# Homeomorphism between different definitions of real projective spaces.

First definition of Projective Space. For same natural $$n\ge 1$$, let $$\mathbb{R}P^n:=\{V\subset\mathbb{R}^{n+1}\;|\; \dim(V)=1\}$$. We introduce a topology on $$\mathbb{R}P^n$$ as follow: define $$\pi\colon\mathbb{R}^{n+1}\setminus\{0\}\to\mathbb{R}P^n$$, $$\pi(x)=\text{span}\{x\}$$, we say $$U\subset\mathbb{R}P^n$$ is open if $$\pi^{-1}(U)$$ is open in $$\mathbb{R}^{n+1}\setminus\{0\}$$, with induced topology of $$\mathbb{R}^{n+1}$$.

Second definition of Projective Space. For same natural $$n\ge 1$$, consider $$\mathbb{S}^n:=\{x\in\mathbb{R}^{n+1}\;|\; ||x||=1\}$$, in $$\mathbb{S}^n$$ we define $$x\sim y$$ iff $$x=y$$ or $$-x=y$$. The map $$q\colon\mathbb{S}^n\to \mathbb{S}^n/\sim$$, $$q(x)=[x]=\{-x,x\}$$ is called projection map, we set $$\mathbb{P}^n:=\mathbb{S}^n/\sim$$. We will say that $$U$$ is open in $$\mathbb{P}^n$$ if $$q^{-1}(U)$$ is open in $$\mathbb{S}^n.$$

Exercise The $$\mathbb{R}P^n$$ is homeomorphic to $$\mathbb{P}^n.$$

My attempt. Let $$\varphi\colon\mathbb{P}^n\to\mathbb{R}P^n$$, $$\varphi(\{-x,x\})=\text{span}\{-x,x\}.$$

$$\varphi$$ is injective: if $$\varphi(\{-x,x\})=\varphi(\{-y,y\})$$ we have that if $$t\in\{\lambda x\;|\;\lambda\in\mathbb{R}\}=\{\mu y\;|\;\mu\in\mathbb{R}\}$$, $$t=\lambda x=\mu y\Rightarrow$$ $$||t||=|\lambda|=|\mu|$$, then $$\lambda=\pm \mu$$. Hence $$\mu x=\mu y$$ or $$-\mu x=\mu y$$ $$\Rightarrow x=y$$ or $$-x=y\Rightarrow$$ $$x\sim y\Rightarrow[x]=[y]\Rightarrow \{-x,x\}=\{-y,y\}$$.

$$\varphi$$ is onto. Let $$r\in\mathbb{R}P^n$$, then $$r=\text{span}(\tilde{x})$$, where $$\tilde{x}\in\mathbb{R}^{n+1}\setminus\{0\}$$.

Let $$x\in r\cap\mathbb{S}^n$$, then $$x\in \text{span}\{\tilde{x}\}\cap\mathbb{S}^n$$, hence $$x=\frac{\tilde{x}}{||\tilde{x}||}$$ or $$x=-\frac{\tilde{x}}{||\tilde{x}||}$$, therefore

$$\varphi(\{-x,x\})=\text{span}\bigg\{-\frac{\tilde{x}}{||\tilde{x}||},\frac{\tilde{x}}{||\tilde{x}||}\bigg\}=\text{span}\bigg\{\frac{\tilde{x}}{||\tilde{x}||}\bigg\}=\text{span}\{\tilde{x}\}=r$$

It's correct?

It remains to show that $$\varphi$$ is an open application, that is it must send open in open. I don't know how to proceed, could anyone give me a hint?

Thanks!

• Here is a hint: First show that $\varphi$ is continuous. Then show that $\mathbb RP^n$ is Hausdorff(see here). Note that a continuous bijection from a compact space to Hausdorff space is a homeomorphism. – Thomas Shelby Jun 28 at 9:12
• @ThomasShelby Thanks for your answer. But I would like to show that the application is open, without using additional properties. – Jack J. Jun 28 at 9:21

Your argument is correct, but you only show that $$\varphi$$ is a bijection. I suggest to do it as follows.
Let us first observe that the maps $$\pi$$ and $$q$$ are quotient maps. Recall that a map $$p : X \to Y$$ is a quotient map if it is surjective and the following holds: $$V \subset Y$$ is open iff $$p^{-1}(V)$$ is open in $$X$$. Quotient maps are continuous and have the following universal property: If $$f: Y \to Z$$ is any function, then $$f$$ is continous iff $$p f : X \to Z$$ is continuous. Now obviously each function $$F : X \to Z$$ such that $$F(x) = F(x')$$ for all $$x,x'$$ with $$p(x) = p(x')$$ induces a unique function $$f : Y \to Z$$ such that $$f p = F$$. Since $$p$$ is a quotient map, $$f$$ is continuous iff $$F$$ is continuous.
Let $$i : S^n \to \mathbb R^{n+1} \setminus \{ 0 \}$$ denote the inclusion map and $$r : \mathbb R^{n+1} \setminus \{ 0 \} \to S^n, r(x) = x/\lvert x \rVert$$. Both maps are continuous. We have $$r i = id_{S^n}$$. The map $$i r$$ is not the identity, but $$i r(x)$$ and $$x$$ span the same one-dimensional subspace whence $$\pi i r = \pi$$.
If $$x \sim y$$ in $$S^n$$, then clearly $$\pi i(x) = \text{span}(x) = \text{span}(y) = \pi i(y)$$. By the universal property of the quotient $$i$$ induces a unique continuous function $$\varphi : \mathbb P^n \to \mathbb RP^n$$ such that $$\pi i = \varphi q$$.
Next consider the map $$q r : \mathbb R^{n+1} \setminus \{ 0 \} \to \mathbb P^n$$. Let $$x,x' \in \mathbb R^{n+1} \setminus \{ 0 \}$$ such that $$\pi(x) = \pi(x')$$. This means that $$x'= \lambda x$$ for some $$\lambda \ne 0$$. Therefore $$r(x') = \lambda x/ \lVert \lambda x \rVert = (\lambda / \lvert \lambda \rvert) (x/ \lvert x \rVert) = (\lambda / \lvert \lambda \rvert) r(x) = \pm r(x)$$, thus $$qr(x') = qr(x)$$. Hence $$qr$$ induces a unique continuous map $$\psi :\mathbb RP^n \to \mathbb P^n$$ such that $$\psi \pi = q r$$.
Now we have $$\psi \varphi q = \psi \pi i = q r i = q = id_{\mathbb P^n} q$$. Since $$q$$ is surjective, we get $$\psi \varphi = id_{\mathbb P^n}$$. Similarly we have $$\varphi \psi \pi = \varphi q r = \pi i r = \pi =id_{\mathbb RP^n} \pi$$ and conclude $$\varphi \psi = id_{\mathbb RP^n}$$.