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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!

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    $\begingroup$ 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. $\endgroup$ – Thomas Shelby Jun 28 at 9:12
  • $\begingroup$ @ThomasShelby Thanks for your answer. But I would like to show that the application is open, without using additional properties. $\endgroup$ – Jack J. Jun 28 at 9:21
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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 $\varphi \psi = id_{\mathbb RP^n}$.

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