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I am reading Lee, Manifolds and Differential Geometry, and I am a bit confused. It is seen that the real projective space $\mathbb{R}P^n$ is homeomorphic to $S^n/ \sim$, where $p \sim q$ iff $p = \pm q$.

However, in one of the exercises (1.48) one is asked to prove that $\mathbb{R}P^1$ is diffeomorphic to $S^1$.

Is this really true (is there a simple proof of this) and does this imply something for the relation between $S^1$ and $S^1/\sim$?

Does this also mean that $\mathbb{R}P^1$ is homeomorphic to $S^1$ (since diffeomorphism is a stronger condition)?

JezuzStardust
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  • Please check whether the answer(s) at the linked question settle things for you. The easiest proof of diffeomorphism I know is to observe that the polar graph $r = \cos\theta$ represents the projective line, and is a Euclidean circle (parametrized at unit speed, no less!). – Andrew D. Hwang Jan 03 '17 at 14:01
  • Ah, so actually the quotient $S^1/\sim$ is actually $S^1$? If that is correct then I understand and it solves my problems. I guess this is a special case and in general $S^n/\sim$ is not homeomorphic to $S^1$. – JezuzStardust Jan 03 '17 at 14:33

2 Answers2

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The unique 1-dimensional compact connected manifold is $S^1$.

To see that $RP^1$ is $S^1$, you view $S^1=\{ e^{it}, t=\in [0,2\pi])\}$. The identification $p\simeq q$ here can be described like so. Consider the interval $I=\{e^{it}, t\in [0,\pi]\}$. If $t\in (0,\pi)$ there exists a unique $t'\in (\pi,2\pi)$ such that $e^{it}\simeq e^{it'}$ and $e^{i0}=1\simeq e^{i\pi}=-1$. Thus the quotient space is obtained by identifying $0$ and $-1$ in $I$. That is you identify, the endpoints of an interval. This is the circle.

Tsemo Aristide
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$f : RP^1 \to S^1, z \mapsto z^2$ is the desired diffeomorphism (hence homeomorphism).