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(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 $\mathbb{S}^n / \sim_2$

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    $\begingroup$ 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 $\lambda\in\Bbb R\setminus\{0\}$? $\endgroup$ Jul 25, 2021 at 7:45
  • $\begingroup$ @GregMartin I have modified my question $\endgroup$
    – Antimony
    Jul 25, 2021 at 7:54
  • $\begingroup$ Maybe this helps: math.stackexchange.com/q/3999019 $\endgroup$
    – Paul Frost
    Jul 25, 2021 at 8:19
  • $\begingroup$ @PaulFrost I don't think I understand that well.Can you help me out from where I left $\endgroup$
    – Antimony
    Jul 25, 2021 at 8:36
  • $\begingroup$ So you're trying to show that $\mathbb{R}P^n \cong S^n/\sim$, right? $\endgroup$
    – jasnee
    Jul 25, 2021 at 8:52

1 Answer 1

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$\mathbb{R}^{n+1}\backslash\{0\}$ 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.

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  • $\begingroup$ can you go through my edit $\endgroup$
    – Antimony
    Jul 25, 2021 at 10:25
  • $\begingroup$ @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. $\endgroup$
    – Paul Frost
    Jul 25, 2021 at 23:08

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