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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!
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 $\varphi : \mathbb{S}^1/\sim \to\mathbb{S}^1$
Another way to do this is to make use of inclusion maps. Consider the following maps:
$$\iota:\mathbb{U}^1\hookrightarrow\mathbb{R}^2$$
$$q:\mathbb{R}^2\to\mathbb{P}^1$$
Where $\mathbb{U}^1$ is the upper semicircle of $\mathbb{S}^1$ and $q$ is the quotient map defined by the relation $x\sim\lambda x$. Note that the map $q\circ\iota$ is a continuous surjection. Because $\mathbb{U}^1$ is compact and $\mathbb{P}^1$ Hausdorff, by the closed map lemma, $q\circ\iota$ is a quotient map. However, note that the quotient map $q’:\mathbb{U}^1\to\mathbb{U}^1/\sim$ induced by the relation $(1,0)\sim (-1,0)$ makes the same assignments as $q\circ\iota$, so $\mathbb{U}^1/\sim\hspace{2mm}\approx\mathbb{P}^1$. But this identification of the “endpoints” of the upper semicircle is identical to attaching the endpoints of the unit interval together, which yields the quotient space $\mathbb{S}^1$. Thus, $\mathbb{P}^1\approx\mathbb{S}^1$.
Alternatively, if you know $\mathbb{P}^1$ is a compact 1-manifold, by the classification theorem it must be homeomorphic to $\mathbb{S}^1$.
