# Proving that $\mathbb{R}/ \sim$ is homeomorphic to $S^1$

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.

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

## 3 Answers

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