$S^n$ is not homeomorphic to $S^{n-1}$

Question

My geometric intuition is clear about the fact that $S^n \ncong S^{n-1}$ $\forall n \geq 2$ . It's very easy to do it for lower dimensions using simple Analysis arguments. ($S^n$ is the n-sphere in $\Bbb R^{n+1}$)

But I really want to learn an Elementary proof for the general case (i.e. $\forall n \geq 2$) without using tools from Algebraic Topology, using basic General Topology arguments.

How to come up with an Elemenatry proof for the fact?

The question linked with this one, I have already visited. It accepts an answer that gives a wiki-link .As I mentioned in the question I was looking for an elementary proof using minimal machineries and hence I've posted this question!

@Wojowu please consider writing an answer. – Aug 15 '18 at 10:54 — , Aug 15 '18 at 10:54

score 0 · Answer 1 · answered Aug 15 '18 at 11:00

0

One could use the cellular approximation theorem: any map $f:S^{n-1}\to S^n$ is nullhomotopic, but no homeomorphism $h:S^n\to S^n$ is nullhomotopic. Thus the two domains $S^{n-1}$ and $S^n$ of $f$ and $h$ cannot be homeomorphic.

answered Aug 15 '18 at 11:00

Arthur

199,419

$S^n$ is not homeomorphic to $S^{n-1}$

1 Answers1