Circle as disjoint union of $\mathbb{R}$ under equivalence relation

Question

I came across a construction of $S^1$ as $\mathbb{R}\sqcup\mathbb{R}/\sim$ where $x\sim y$ when $x=y^{-1}$ when $x,y\neq 0$, but i can't see why this is true.

Answer 1

Here's the intuition:

Each circle with a missing point is homeomorphic to $\mathbb R$. The equivalence relation identifies a point in one circle with a point in the other whenever the two points lie on the same ray extending from the origin.