Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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.


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.

share|cite|improve this answer

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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.