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I feel like the proof is very similar to the one in this link:

$S^n \backslash S^m $ homotopy equivalent to $ S^{n-m-1} $

But I could not spot out where will be the difference? could anyone help me with this, please?

Thanks!

N. Owad
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Intuition
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  • yes it is the join @freakish yes I know this property but then how can I use it? , Also can not I proceed by induction on n as I have already proved the statement for n=m=1? – Intuition Sep 13 '19 at 07:54
  • Do you know also how can I prove the property you mentioned? @freakish and is S^n a CW complex?? – Intuition Sep 13 '19 at 07:56

2 Answers2

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In fact $S^m *S^n$ and $S^{m+n+1}$ are homeomorphic. This is proved as in https://math.stackexchange.com/q/3354659:

$S^m \approx S^0 * \ldots * S^0$ witk $m+1$ factors, $S^n \approx S^0 * \ldots * S^0$ witk $n+1$ factors, thus $S^m * S^n \approx S^0 * \ldots * S^0$ witk $m+n+2$ factors, the latter being homeomorphic to $S^{m+n+1}$.

Paul Frost
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  • I want to prove that they are homotopy equivalent only as I do not understand the proof of homeomorphism ..... so I am behaving as if I do not know this homeomorphism – Intuition Sep 13 '19 at 07:58
  • Can I proceed by induction on n as I have already proved the problem for $m=n=1$? – Intuition Sep 13 '19 at 08:00
  • The proof of $S^m \approx S^0 * \ldots * S^0$ is "hidden" inductive. I do not know any better proof than that in my answer. – Paul Frost Sep 13 '19 at 08:11
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Here are some facts that you can assemble:

  1. For any CW complexes $A,B$ the join $A*B$ is homotopy equivalent to $\Sigma(A\wedge B)$
  2. For any $n,m$ we have a homeomorphism $S^n\wedge S^m\simeq S^{n+m}$
  3. For any $n$ we have a homeomorphism $\Sigma S^n\simeq S^{n+1}$
  4. If $A\simeq B$ then $\Sigma A\simeq \Sigma B$
  5. Spheres are CW complexes (the simpliest there are)

Can you combine them together to get the result?

freakish
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  • So we can not proceed by induction on n .... correct? as I have proved it for the case $m=n=1$ – Intuition Sep 13 '19 at 08:14
  • I'm not saying this. Maybe we can but I wouldn't know how. This just looked like a simpliest way too me. – freakish Sep 13 '19 at 08:14
  • from where in Hatcher you get property 4? – Intuition Sep 13 '19 at 08:17
  • I'm not sure if property 4 is in Hatcher. But you can prove it yourself. First of all $\Sigma$ also acts on maps, turning $f:X\to Y$ into $\Sigma f:\Sigma X\to \Sigma Y$ via $\Sigma f([x,t])=[f(x), t]$. And this is functorial, meaning $\Sigma (f\circ g)=\Sigma f\circ \Sigma g$. And so (as with any functor) it maps isomorphisms to isomorphisms. In this case "isomorphism" is a "pointed homeomorphism". – freakish Sep 13 '19 at 09:16