I have the following exercise:
Show that if $I = [0,1] \subset \mathbb{R}$ and $\sim$ is the equivalence relation $x \sim x´$ if and only if $\{x, x´\}$ = $\{0,1\}$ or $x = x´$, then $I / \sim$ is homeomorphic to $S^{ 1}$.
My attempt:
My first idea was to use the parametrization of the unit circle, that is, $f (x) = (\cos (2 \pi x), \sin (2 \pi x))$ and in this publication they also "construct" it.
The problem I have is that we know that a homeomorphism is a function between two spaces, bijective, continuous and with continuous inverse. From the problem clearly the $f$ that I provide is bijective, but how do I know that the inverse is continuous, or how do I obtain said inverse?
Another way to attack the problem would also be to see $\mathbb{R}^2$ as the complex plane and use the function $f (x) = e^{2i \pi x}$, but it seems to me that in this case we would have to make the composition of $f$ with another function than give us a homeomorphism of $\mathbb{R^{2}}$ with the complex plane.
I do not have much experience in homeomorphisms, for that reason the doubts that I present previously have arisen, in advance, thank you.
Note: $S^{1}=\{x\in \mathbb{R^{2}} : \|x\|=1\}$.
Consider the map $f:I/\sim \rightarrow S^1$ given by
$$f(\theta) = (\cos(2\pi\theta), \sin(2\pi\theta)) \in \mathbb{R}^2.$$
It is continuous because each component function is continuous (i.e. cosine and sine are continuous functions). Consider the function $g:S^1 \rightarrow I / \sim$ given by
$$ g(x, y) = \begin{cases} \frac{\arccos(x)}{2\pi}, & y \geq 0, \\ \frac{-\arccos(x)}{2\pi} \mod 1, & y < 0 \\ \end{cases} $$
Here, we are thinking of $\arccos$ as a continuous bijective map $[-1, 1] \rightarrow [0, \pi]$.
One can check that $g$ is continuous and that it is the inverse of $f$. Therefore, $f$ is a homeomorphism.
Note: By "mod 1" I mean that $-1.05 \equiv -0.05 \equiv 0.95 \mod 1$.