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I was trying to show that "Every manifold has a basis of coordinate balls". My approach was like this: Given a manifold $M$, it's well-known that for every point $\mathbf{x}\in M$ there exists a neighborhood of $\mathbf{x}$ homeomorphic to an open ball in $\mathbb{R}^n$. At first, I thought that I could obtain this basis using the set of all balls in $\mathbb{R}^n$ that are homeomorphic to some neighborhood of every point in $M$, but this became confusing to me. Is this the right approach? If yes, how can I go on?

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I forgot to ask something else... I have to show that there is a coordinate ball in S^n whose closure is equal to all S^n. The open ball of radius 1 is what we're looking for, isn't? –  Br09 May 14 '12 at 23:31
    
You can edit your question –  matgaio May 14 '12 at 23:31
    
Perhaps, you can show that the coordinate charts constitute a basis for the topology of the manifold. Remember that a set of open sets $\beta$ is a basis if any open set of $M$ can be described as union of elements of this $\beta$ –  matgaio May 14 '12 at 23:35
    
Yes, the question added is that. –  matgaio May 14 '12 at 23:36

1 Answer 1

up vote 1 down vote accepted

Hints:

Question 1:

Yes, that's probably the right approach. Here's a generous hint to help you get unstuck: Use the definition of a basis for a topology, in conjunction with some fact about restriction of coordinate charts.

Question 2:

As for your second question, what are the standard coordinate charts on $\mathbb{S}^n$? To be concrete, work with $n=2$. What is the closure of such a chart?


Answers follow.

Question 1:

Let $U$ be an open set in $M$. Let $x\in U$. About $x$ there exists some open neighborhood $C$ which admits a diffeomorphism $\phi_C$ to some open neighborhood of the origin of $\mathbb{R}^n$. Let $B$ denote the component of $U\cap C$ which contains $x$. Then $\phi_C|_B$ is a diffeomorphism onto its image, which is again a connected open neighborhood of the origin of $\mathbb{R}^n$. (By possibly postcomposing with a translation, we may WLOG assume that $\phi_C(x) = 0$.) Now take a small open ball $S$ about $0$ in $\mathbb{R}^n$ which has $0\in S\subseteq \phi_C(B)$. Its preimage $\phi_C^{-1}(S)$ is diffeomorphic to $\mathbb{R}^n$ via $\phi_C|_{\phi_C^{-1}S}\circ \varphi$, where $\varphi$ is a diffeomorphism between an $\epsilon$-ball and all of $\mathbb{R}^n$, so $\phi_C^{-1}(S)$ together with the restriction of $\phi_C$ is a coordinate chart. Finally we have $x\in\phi_C^{-1}(S)\subseteq U$.

Therefore, for every open set $U$ of $M$ and for every point $x\in U$ we may find a coordinate chart $S$ which has $x\in S\subseteq U$, so coordinate neighborhoods form a basis for the topology of $M$.

Question 2:

Use stereographic projection to get coordinates. Equivalently, $\mathbb{S}^n$ is the one-point compactification of $\mathbb{R}^n$.

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Is S really contained in B (line 6)? You meant S contained in the image of B via (phi), no? –  Br09 May 16 '12 at 22:26
    
@ItamarSales Yes, thank you. After all, the whole point of $\phi$ is to let one identify $C$ and $\mathbb{R}^n$, so that is the cause of the mistake :) –  Neal May 18 '12 at 0:49

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