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I was reading in a lecture notes the proof which says that the real projective space $\mathbb{R}P^n$ is homeomorphic to $S^n/ \sim$, where $p \sim q$ iff $p = \pm q$.

Just after this proof, there is an exercise which ask to prove $\mathbb{R}P^1$ is diffeomorphic to $S^1$.

I know how to prove this without using the claim proved just before, but if one wanted to use the fact that $\mathbb{R}P^1$ is homeomorphic to $S^1/ \sim$, then he must prove first that $S^1/ \sim$ is homeomorphic to $S^1$. Is really $S^1/ \sim$ homeomorphic to $S^1$? (My guess is yes, by identifying both $S^1$ and $S^1/\sim$ with the suitable intervals in $\mathbb{R}$ and then using the fact that all open intervals in $\mathbb{R}$ are homeomorphic. Is this correct?)

Asma
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Consider the map $z \mapsto z^2$ on $\mathbb{S}^1$. It descends to a continuous bijection from $\mathbb{S}^1 / \sim$ to $\mathbb{S}^1$. Since $\mathbb{S}^1$ is compact and Hausdorff, this map is a homeomorphism.

Orange Mushroom
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  • Thanks a lot! But I still wanted to know was my idea true or not ? – Asma Jan 07 '24 at 21:18
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    It's unclear how you want to identify the sphere and its quotient with intervals on the line. If you fleshed out your argument, I'm sure people would be able to provide more detailed feedback. – Orange Mushroom Jan 07 '24 at 21:22
  • Now that I wanted to write down my idea I realized it was false ( I thought we can identify $S^1$ with an interval which is symmetric with respect to $0$ and then identify the quotient with the positive part of the interval but No it is not true, the circle is identified with an interval like $]0,1]$ which doesn't contain negative numbers!). – Asma Jan 07 '24 at 21:31