The relation $\sim$ is as follows: $x \sim y$ if $x-y \in \mathbb{Z}$ and I can't really figure out what are the classes of equivalence from that relation. Help please.

  • 5
    $\begingroup$ Hint: try winding a piece of string about your finger while you think about this question. $\endgroup$ – Rob Arthan Oct 30 '16 at 23:22
  • $\begingroup$ For the future: It's generally preferable if you put the entire question in the post, instead of just the title. $\endgroup$ – AJY Oct 30 '16 at 23:28
  • $\begingroup$ See also this post: Show: Quotient space is homeomorphic to unit sphere $\endgroup$ – Martin Sleziak Oct 31 '16 at 9:16

The class of $x$ is simply $\xi (x) = [x]=x+\mathbb Z = \{ x+n : n\in\mathbb Z\}$. The map $\xi : {\Bbb R} \mapsto {\Bbb R}/{\Bbb Z}$ associates to $x$ its class. Perhaps more important, a set $A$ in the quotient is open iff $\xi^{-1}(A)$ is open in ${\Bbb R}$.

You may then show that $\cos$ and $\sin$ (of the reals) gives rise to a well-defined continuous bijective function $$\phi\in {\Bbb R}/{\Bbb Z}\mapsto (\cos 2\pi \phi,\sin 2\pi \phi) \in S^1\subset {\Bbb R}^2$$ which provides the wanted homeomorphism.


Hint: Remember $S^1 = \{\exp{i 2\pi s} : s \in [0, 1)\}$

Define $\Theta : S^1 \rightarrow \mathbb{R}/\sim$ as $\Theta(\exp{i2\pi s}) = \{s + k : k \in \mathbb{Z}\}$ and show that $\Theta$ is an homeomorphism.


Since my original approach was flawed, I feel somewhat obligated to not only correct myself, but also think through more of the details.

First of all, for every $[x] \in \mathbb{R}/ \sim$ there is a unique $\tilde{x} \in [0,1)$ such that $x \sim \tilde{x}$, namely $\tilde{x} := x - \lfloor x \rfloor$, where $\lfloor x \rfloor := \max \{ z \in \mathbb Z \mid z \le x \}$.

Now consider the map $$ j: \mathbb R / \sim \to S^{1}, [x] \mapsto e^{i \tilde{x} 2 \pi}. $$ Note that $i(x) = i(y)$ iff $x-y \in \mathbb Z$ iff $x \sim y$ iff $\tilde{x} = \tilde{y}$, so the above map can also be expressed as $$ j: \mathbb R / \sim \to S^{1}, [x] \mapsto e^{i x 2 \pi} $$ if for every $a \in \mathbb R / \sim$ we fix some (any) $x \in \mathbb R$ with $a = [x].

By this argument, $j$ is injective and clearly surjective.

To see that $j$ is a homeomorphism, let us further consider the projection $$ p \colon \mathbb R \to \mathbb R / \sim, x \mapsto [x]. $$ and $$ c: \mathbb R \to S^{1}, x \mapsto e^{ix \pi}. $$

$c$ is continous and hence, for any open $O \subseteq S^{1}$, the set $c^{-1}"[O] \subseteq \mathbb R$ is open. It follows, from the definition of the quotient space topology that $j$ is continuous.

On the other hand, if $O \subseteq \mathbb{R} / \sim$ is open, then $U := p^{-1}"O \subseteq \mathbb R$ is open and in fact $j"[O] = c"[U]$. Since $c$ is an open map, it follows that $j$ is open as well.

  • $\begingroup$ Uh...? No, as the OP noted the quotient is the circle. What you might be trying to say is that every real number has a unique representative in $[0,1)$ under this equivalence relation. $\endgroup$ – Pedro Tamaroff Oct 30 '16 at 23:22
  • $\begingroup$ You're writing that $[0,1)\simeq S^1$. This is false. $\endgroup$ – Pedro Tamaroff Oct 30 '16 at 23:25
  • $\begingroup$ @PedroTamaroff Shoot, you're right. $[0,1)$ isn't compact. $\endgroup$ – Stefan Mesken Oct 30 '16 at 23:26

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.