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Let $U$ be a open subset of $\mathbb{R}^n$ where $n\geq 2$ and let $x\in U$. Show that $H_{n-1}(U\setminus\{x\})$ is not the trivial group.

What I know is that $H_{n-1}(\mathbb{S}^{n-1})=\mathbb{Z}$ and that $U$ is homeomorphic to an open subset of the sphere $\mathbb{S}^{n}$. Can these two facts help?

UserA
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  • Do you know that homology groups are invariant under deformation retracts? – Steve D May 14 '18 at 19:43
  • @SteveD of course – UserA May 14 '18 at 19:44
  • Well, that was my hint as well... Can you do the case where $U$ is an open unit ball, with center $x$? – Steve D May 14 '18 at 19:45
  • @SteveD yes, in this case $U$ deformation retracts to $S^{n-1}$ right? – UserA May 14 '18 at 19:47
  • @SteveD but for an arbitrary open set $U$, $U\setminus{x}$ deformation retracts to what exactly? – UserA May 14 '18 at 19:49
  • Well, you can restrict to the connected case, right, since the homology group is just the direct sum of the homology groups of the components. Now since $U$ is open, it contains a ball around every point... – Steve D May 14 '18 at 20:17

1 Answers1

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Let me outline a slightly different solution.

If $A=U$ and $B=\mathbb{R}^n-\{x\}$, then

  • $A\cup B=\mathbb{R}^n$
  • $B$ deformation retracts to $S^{n-1}$
  • $A\cap B=U-\{x\}$

Now what does Mayer-Vietoris tell you about $H_{n-1}$?

Steve D
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  • What was your original answer though? – UserA May 15 '18 at 06:50
  • Also, intuitively what does it mean that the n-1 homology group of a connected open set minus a point is at least $\mathbb{Z}$? – UserA May 15 '18 at 07:02
  • Intuitively, there is homology contained in the punctured open ball around the missing point, and the dimension is high enough that it won't be canceled by anything else. That is, there's a sphere in that punctured open ball, and it is not the boundary of any subcomplex that U contains (or even that the whole Euclidean space contains). – Steve D May 15 '18 at 12:19
  • My original approach was via excision/local homology, but I realized that was more complicated than it needed to be. – Steve D May 15 '18 at 12:26