# How to show that the unit sphere is a topological manifold?

Sorry for this basic question, but I´m not sure of something. I want to see one example. The definition of a n-manifold is a Hausdorff space, such that each point has an open neighborhood homeomorphic to the open n-dimensional disc. How can i prove that $$S^n = \left\{ {x \in \mathbf{R}^{n + 1} :\left| x \right| = 1} \right\}$$ is an n-manifold

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Try for $n=1$ first. Then $n=2$. – lhf Sep 1 '11 at 1:21

Alternatively, you can use the local form of a submersion to proove tha $S^n$ is a submanifold of $\mathbb{R}^{n+1}$

http://en.wikipedia.org/wiki/Submersion_%28mathematics%29

$f:\mathbb{R}^{n+1}\rightarrow\mathbb{R}$, $f(x_1,...,x_{n+1})=x_1^2+...+x_{n+1}^2$ is a $C^\infty$-submersion on $S^n=f^{-1}(1)$.

Another nice method is to use steregraphic projections

http://en.wikipedia.org/wiki/Stereographic_projection

$S^n=U_N\cup U_S$ where $U_N=S^n\setminus\{(0,...,0,1)\}$ and $U_S=S^n\setminus\{(0,...,0,-1)\}$ are opens for the induced topology of $\mathbb{R}^{n+1}$ on $S^n$.

One can check that : $\varphi_N : U_N\rightarrow\mathbb{R}^n$, $\varphi_N(x_1,...,x_{n+1})=\big(\dfrac{x_1}{1-x_{n+1}},...,\dfrac{x_n}{1-x_{n+1}}\big)$ and $\varphi_S : U_S\rightarrow\mathbb{R}^n$, $\varphi_S(x_1,...,x_{n+1})=\big(\dfrac{x_1}{1+x_{n+1}},...,\dfrac{x_n}{1+x_{n+1}}\big)$ are homomorphisms.

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One of the things that has always mystified me about this approach is how one actually goes about finding the inverses of the projections. Usually, the inverses are given and it can be shown through composition that the supplied inverse actually works. I guess the algebra of finding an inverse for a map between multidimensional spaces eludes me. – ItsNotObvious Sep 1 '11 at 13:21
@3Shpere. I had always thought that inverses are computed, not given (by some sort of "god"?). In both cases, the stereographic projection and my answer, draw a picture and you'll be able to compute inverses. With the stereographic projection, Wikipedia provides you the picture. If you understand how the formula amine writes for it is found from the picture, then you can also compute the formula for its inverse. – a.r. Sep 1 '11 at 14:05

$S^n$ is a Hausdorff space because it's a subspace of a Hausdorff space ($\mathbb{R}^{n+1}$).

As for an open neighbourhood for every point homeomorphic to some open subset of $\mathbb{R}^n$ (you don't need this open subset to be the open $n$-dimensional disk, but , in this case, you naturally get a disk), you can assume the point is the North pole $(0,\dots, 0, 1) \in S^n$. (Why? -Well... Think about it. :-) ). Then, for instance, consider the open North hemisphere:

$$S^n_+ = \left\{ (x_1, \dots , x_n , x_{n+1}) \in S^n \ \ \vert \ \ x_{n+1}> 0 \right\}$$

with the natural projection onto the $x_{n+1} = 0$ hyperplane of $\mathbb{R}^{n+1}$, which happens to be homeomorphic to $\mathbb{R}^n$:

$$p: S^n_+ \longrightarrow \mathbb{R}^n \ , \qquad p(x_1,\dots , x_n, x_{n+1} )= (x_1, \dots , x_n)$$

Exercises:

1. Why is $S^n_+$ an open neighbourhood of the North pole inside the sphere $S^n$?
2. Who is the image of $p$?
3. Find an inverse for $p$, from its image to $S^n_+$.

Hint. As lhf suggests you, try first with $n=1, 2$. Draw some pictures.

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