I was reading in a lecture notes the proof which says that the real projective space $\mathbb{R}P^n$ is homeomorphic to $S^n/ \sim$, where $p \sim q$ iff $p = \pm q$.
Just after this proof, there is an exercise which ask to prove $\mathbb{R}P^1$ is diffeomorphic to $S^1$.
I know how to prove this without using the claim proved just before, but if one wanted to use the fact that $\mathbb{R}P^1$ is homeomorphic to $S^1/ \sim$, then he must prove first that $S^1/ \sim$ is homeomorphic to $S^1$. Is really $S^1/ \sim$ homeomorphic to $S^1$? (My guess is yes, by identifying both $S^1$ and $S^1/\sim$ with the suitable intervals in $\mathbb{R}$ and then using the fact that all open intervals in $\mathbb{R}$ are homeomorphic. Is this correct?)
Consider the map $z \mapsto z^2$ on $\mathbb{S}^1$. It descends to a continuous bijection from $\mathbb{S}^1 / \sim$ to $\mathbb{S}^1$. Since $\mathbb{S}^1$ is compact and Hausdorff, this map is a homeomorphism.