Let $I=[0,1]$ and the equivalence relation $\sim$ that identifies points $0$ and $1$.
I used this theorem :
$\textbf{Theorem:}$ Let $g: X \rightarrow Z$ continuous and surjective. Consider the equivalence relation in $X: x\sim y \Leftrightarrow f(x)=f(y)$ and consider the quotient topology in $\dfrac{X}{\sim}$. Then $q$ induces a bijective continuous function $f: \frac{X}{\sim} \rightarrow Z$ such that
- $f$ is homeomorphism $\Leftrightarrow$ $g$ is a quotient map.
$\textbf{My attempt:}$
Consider $g: I \rightarrow \mathbb{S}^1$ where $g(t)=(\text{cos}(2\pi t), \text{sin}(2 \pi t))$. Is continuos and surjective, by theorem the function
$$ f: \dfrac{I}{\sim} \rightarrow \mathbb{S}^1 $$
defined by $f([x])=(\text{cos}(2\pi x), \text{sin}(2 \pi x))$ is a continuous and bijective function. To see that it is a homeomorphism it would be enough to see that g is a quotient application (or that it is open or closed). Any hint?
You can consider the map
$exp : \mathbb{R}\to \mathbb{S}^1$
Such that $exp(t)=(cos(2\pi t), sin(2\pi t))=e^{2\pi it}$. This map is a covering map, so it is open. In other words it is the quotient induced by the action of $\mathbb{Z}$ on $\mathbb{R}$.
Your map is the restriction of this map on $I/\sim$, and so it is open.