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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.

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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 $S^{n-1}$.

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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 $Y$.

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