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I am studying the fundamental group and I am facing a problem with the projective space: we know that, by definition, the real projective space $P^m(\mathbb{R})$ is the quotient of $R^{m+1}-\{0\}$ under the equivalence relation $x\sim \lambda x\, \forall \lambda \neq 0$, that is the action of the multiplicative group of the non null real numbers. But the projective space is also homeomorphic to the quotient of the sphere $S^m$ under the equivalence relation induced by the antipodal map $\alpha(x)=-x$, that is the action of the group $\mathbb{Z}_2$ on the sphere. However, I know that the projective real line is homeomorphic to $S^1$, because the projective line is homeomorphic to the Alexandroff compactification of $\mathbb{R}$, which is homeomorphic to $S^1$. Thus, my question is: why the projective real line is not the quotient of $S^1$ under the equivalence relation induced by the antipodal map? Why this holds just for $m\geq 2$?

Thanks for any answer and help!

Jyrki Lahtonen
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Steppenwolf
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Why is the projective real line not the quotient of $S^1$ under the equivalence relation induced by the antipodal map?

It is! It is also homeomorphic to $S^1$, as you correctly observe. Perhaps it is surprising at first sight that the quotient of $S^1$ by the antipodal action is just $S^1$ again, in particular since this is not true for all higher-dimensional spheres, but showing that this is the case is actually rather straightforward point-set topology. In case you don't want to work this out yourself (although I recommend it), this has been done a number of times on this site alone. Here's an answer of mine that gives a fair amount of detail, for instance.

Ben Steffan
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