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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 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 of $\mathbb S^1 / \{x \sim -x\} \simeq\mathbb S^1$ (cf. Alessandro Bigazzi's comment below):

Let $\mathbb{S}^1=\{z\in\mathbb{C}: ||z||=1\}$ 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 property of quotient topology, there exists an unique homeomorphism $\varphi : \mathbb{S}^1/\sim \to\mathbb{S}^1$.

Harvey Peng
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Lucas Kaufmann
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    Indeed $S^1/\sim$ is homeomorphic to a semicircle with endpoints identified. – Cheerful Parsnip Feb 22 '13 at 14:34
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    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 02:46
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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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    @AlessandroBigazzi I think your comment should be the answer, as it seems the most simple, yet most concise. – polynomial_donut May 27 '16 at 09:10
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    I'd like to add that the universal property on it's own doesn't guarantee the existence of a homeomorphism, merely a continuous bijection. But here it is a continuous bijection from a compact space to a hausdorff one, and thus must be open as well. – Tim The Enchanter May 23 '20 at 14:46
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    Lucas Kaufmann, why is $RP^1$ homeomorphic to $S^1$ modulo the equivalence? – user652838 Oct 02 '20 at 12:19
  • How to show $f:S^1\rightarrow S^1$ is a quotient map? – Infinity Feb 01 '23 at 07:21
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    @Spectrum I don't think we can show $f$ is a quotient map. The antipodal points on $S^1$ obviously show it's not injective. However, the map induced by $f$ on $S^1$/~ is the desired homeomorphism. Suppose the map is $\phi : S^1\text{/ ~} \to RP^1 $. You can verify the well-definedness, bijectivity and continuity of $\phi$. Since compactness is preserved under quotient map, then $\phi$ is a bijection from a compact space to a Hausdorff space, hence a homeomorphism. – Stack_Underflow May 02 '23 at 13:57
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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 $\mathbb{S}^1$.

user_not_found
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