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

analysisgeneral-topology

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

Best Answer

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