Q: Is $S^1$ homeomorphic to $\mathbb{R}P^1$? I am supposed to construct a homeomorphism of $S^1$ and $\mathbb{R}P^1$ but I am not toally sure that this is even possible. I think I have learned at some point that
$$\mathbb{R}P^1=S^1/\{x=-x\}$$
Thank you.

A: As $z\mapsto z^2$ (or $e^{it}\mapsto e^{2it}$) shows, we have $S^1\approx S^1/\{x=-x\}$.

A: Let $\sim$ be the equivalence relation on $S^1$ identifying antipodal points.  Clearly $S^1/\sim$ is homeomorphic to $\mathbb{R}P^1$, what isn't clear is that $S^1$ is homeomorphic to $S^1/\sim$.  But notice the latter space is homeomorphic to a semicircle with endpoints identified, which is clearly homeomorphic to $S^1$, hence $S^1\simeq S^1/\sim\simeq\mathbb{R}P^1$.
Edit: Thanks for fix Kevin, I didn't notice my typos.

A: It is possible, as the two spaces are indeed homeomorphic.
