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Is there an easy way to calculate the homology groups

$$H_k(\mathbb{R}\setminus \{0\}), k\geq 0$$

I was able to calculate $H_k(\mathbb{R}^m\setminus \{0\}), m > 1$ because we have a homeomorphism onto the sphere $S^{m-1}$ whose homology groups I know, but we do not have $\mathbb{R}\setminus \{0\} \cong S^0 = \{-1,1\}$ so for $m=1$ this approach fails.

Of course, $H_0(\mathbb{R}\setminus \{0\}) \cong \mathbb{Z} \oplus \mathbb{Z}$ since there are two path components.

  • Note that $\Bbb R\backslash {0}\simeq (-\infty,0)\sqcup (0,\infty)$. Hence, $H_k(\Bbb R\backslash {0})=H_k(-\infty,0)\oplus H_k(0,\infty)=0\oplus 0$ text for $k>0$. As convex sets are contractible. – Sumanta Jun 24 '20 at 10:03
  • Let $m=1$? Anyway $\Bbb R\setminus{0}$ is homotopy equivalent (not homeomorphic) to two points. – Angina Seng Jun 24 '20 at 10:03

1 Answers1

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Hint: $\mathbb{R}\backslash\{0\}$ deformation retracts to two discrete points.

Also, $\mathbb{R}^m\backslash\{0\}$ is not homeomorphic to a sphere (again, deformation retract).

Daniel Robert-Nicoud
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  • How is $\mathbb{R}^m \setminus {0}$ not homeomorphic to the sphere? Don't we have $\mathbb{R}^m \setminus {(0, \dots, 0,1}$ is homeomorphic to the sphere via stereographic projection? –  Jun 24 '20 at 10:10
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    @user745578 $\Bbb R^m\backslash{0}$ is non-compact, but sphere is always is compact. So, $\Bbb R^m\backslash{0}$ is not homeomorphic to sphere as continuous image of a compact set is compact. But $\Bbb S^m-{\text{North pole}}$ is homeomorphic to $\Bbb R^m$ via stereographic projection. – Sumanta Jun 24 '20 at 10:19