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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!

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The circle is given the subspace topology from the plane, and the projective line is given the quotient topology. – Olivier Bégassat Feb 22 '13 at 13:40

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 $S^1$ we get a continuous open mapping from $S^1 \to R P^1$ that identifies two antipodal points. We see then that $\mathbb RP^1$ is homeomorphic to $S^1$ modulo the equivalence $x \sim -x$ (again with the quotient topology).

What remains to be shown is that $S^1 / x \sim -x$ is homeomorphic to $S^1$, which is not difficult.

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Indeed $S^1/\sim$ is homeomorphic to a semicircle with endpoints identified. – Grumpy Parsnip Feb 22 '13 at 14:34
    
I know you said showing $S^1 / x \sim -x$ is homeomorphic to $S^1$ is not difficult, but could you perhaps elaborate on that? I've been dealing with a similar problem, and have been trying to show just that with no luck. – Ryker Mar 26 '13 at 2:46
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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 $\varphi : \mathbb{S}^1/\sim \to\mathbb{S}^1$. – Alessandro Bigazzi Jul 11 '13 at 12:41

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