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

Let $H=\{(x_{1}, ..., x_{n+1})∈R^{n+1}|x_{n+1}=1\}$. Prove $R^{n}/S^{n-1}≈S^{n}∪H$? Any kind of help is welcome.

share|cite|improve this question
up vote 1 down vote accepted

Im gonna try to give a proof without stating the explicit homemorphism.

Notice that since $H$ is an hyperplane in $\mathbb{R^{n+1}}$ (it is defined by one equation), then $H \simeq \mathbb{R^n}$.

We can think of $\mathbb{R}^{n}$ as $D^{n} \bigcup_{S^{n-1}} \{ \mathbb{R}^n - B^n \} \ $, where $S^{n-1} = \partial D^{n}$. This notation means seeing $\mathbb{R^n}$ as the n-dimensional disk with the complementary of the n-dimensional ball joint by the sphere $S^{n-1}$.

Now it is easy to see that $D^{n} / \partial D^{n} \simeq S^{n}$ and that $ \{ \mathbb{R}^n - B^n \}) / S^{n-1} \simeq \mathbb{R}^n$

So finally:

\begin{equation} \begin{split} \mathbb{R}^n / S^{n-1} &\simeq (D^{n} \bigcup_{S^{n-1}} \{ \mathbb{R}^n - B^n \}) / S^{n-1}\\ &\simeq D^n / \partial D^n \bigcup_p \{ \mathbb{R}^n - B^n \}) / S^{n-1}\\ &\simeq S^{n} \bigcup_p \mathbb{R}^n \end{split} \end{equation}

Where $p$ represents the point to which the quotient map sends $S^{n-1}$.

share|cite|improve this answer

I think what follows should work, but I haven't written all the details.

Set $X=\mathbf R^n/S^{n-1}$ and $Y=S^n\cup H$. The space $Y$ is connected. If you remove the point $\{Q\}=S^n\cap H$, it's not anymore. Therefore the pre-image of $Q$ by any homeomorphism $f:X\to Y$ is a point $P\in X$ such that $X\setminus\{P\}$ is not connected. There is an obvious guess as to which point $P$ that must be: the image of $S^{n-1}$ in the quotient $X$.

Now look at the connected components. $X\setminus\{P\}$ has two connected components, one of which is the unit $n$-ball $B^n$ with its boundary collapsed into the point $P$, minus $P$; that's homeomorphic to $S^n\setminus\{Q\}$, which is one of the connected components of $Y\setminus\{Q\}$. Then you can show that the other connected component of $X\setminus\{P\}$, call it $A\setminus\{P\}$, is homeomorphic to the other connected components of $Y\setminus\{Q\}$, which is $H\setminus\{Q\}$. Notice that you can map $\{x\in\mathbf R^n:\|x\|>1\}$ homeomorphically into $B^n\setminus\{0\}$, so when you take the quotient by $S^{n-1}$ you end up sending $A\setminus\{P\}$ into an $n$-sphere minus $P$ and another point. Then use the stereographic projection to show that it's $\simeq H\setminus\{Q\}$.

If you've done all that, you should be able to piece it all together into a homeomorphism between $X$ and $Y$.

share|cite|improve this answer

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.