homeomorphism between the real projective line and a circle I'm currently following an introductory course in geometry and
it was mentioned that the real projective line is homeomorphic to a circle.
Could someone please state the topologies on both the real projective line and the circle
and a corresponding homeomorphism? 
Thanks a lot!
 A: The real projective line is the set of lines through the origin in $\mathbb R^2$. It can be seen as the quotient of $\mathbb R^2 \setminus \{0\}$ by the relation $x \sim y$ iff $x=\lambda y$ for some nonzero $\lambda \in \mathbb R$. The topology here is the quotient topology for the map $\mathbb R^2 \setminus \{0\} \to \mathbb R P^1$ sending a point to its equivalence class.
Restricting this map to $\mathbb S^1$ we get a continuous open mapping from $\mathbb S^1 \to \mathbb R P^1$ that identifies two antipodal points. We see then that $\mathbb RP^1$ is homeomorphic to $\mathbb S^1$ modulo the equivalence $x \sim -x$ (again with the quotient topology).
What remains to be shown is that $\mathbb S^1 / \{x \sim -x\}$ is homeomorphic to $\mathbb S^1$, which is not difficult.
Proof that $\mathbb S^1 / \{x \sim -x\} \simeq\mathbb S^1$ (cf. Alessandro Bigazzi's comment below):
Let be $\mathbb{S}^1=\{z\in\mathbb{C}\mid ||z||=1\}$ and consider the map $f:\mathbb{S}^1\to \mathbb{S}^1$ defined as $f(z)=z^2$. You can check easily that $f$ is a continuous surjective function and such that $f(z)=f(-z)$, proving that $f$ passes to quotient under equivalence relation $\sim$ identifying antipodal points on $\mathbb{S}^1$. By universal propriety of quotient topology, there exists an unique homeomorphism $\varphi : \mathbb{S}^1/\sim \to\mathbb{S}^1$
