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The unit circle $$\mathbb S^1 := \{z: |z|=1\}$$ is a circle on the complex plane.

The Real Projective Line $\mathbb{RP}^1$ is defined as pairs such as $\pm z$ on $\mathbb S^1$.

Function $z \mapsto z^2$ maps the the Real Projective Line $\mathbb{RP}^1$ to $\mathbb S^1$, futhermore, such mapping under polar coordinate is 1-1, so I would like to think that the two groups are isomorphic, ie $$\mathbb{RP}^1 \cong \mathbb S^1$$ Am I right to say so?

Sammy Black
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athos
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    Yes, they're isomorphic, although I'd be a little careful using the term "group" here. There is a group structure here but it doesn't generalize to either higher-dimensional projective spaces or to the complex projective line so it's sort of accidental. – Qiaochu Yuan Oct 08 '22 at 23:37
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    Indeed, the group theory tag doesn't belong here, as projective spaces are not groups. You should be discussing homeomorphism or diffeomorphism and tagging more topologically. – Ted Shifrin Oct 09 '22 at 00:03
  • @TedShifrin Those are groups, so it's ok to discuss isomorphisms and use the tag group theory. – jjagmath Oct 09 '22 at 00:09
  • @jjagmath Really? Define the group structure on $\Bbb RP^1$ in a natural way without using the homeomorphism. – Ted Shifrin Oct 09 '22 at 00:55
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    @TedShifrin The OP defined $\Bbb R\Bbb P^1$ as pairs such as $\pm z$ is on $\Bbb S^1$. That's the quotient of the group $\Bbb S^1$ by the subgroup ${1,-1}$. It's a purely algebraic definition. No topology involved. – jjagmath Oct 09 '22 at 01:48

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