few refinements.

First of all, there's no need for the absolute value bars--we can simply say $u-t=k\cdot l$, instead, and it will make it easier to prove transitivity, without making symmetry particularly difficult to prove.

Second of all, there's no need to leave $l$ general, nor even think of it as a length. Any non-zero real number will do, once we've dropped the absolute value bars, but I'd pick $l=1$ for simplicity. That reduces the condition to $u-t=k.$

Cameron Buie
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