Help understand the construction of the real projective line $\mathbb{R}P^1$.

Question

So I am trying to get a better understanding of real projective spaces $\mathbb{R}P^n$. I'm starting off by looking at $\mathbb{R}P^1$ and trying to construct it from first principles and showing that it is equivalent to the circle $S^1$. By "equivalent", we mean that there exists a homeomorphism $f:S^1\to\mathbb{R}P^1$, a.k.a a bicontinuous bijection, between the spaces.

Starting off with the actual construction of $\mathbb{R}P^1$, I know it is the quotient space $S^1/\sim$ where for $x,y\in S^1$, $x\sim y$ if and only if $x=\lambda y$ where $\lambda\in\{-1,1\}$. In other words, $x$ and $y$ are similar if they are antipodal to each other. Then $\mathbb{R}P^1=\{[x]:x,-x\in [x]\;\forall\;x\in S^1\}$, i.e. the set of equivalence classes of antipodal points on the circle.

So far so good.

The trouble comes for me in showing that this is homeomorphic to the circle. Taking the canonical projection map from $S^1$ to $\mathbb{R}P^1$ defined by $x\mapsto [x]$. To show this is a homeomorphism: Surjectivity is obvious, as is bicontinuity from the definition of quotient spaces. However, suppose $[x],[y]\in\mathbb{R}P^1$ and $[x]=[y]$. Then $x\in [y]$ and $y\in [x]$. This implies that either $x=y$, in which case huzzah, or $x=-y$, in which case we have a problem with injectivity. Am I missing something here? Is there some basic logic I'm not applying? Or do I need to try a different function as a homeomorphism? Any help would be much appreciated!

Answer 1

You are correct that the quotient map $S^1\to\mathbb{R}P^1$ is not a homeomorphism because it fails to be injective. In fact it is a double cover. What you need to do is find a different map from $S^1$ to $\mathbb{R}P^1$ which witnesses the fact that they are homeomorphic.

Identify $S^1$ with the unit circle in $\mathbb{C}$ and parameterise the points by $e^{i\theta}$ for $\theta\in[0,2\pi)$. Now $\mathbb{R}P^1=\{[e^{i\theta}]\mid\theta\in[0,2\pi)\}$, where $[e^{i\theta}]=\{e^{i\theta},e^{i(\theta+\pi)}\}$. Define the map $$S^1\to\mathbb{R}P^1\colon e^{i\theta}\mapsto[e^{i\theta/2}].$$

You should be able to check this is a homeomorphism. In particular note that $$\lim_{\theta\to 2\pi^-}[e^{i\theta/2}]=[e^{i\pi}]=[e^{i2\pi}]=[e^{i0}],$$ which ensures that the map is continuous and surjective.

Answer 2

It is easier to construct a homeomorphism $h : \mathbb RP^1 \to S^1$. Define $$\phi : S^1 \to S^1, \phi(z) = z^2 .$$ This is a continuous map. Since each complex number $w$ has a square root (with absolute value $\sqrt{\lvert w \rvert}$), $\phi$ is surjective. Moreover we have $$\phi(z) = \phi(z') \iff z^2 = (z')^2 \iff z = \pm z' . \tag{1}$$

By $(1)$ $\phi$ induces a unique continuous $h : \mathbb RP^1 \to S^1$ such that $h \circ \pi = \phi$, where $\pi : S^1 \to \mathbb RP^1$ is the quotient map. Since $\phi$ is surjective, also $h$ is surjective. Again by $(1)$ $h$ is injective.

Thus $h$ is a continuous bijection. Since $ \mathbb RP^1$ is compact and $S^1$ is Hausdorff, $h$ is a homeomorphism.