Projective real line is homeomorphic to $S$.

Question

Let $S = S^1 / \sim$, being $x \sim y$ if $x = - y$. To obtain the compactness and the connectess of the projective real line, $\mathbb{R} P^1$, I need to prove that $S$ is homeomorphic to $\mathbb{R} P^1$. I can show that exists a biyection $h : \mathbb{R} P^1 \to S$, but I don't know prove that it is an homeomorphism too. Thank you very much.

score 2 · Answer 1 · answered Apr 10 '17 at 16:04

Actually, a quotient of a compact space is again compact, and the same holds for connectedness. So no doubt $\mathbb{R}P^1$ is connected and compact. However, the homeomorphism you are referring to is not hard to construct.

Thinking of $S^1$ as $\mathbb{R}/2\pi\mathbb{Z},$ define a map$$f:S^1\to S^1,\quad t\mapsto2t.$$This is clearly continuous. Furthermore, for $x,y\in S^1$ we have $f(x)=f(y)\Leftrightarrow x\sim y$. Hence, $f$ induces a continuous bijection $\mathbb{R}P^1\to S^1$. As $\mathbb{R}P^1$ is compact and $S^1$ is Hausdorff, the induced map is in fact a homoemorphism.

Projective real line is homeomorphic to $S$.

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