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On page 35 Hatcher writes that $\mathbb{R}^n - \{x \}$ is homeomorphic to $ S^{n-1} \times \mathbb{R}$. I know e.g. $\mathbb{R}^2 - \{x\}$ is homotopy equivalent to $S^1$ and also to $S^1 \times \mathbb{R}$. I don't see though how they are homeomorphic. Is this a typo? Thanks for your help!

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They are homeomorphic - think about $\mathbb{R}^2$ case, there's an obvious map from $\mathbb{R}^2 - 0 \to S^1 \times (0, \infty)$ (hint: polar coordinates), and it's easy to show that's a homeomorphism. Same thing happens for higher dimensions. –  Soarer Sep 1 '11 at 9:41
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2 Answers 2

up vote 13 down vote accepted

After a translation we can assume $x = 0$ and consider the function $f(\xi, t) = e^{t}\xi$ for $(\xi,t) \in S^{n-1} \times \mathbb{R}$. This is obviously a bijection and it is not hard to show that it is a diffeomorphism.


Later:

As symbo'leon pointed out in the other answer, you can write down the inverse explicitly: For $x \in \mathbb{R}^{n} \smallsetminus \{0\}$ put $\varphi(x) = \left( \frac{x}{\|x\|},\, \log{\|x\|} \right)$ and check that $f \circ \varphi = \operatorname{id}_{\mathbb{R}^{n} \smallsetminus \{x\}}$ and $\varphi \circ f = \operatorname{id}_{S^{n-1} \times \mathbb{R}}$ and that both maps are smooth.

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An explicit diffeomorphism is: for $y\in\mathbb{R}^{n}\backslash\{x\}$ set

$$\varphi(y)=(\frac{y-x}{\left\Vert y-x\right\Vert },\ln\left\Vert y-x\right\Vert )$$

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You could have said that this is essentially the inverse of what I wrote. –  t.b. Sep 1 '11 at 11:43
    
@ THeo: OK, agreed :) –  t22 Sep 1 '11 at 11:58
    
Okay, very good then :) –  t.b. Sep 1 '11 at 12:01
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