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I want to show that $\Bbb R \Bbb P^1 \simeq \Bbb S^1$, So for this I consider the quotient $\Bbb R^2 / \{(0, 0) \}$ induced by the equivalence relation $x \sim y$ iff $x= \lambda y$ with $\lambda \in \Bbb R -\{0\}$

We will say that $\Bbb R \Bbb P^1$ is the map quotient of $\Bbb R^2 / \{(0, 0) \}$ induced by $\pi:\Bbb R^2 / \{(0, 0) \} \to \Bbb R \Bbb P^1$ given by $\pi(x)=\overline{x}$, we can consider the restriction $$\varphi: \Bbb S \to \Bbb R \Bbb P^1$$ which identifies the antipodal points, i.e. we can describe the projective plane as

$$\Bbb R \Bbb P^1= \{\overline{x}:x, -x \in \overline{x}, x \in \Bbb S^1 \}$$

Now, related to the answer given in this post I understand that the idea is to consider the circle cocient with the equivalence relation that identifies all the antipodal points and to see that it is homeomorphic to the projective plane.

This means, that from this application I identify the projective plane, which corresponds precisely to all the antipodal points?. I do not understand very well because everything is based on proving that the circle cocient with the equivalence relation is homeomorphic to the same circle.

Alex
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Wrloord
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  • Think about the unit circle with antipodal points identified. We can erase the bottom half (where $y<0$) since all of those points are identified with some point in the top half. But then the left and right points $(-1,0)$ and $(1,0)$ still need to be identified. So you get something homeomorphic to a line segment with its endpoints identified, which is homeomorphic to a circle. – Cheerful Parsnip Mar 09 '24 at 20:22
  • Hint: first show that $S^1/(x\sim -x)$ is homeomorphic to $\mathbb {RP}^1)$, then show that $S^1$ is homeomorphic to $S^1/\sim$. – Chris Mar 09 '24 at 20:24
  • Why do you say projective plane? This is the projective line. – Ted Shifrin Mar 09 '24 at 22:48

1 Answers1

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Let $\newcommand{\S}{\mathbb{S}}\newcommand{\RP}{\mathbb{R}\mathrm{P}} \S^1_+ = \{e^{i\pi t} \mid t \in [0, 1]\} \subseteq \S^1$ be the upper half-circle. Note that $\S^1_+$ contains only a single pair of antipodal points, namely $1$ and $-1$. Clearly the restriction $\varphi|_{\S^1_+}\colon \S^1_+ \to \RP^1$ is surjective as $\S^1_+$ contains at least one representative of each class, and by the preceding remark it is also injective except for $\varphi(1) = \varphi(-1)$. Thus, it factors over the quotient $\tilde{\S}^1 = \S^1_+ / \{-1, 1\}$ to yield a homeomorphism $\tilde{\S}^1 \to \RP^1$ (by the compact-Hausdorff lemma), so all that's left to show is that $\tilde{\S}^1$ is homeomorphic to $\S^1$.

For this, consider the map $\psi\colon \S^1_+ \to \S^1$ given by $\psi(x) = x^2$. Again this is clearly surjective since $\S^1 = \{e^{2 i \pi t} = (e^{i \pi t})^2 \mid t \in [0, 1]\}$, and again we find that the map is injective except for $\psi(1) = \psi(-1) = 1$. By the same argument as above, it factors over $\tilde{\S}^1$ to give a homeomorphism $\tilde{\S}^1 \to \S^1$, completing the proof.

Ben Steffan
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  • Thanks for your help, but I'm not getting clear why the application is continuous? Well to use the lemma it should be a continuous bijection. Could you enlighten me a little? – Wrloord Mar 10 '24 at 15:18
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    @Wrloord That's just the universal property of the quotient. If you have a continuous map $f\colon X \to Y$ that is constant on a subspace $X' \subseteq X$, then there is a unique continuous map $f'\colon X / X' \to Y$ such that $f = f' \circ \pi$ where $\pi\colon X \to X / X'$ is the quotient projection. You should be able to find a proof of this in any general topology textbook. – Ben Steffan Mar 10 '24 at 17:22