Quotient topology R/Z homeomorphic to [0,1]/{0,1}

Question

I want to show that the quotient topology on $\mathbb{R}/\mathbb{Z}$ is homeomorphic to [0,1]/{0,1}. I have seen a few similar questions but none have provided a rigorous proof.

I know that the projection map $\pi: \mathbb{R} \rightarrow \mathbb{R}/\mathbb{Z}$ sending $x$ to its equivalence class $[x]$ is continuous. I can define $g: \mathbb{R} \rightarrow [0,1]/\{0,1\}$ by $$g(x) = \min\{x-n: x-n\geq 0, n \in \mathbb{Z}\}$$ It is clear that $g(x) = g(y) \Leftrightarrow x-y \in \mathbb{Z}$ ($g$ is constant on equivalence classes) and that $g$ is surjective. So $\exists! f:\mathbb{R}/\mathbb{Z} \rightarrow [0,1]/\{0,1\}$ with $f([x]) = g(x)$ is well-defined and bijective.

I now want to show that $f$ is a homeomorphism, so I need to show that it is continuous with a continuous inverse. I am looking for an easy method to show this that is rigorous (this may involve a different choice of $g$).

My attempt so far has been as follows; given $A$ open in $[0,1]$, we know A is a union of open intervals $(a,b) \ a \leq b \ a,b \in [0,1] (\text{or} \ A = [0,1])$ and $g^{-1}((a,b)) = \cup_{n \in \mathbb{Z}} (a+n, b+n)$ which is open in $\mathbb{R}$ so we have that $g$ is continuous. This implies that $f$ is also continuous. (is this assertion of the open sets on [0,1] correct?)

Also, given an open interval $(c,d) \in \mathbb{R}$, if $d-c \geq 1$ then $g((c,d)) = [0,1]$ which is open in [0,1]. If $d-c <1$ then $g((c,d)) = (c-n, d-n)$ for some $n \in \mathbb{Z}$ which is open in [0,1]. This implies that $g$ is a quotient map and so $f$ is a homeomorpihsm.

Is this proof correct and is there a simpler one?

Answer 1

Some thoughts...

$f$ is continuous if $g$ is. This is an instance of the universal property characterizing quotient maps ($\pi$ is a quotient map, as you point out) .

Then you still need to show $f^{-1}$ is continuous... (should be straightforward).

Note: open intervals $(a,b)$ do form a basis for the (standard) topology on $\mathbb R$; but for $A$ open in $[0,1]$, $A$ is a union of intersections of intervals $(a,b)$ with $[0,1]$...