# 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!

• The circle is given the subspace topology from the plane, and the projective line is given the quotient topology. Feb 22, 2013 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 $$\mathbb S^1$$ 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 $$\varphi : \mathbb{S}^1/\sim \to\mathbb{S}^1$$

• Indeed $S^1/\sim$ is homeomorphic to a semicircle with endpoints identified. Feb 22, 2013 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. Mar 26, 2013 at 2:46
• 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 $\varphi : \mathbb{S}^1/\sim \to\mathbb{S}^1$. Jul 11, 2013 at 12:41
• @AlessandroBigazzi I think your comment should be the answer, as it seems the most simple, yet most concise. May 27, 2016 at 9:10
• 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. May 23, 2020 at 14:46

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 $$\mathbb{S}^1$$.