2
$\begingroup$

I am trying to write rigorous proof of the statement in title, since this is really a simple statement but this will check much about my understanding of quotient spaces.

Let $B$ denote open unit disc in $\mathbb{R}^n$ and $D$=closed unit disc in $\mathbb{R}^n$. Then we have disjoint union: $$D=B\cup \partial B.$$ (1) Let $B\cup\{\infty\}$ denote the one-point compactification of $B$; it is homeomorphic to one point compactification of $\mathbb{R}^n$ (since $B\cong \mathbb{R}^n$), which is $S^n$.

(2) Define a map $f:(B\cup \partial B)\rightarrow B\cup\{ \infty\}$ (keeping in mind $B\cup\{\infty\}\cong S^n$) by $$f(x)=x \mbox{ if }x\in B \mbox{ and } f(\partial B)=\infty.$$

It is easy to prove that $f$ is continuous; obviously it is surjective.

(3) Next I want to prove that $f$ is open (or closed) map. This ensures that the topology of $B\cup \{\infty\}$ is the quotient topology through $f$. This I was unable to prove. I was trying to prove it for basic open (closed) sets in the closed disk; but still couldn't succeed. Any hint for this?


There could be different proofs posted on this site of statement in the title; but the above three steps describe a way I was trying to prove myself, and where I get stucked. Here in (3) I was lacking in understanding of quotient topology. For proof of (2), one can use structure of open sets in one-point compactification; am I right?

$\endgroup$
10
  • $\begingroup$ Your title says that you want to describe $S^n$ as a quotient, but your reasoning shows that you want to describe it as the compactification. I feel like the correct way to view the sphere as a quotient, namely by identifying certain points in $R^n$ whould be the identification in $D$ as $x\sim y$ iff $x,y \in \partial D$ . $\endgroup$
    – Nick A.
    Jan 9, 2018 at 8:26
  • $\begingroup$ sphere is one point compactificatin of $R^n$, which is one point compactification of open unit disc. So I restricted to open disc only; am I right? $\endgroup$
    – Beginner
    Jan 9, 2018 at 8:27
  • $\begingroup$ *Related to the approach you have described, maybe try working with the stereographic projection. $\endgroup$
    – Nick A.
    Jan 9, 2018 at 8:28
  • $\begingroup$ Yes, you can compactify the open disc and the whole $\mathbb{R^n}$ without problem ! $\endgroup$
    – Nick A.
    Jan 9, 2018 at 8:29
  • 1
    $\begingroup$ @Beginner yes because closed subset of compact space is compact. $\endgroup$
    – freakish
    Jan 9, 2018 at 9:36

1 Answer 1

1
$\begingroup$

$\mathbb{S}^n \simeq \alpha\mathbb{R}^n$ where $\alpha X$ denotes the Aleksandrov (one-point) compactification of a locally compact space $X$. This is a classical fact proved by the stereographic projection.

A handy fact I will use (proof here):

Theorem If $X$ is locally compact and Hausdorff and $Y$ is compact Hausdorff such that for some $p \in Y$, $X \simeq Y \setminus \{p\}$, then $\alpha X \simeq Y$.

Also if $D^n = \{x \in \mathbb{R}^n : \|x\| \le 1\}$ is the closed unit ball (disk I would reserve for $n=2$, really), then $U=\operatorname{int}(D^n) \simeq \mathbb{R}^n$ and $A = \partial D^n = \mathbb{S}^{n-1}$ is a closed subset, and if $q: D^n \to Y= D^n / A$ is the quotient map that identifies $A$ to a point, then $Y$ is compact and Hausdorff (in the quotient topology) and $q|U$ is a homeomorphism between $U$ and $q[U] \subseteq Y$. So by applying the above theorem to the identified point $q[A]$ of $Y$ we see that $$Y \setminus \{q[A]\} \simeq U \simeq \mathbb{R}^n$$ so

$$Y \simeq \alpha \mathbb{R}^n \simeq \mathbb{S}^n$$ and so the $n$-sphere is a quotient of $D^n$, as claimed.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .