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.
Best Answer
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.