Stuck on last step showing that $\mathbb{R}/\mathbb{Z}$ is homeomorphic to $S^1$, and stating what the homomorphism is. Let $q:\mathbb{R}\to\mathbb{R}/\mathbb{Z}$ be the quotient map, and let $\phi:\mathbb{R}\to S^1$ be the map given by $\phi(\theta)=(\cos(2\pi\theta),\sin(2\pi\theta)).$
I proved in a previous problem that this will induce a unique continuous bijection, $\hat{\phi}:\mathbb{R}/\mathbb{Z}\to S^1$ that satisfies $\phi = \hat{\phi}\circ q$. I've already shown that the mapping $$[\theta]\mapsto (\cos(2\pi\theta),\sin(2\pi\theta))$$ satisfies this property, and thus this mapping must be equal to $\hat{\phi}$.
So what I've currently shown is that $\hat{\phi}([\theta]) = (\cos(2\pi\theta),\sin(2\pi\theta))$ is a well defined continuous bijection between the quotient and $S^1$. But to show that this is a homeomorphism, I need to show that its inverse is continuous, and that's where I'm stuck
I wanted to say something like;
Since we need to prove the inverse is continuous, we pick some open $U$ in $\mathbb{R}/\mathbb{Z}$, then we need to show that $\hat{\phi}^{-1}(U)$ is open in $S^1$, but I can't see why this should be true. I'd like to use the openness of $U$ in $\mathbb{R}/\mathbb{Z}$ along with $q$ to move this into an open set in $\mathbb{R}$, then apply $\phi$ and use its continuity to say something about how $U$ maps back into $S^1$ in under $\hat{\phi}$. But I can't quite get that to work; unfortunately the image of an open set isn't open under a continuous function, so I'm stuck on this last part.
Any thoughts on how to prove that $\hat{\phi}^{-1}$ is continuous would be greatly appreciated.
Thanks in advance.
 A: I think conceptually the best approach is to show that $\phi : \mathbb R \to S^1, \phi(t) = e^{2 \pi i t}$, is a quotient map. Then the following applies:

Let $f : X \to Y$ and $g : X \to Z$ be quotient maps and $h : Y \to Z$ be a bijection such that $h \circ f = g$. Then $h$ is a homeomorphism.

Let us show that $\phi$ is an open map (which implies that it is a quotient map). Since the open intervals $(a,b)$ form a basis for the topology on $\mathbb R$, it suffices to show that all $\phi((a,b))$ are open in $S^1$.
This is trivial for $b - a > 1$ because then $\phi((a,b)) = S^1$.
For $b - a \le 1$ the closed interval $[a,a+1]$ contains $(a,b)$. The set $K = [a,a+1] \setminus (a,b)$ is compact, hence $\phi(K) \subset S^1$ is compact, thus closed in $S^1$. Therefore $S^1 \setminus \phi(K)$ is open in $S^1$.  We have $S^1 = \phi([a,a+1]) = \phi(K \cup (a,b)) = \phi(K) \cup \phi((a,b))$. But $K$ and $(a,b)$ are disjoint, thus $s \in K$ and $t \in (a,b)$ cannot have the same image under $\phi$ (note that the only two distinct points in $[a,a+1]$ having the same image under $\phi$ are $a$ and $a+1$ which are both contained in $K$). We conclude that $\phi(K)$ and $\phi((a,b))$ are disjoint, hence $\phi((a,b)) = S^1 \setminus \phi(K)$ which is open in $S^1$.
See also the following related questions:

*

*Is it an open function?


*Open sets on the unit circle $S^1$
