Realizations of Circle

Question

I really don't know how to prove without brute force that: $\mathbb{R}/{\sim}\cong[0,1]/{\sim}$

I know already that: $[0,1]/{\sim}\cong\mathbb{S}^1$
(simply use closed map lemma and uniqueness of quotient spaces)

The point is that I cannot apply the closed map lemma on wrapping real line around circle to deduce that this is indeed a quotient map...

Answer 1

Define the map $i: [0,1]/{\sim} \rightarrow \mathbb{R}/{\sim}$ by $i([x]) = [x]$, where $[x]$ denotes the class of $x$ in the respective quotient. Now as every point $x \in \mathbb{R}$ has a representative in $[0,1]$, namely $x - \lfloor x \rfloor$, the map $i$ is onto. And this map is clearly continuous: $i \circ q$, where $q$ is the standard quotient map from $[0,1]$ onto $[0,1]/{\sim}$ is just the map sending $x \in [0,1]$ to $[x]$ in $\mathbb{R}/{\sim}$, so just the restriction of the other quotient map between $\mathbb{R}$ and $\mathbb{R}/{\sim}$. Continuity now follows from the standard universal property for quotients.

So your quotient $\mathbb{R}/{\sim}$ is compact and you can apply this theorem again.

Answer 2

$\sim$ is defined here on $[0,1]$ as follows :$s\sim t$ iff $s=t+n,\ n\in\mathbb{Z}$ which in $[0,1]$ is possible only when $0\sim 1$ and for every other $x\in [0,1]$, $x\sim x$. Now define $p:[0,1]\longrightarrow \mathbb{S}^1$ as $f(x)\mapsto e^{2\pi ix}$. Check that this is a continuous surjection such that $f(x)=f(y)\Longleftrightarrow x\sim y$, thus deducing that $[0,1]/\sim \ \approxeq \mathbb{S}^1$.