# Prove that $(\mathbb{R}^{n+1} \backslash \{0\}) / \sim_1$ and $\mathbb{S}^n/ \sim_2$ are homeomorphic.

(a) Take the unit sphere $$S^n$$ in $$\mathbb{R}^{n + 1}$$ and partition it into subsets which contain exactly two points, the points being antipodal (at opposite ends of a diameter). $$P^n$$ is the resulting identification space. We could abbreviate our description by saying that $$P^n$$ is formed from $$S^n$$ by identifying antipodal points.

(b) Begin with $$\mathbb{R}^{n + 1} \backslash \{0\}$$ and identify two points if and only if they lie on the same straight line through the origin. (Note that antipodal points of $$S^n$$ have this property.)

Prove that $$[b]$$ and ($$\mathbb{S}^n/ \sim$$ )(...[a]) are homeomorphic where $$\sim$$ denotes the identification of the antipodal points.

$$\require{AMScd}$$ $$\begin{CD} \mathbb{R}^{n+1}\backslash\{0\} @>g(x) = x /||x||>> S^{n} \\ \ @VV\pi(x) = cl(x)V \\ \ @. \mathbb{S}^{n} / \sim \\ \end{CD}$$

We see that $$g(x)$$ is a continuous surjective map.Also we know that $$\pi(x)$$ is a surjective map from the compact space($$\mathbb{S}^n$$) to the hausdroff space ($$\mathbb{S}^n/ \sim)$$ so it is an identification map.

I know of the theorem that ,

$$\require{AMScd}$$ $$\begin{CD} \mathbb{X} @>g(x)>> Y \\ @VV p V \\ \ \mathbb{X} / \sim \\ \end{CD}$$

If $$g(x)$$ is an identification map and $$p$$ is the projection map then we know that $$Y$$ is homeomorphic to $$\mathbb{X} / \sim$$.

This is what I could come close to. Can someone help me out from here instead of suggesting some other answer? I did go through the various answers on stackexchange and nothing seems to help me as I dont really get the intuition.

edit1:After PaulFrost's answer I proceeded in the following way :

Let $$\require{AMScd}$$ $$\begin{CD} \mathbb{S^{n}} @>h(x)=x>> \mathbb{R}^{n+1} \backslash \{0\} @>\pi(x)=cl(\{x\})>> (\mathbb{R}^{n+1} \backslash \{0\})/\sim_1 \\ @. @. @. \\ \end{CD}$$

where $$\pi(x)$$ is a surjective map so $$\pi \circ h(x)$$ is surjective and since $$\mathbb{S}^n$$ is compact and $$(\mathbb{R}^{n+1} \backslash \{0\})/ \sim_1$$ is Hausdorff so we can conclude that $$\pi \circ h(x)$$ is the identification map.

Now,

$$\require{AMScd}$$ $$\begin{CD} \mathbb{S^{n}} @>h(x)=x>> \mathbb{R}^{n+1} \backslash \{0\} @>\pi(x)=cl(\{x\})>> (\mathbb{R}^{n+1} \backslash \{0\})/\sim_1 \\ @VV \pi_1(y) = cl(\{y\}) V @. @. \\ \mathbb{S}^n / \sim_2 \\ \end{CD}$$

So we can conclude that $$(\mathbb{R}^{n+1} \backslash \{0\} ) / \sim_1$$ is isomorphic to $$\mathbb{S}^n / \sim_2$$

• Those spaces are not homeomorphic. Do you mean homotopy equivalent? Or is there also an equivalence relation on $\Bbb R^{n+1}\setminus\{0\}$, perhaps identifying $v$ with $\lambda v$ for all $v\in \Bbb R^{n+1}\setminus\{0\}$ and all $\lambda\in\Bbb R\setminus\{0\}$? Jul 25, 2021 at 7:45
• @GregMartin I have modified my question Jul 25, 2021 at 7:54
• Maybe this helps: math.stackexchange.com/q/3999019 Jul 25, 2021 at 8:19
• @PaulFrost I don't think I understand that well.Can you help me out from where I left Jul 25, 2021 at 8:36
• So you're trying to show that $\mathbb{R}P^n \cong S^n/\sim$, right? Jul 25, 2021 at 8:52

$$\mathbb{R}^{n+1}\backslash\{0\}$$ and $$\mathbb RP^n = S^n/\sim$$ are definitely not homeomorphic. As Greg Martin comments, you have to take $$P = (\mathbb{R}^{n+1}\backslash\{0\})/\sim$$ where $$x \sim y$$ if there exists $$\lambda \in \mathbb R$$ such that $$x = \lambda y$$. Then in fact $$P \approx \mathbb RP^n$$.
Your map $$g$$ is a retraction, hence a quotient map. See 2. in my answer to When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map? Thus $$\pi \circ g : \mathbb{R}^{n+1}\backslash\{0\} \to \mathbb RP^n$$ is a quotient map and your theorem applies.
• @Antimony It is correct. But doing it that way, you have to use that $\mathbb{R}^{n+1} \backslash \{0\})/\sim_1$ is Hausdorff. This is true and well-known, but not trivial. In my answer we do not need that fact. Jul 25, 2021 at 23:08