# Universal covering space of the real projective line?

I´m thinking about universal covering spaces. I´ve seen a lot of examples and authors ever say "the sphere $$S^n$$ is the universal covering space of the $$n$$-dimensional projective space $$\mathbb{R}P^n$$ for $$n \geq 1$$.

So my question is: and what about the real projective line $$\mathbb{R}P^1$$? Has it universal covering space?

Thanks!

$$\mathbb{R}P^1$$ is homeomorphic to $$\mathbb{S}^1$$. To see this, first note that more generaly $$\mathbb{R}P^n\simeq\mathbb{S}^n/_{\pm id}$$ and in the case $$n=1$$ you have $$\mathbb{S}^1/_{\pm id}\simeq \mathbb{S}^1$$ (just factorize the map $$z\mapsto z^2$$).

From here you can conclude.

• Sure! I had the homeomorphism but I missed it! Lol. Thanks! :D – user183002 Apr 20 at 17:19

The real projective line is just a circle, so the universal covering space is the real line.

• I don´t agree. Even it´s not a circumference. It´s a quotient of a circumference, yeah? – user183002 Apr 20 at 17:10
• @user A cheap way to see it is that the unique closed one dimensional manifold is a circle. There's nothing else it can be. – Matt Samuel Apr 20 at 17:15
• Important to note is that it isn’t the universal cover. – Connor Malin Apr 20 at 18:47
• What I mean is that $S^1 \rightarrow \mathbb{R}P^1$ is a covering space, but not a universal one. – Connor Malin Apr 20 at 19:29
• @Connor I agree. Note my answer says the real line is the universal cover. – Matt Samuel Apr 20 at 19:31