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An equivalence relation on $\mathbb{R}$ is given by $$ x\sim y\Leftrightarrow x-y\in\mathbb{Z}. $$ Show that the quotient space $(\mathbb{R}/{\sim},\tau_1)$ is homeomorphic to $(S^1,\tau_2)$, where $\tau_1$ is the quotient topology and $\tau_2$ the induced topology.

I have to find a bijective continuous function $$ f\colon \mathbb{R}/{\sim}\to S^1 $$ with $f^{-1}$ continuous.

Do you have an idea how to find such a function?

Stefan Hamcke
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4 Answers4

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Let $S^1=\{z\in\mathbb{C}\mid|z|=1\}$. This is a common definition for $S^1$ but may not be the one you've been given.

Let $f\colon\mathbb{R}/{\sim}\rightarrow S^1$ be given by $f([t])=e^{2\pi it}$.

Can you show that this is well defined? (That is, show that if $t_1\in[t]_{\sim}$ and $t_2\in[t]_{\sim}$ then $e^{2\pi it_1}=e^{2\pi it_2}$).

Can you show that this is a homeomorphism?

Michael Hardy
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Dan Rust
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  • Why not taking $[t]_{\sim}\mapsto e^{it}$? Because then it is not well defined, right? Then it depends on what t I take as representer. –  Aug 18 '13 at 17:10
  • The $2\pi$ is in there because a full revolution of a circle is $2\pi$ radians. You want $f$ to 'wrap' the real line around the circle once for every integer so for every interval between two integers that you move on $\mathbb{R}$, you want to move one circumference's length around the circle, i.e. $2\pi$. Think of $t$ as measuring the angle around the circle. The fractional part of $t$ is the fraction of the circle's circumference that it has gone around. – Dan Rust Aug 18 '13 at 17:26
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    @DanielRust : If you type a\sim b you get $a\sim b$, but if you type a{\sim}b you get $a{\sim}b$, which looks wrong because it lacks the spacing before it and after it. But when you type "\mathbb R/\sim", then that spacing is inappropriate because in that context "$\sim$" is not being used to mean that what precedes it is related to what follows it. So I changed it to "\mathbb R/{\sim}", and as you can see, it looks different that way. – Michael Hardy Aug 18 '13 at 18:25
  • To show that the function is continious one can show that the function $h\colon\mathbb{R}\to S^1, x\mapsto e^{2\pi ix}$ is continious (universal feature of final topology), right? And I guess one can simply say here: the complex exponential function is continious? So it remains to show the bijectivity of $f$ and the continity of $f^{-1}$? –  Aug 18 '13 at 18:38
  • @MichaelHardy thanks for the LaTeX tip and the edit. – Dan Rust Aug 18 '13 at 19:08
  • @math12 The universal property of the quotient is definitely the easiest way to show that $f$ is continuous. This map is also simple enough that you can prove it directly from the open preimage definition of continuity. Bijectivity is easy enough (first show that for all $t\in\mathbb{R}$, there is a $t_0\in[0,1)$ such that $t_0\sim t$) and the continuity of the inverse can be shown from the open preimage definition. – Dan Rust Aug 18 '13 at 19:15
  • @Daniel Rust In order to use the universal property is it enough to say that $\exp\colon\mathbb{C}\to\mathbb{C}$ is continious and so $\mathbb{R}\ni x\longmapsto e^{2\pi xi}\in\mathbb{C}$ is continious? –  Aug 18 '13 at 19:27
  • It depends on what you've proved in class and what your teacher expect, but I would say using the theorem 'the restriction of a continuous function to a subspace of the domain is a continuous function' is good enough here. In my opinion, showing that the preimage of an open ball in $S^1$ is open in $\mathbb{R}$ under $\exp\colon\mathbb{R}\rightarrow S^1$ is easier. – Dan Rust Aug 18 '13 at 19:36
  • It's easier? I have to show that for $h\colon x\longmapsto e^{2\pi ix}$ and for $O=S^{-1}\cap O'$, $O'$ open ball in $\mathbb{R}^2$ it is $h^{-1}(O)\in\mathcal{O}_n$, where $\mathcal{O}_n$ is the natural topology on $\mathbb{R}$. But how to do so? -- $h^{-1}(O)=h^{-1}(S^1)\cap h^{-1}(O')=\mathbb{R}\cap h^{-1}(O')$. That's what i get. –  Aug 18 '13 at 19:43
  • +1 If you have time, how do you show that if $t_1\in[t]{\sim}$ and $t_2\in[t]{\sim}$ then $e^{2\pi it_1}=e^{2\pi it_2}$ I understand it intuitively but i am not sure how to show it. – JKnecht Mar 10 '16 at 23:22
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I'm going to give a less technical answer than what others have written here, because I think you must have missed something other than that part if you didn't think of $x\mapsto e^{2\pi ix}$.

You have $x\sim y\Leftrightarrow x-y\in\mathbb{Z}$. That means $0$ becomes the same point as $1$, while everything between $0$ and $1$ is a different point from the one point that is $0\sim1$. And $0.1$ becomes the same point as $1.1$, and $0.2$ becomes the same as $1.2$, and so on. In other words, moving along the interval from $0$ to $1$, you return to your starting point and start over, just as with the circle. Therefore, the interval $[0,1]$ should get wrapped around the circle, with $0$ and $1$ getting mapped to the same point. In then the interval $[1,2]$, being the same as $[0,1]$ in the quotient space, gets wrapped around the circle in the same way, with $2$ being mapped to the same point on the circle to which $0$ and $1$ were mapped, and so on.

When you learned trigonometry, you saw $(\cos(2\pi x),\sin(2\pi x))$ wrapping around the circle in that same way. Since those are periodic functions with period $1$, it follows that if $x\sim y$ then $(\cos(2\pi x),\sin(2\pi x))=(\cos(2\pi y),\sin(2\pi y))$, so that is the mapping you need.

Michael Hardy
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  • This is a nice, detailed explanation for why the exponential map can be seen geometrically as 'wrapping' the real line around the circle. – Dan Rust Aug 18 '13 at 19:17
  • And now I understood. I repeat it in my words: For example $[1]_{\sim}=\left{...,0,1,2,3,...\right}$ and all these points are "equal". So one has to give all these points the same point on the uni sphere. That means the same point as $1$ has on the uni sphere and this is the point (1/0) on the cirlce. And so one has to multiplicate with the factor $2\pi$ in order to go a full circle, two full cirlces and so on to the same point. –  Aug 18 '13 at 19:23
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Try $f(t)=e^{2\pi i t}$ and find its kernel.

Mikhail Katz
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What about $f(x) = e^{2\pi i x}$? It factors through the quotient.

Vishal Gupta
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