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If $M$ is a $k$-dimensional manifold, show that every point of $M$ has a neighborhood homeomorphic to all of $\mathbb{R}^k$ . Therefore, charts can always be chosen with all of Euclidean space as their co-domains.

I'm confused because I thought that the first sentence was part of the definition of a $k$-dimensional manifold. I don't understand what I'm being asked to prove.

user394412
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    how did your textbook define $k$-dimensional manifolds exactly? sometimes the locally euclidean property is interpreted as: every point has a neighbourhood homeomorphic to all $\mathbb{R}^k$, and sometimes as having a neighbourhood homeomorphic to an open subset of $\mathbb{R}^k$. I'm assuming it's the latter – Thiago Nov 30 '17 at 21:34
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    of course, they're equivalent statements because of this fact you want to prove – Thiago Nov 30 '17 at 21:35
  • It says, "An $n$-dimensional topological manifold $M$ is a Hausdorff, topological space with a countable basis for its topology, and which is locally homemorphic to $\mathbb{R}^n$. This means for every $p\in M$, there exists open neighborhood $U$ of $p$ and a homeomorphism $h:U\rightarrow U'$ where $U'\subseteq\mathbb{R}^n$ is open." – user394412 Nov 30 '17 at 21:40
  • So, yes the latter. So, how do I prove the two are equivalent? – user394412 Nov 30 '17 at 21:41
  • i think the stantdard way to go around this is to prove that the open ball in $\mathbb{R}^k$ is homeomorphic to the whole space. since $U'$ is open you can restrict the homeomorphism to an open ball, which then follows that this point in the manifold has a neighbourhood of the type you want – Thiago Nov 30 '17 at 21:48
  • I think you are on to something because the question before this one was "suppose $e_{a}^{k}\subset \mathbb{R}^k$ is the open ball $e_{a}^{k} = { x\in \mathbb{R}^k | |x|^2<a}$. Show that $e_{a}^{k}$ is homeomorphic to $\mathbb{R}^k$. – user394412 Nov 30 '17 at 21:52

2 Answers2

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It is obvious that if the definition ( homeomorphic to $\mathbb{R}^k$ ) implies your definition.

Now suppose for every point $p \in M$, there exists homeomorphism $h \colon U \to U'$ where $U \subseteq M$ and $U' \subseteq \mathbb{R}^k$ are open. Take an open ball $V'$ centered at $h(p)$ and $V' \subset U'$. The restriction $h|_{h^{-1}(V')} \colon h^{-1}(V') \to V'$ is an homeomorphism. Then we know an open ball $V'$ is homeomorphic to $\mathbb{R}^k$.

user1101010
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  • How do you prove an open ball is homeomorphic to $\mathbb{R}^k$ – user394412 Nov 30 '17 at 22:00
  • There are other answers to this question on SE, but I was wondering if there was one that used more vector bundle/differential topology. – user394412 Nov 30 '17 at 22:07
  • The homeomorphism can be constructed explicitly. Take a look at this answer https://math.stackexchange.com/questions/1072741/is-an-open-n-ball-homeomorphic-to-mathbbrn. Here you probably need to scale your ball and scaling is a homeomorphic operation. – user1101010 Nov 30 '17 at 22:07
  • While I have you here, do these two relate to how I could prove: Suppose $N$ is a non-empty, $n$-dimensional manifold, with $k \leq n$. Show there exists an embedding of $\mathbb{R}^k$ into $N$. – user394412 Nov 30 '17 at 22:10
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Let $p\in M$,

then by definition of manifold there exist co-ordinate chart $(\phi,U)$ around $p$. Hence $\phi$ is homeomorphism from $U$ to some open subset $V$ of $\mathbb R^n$. As $V$ is open there exist $r>0$ such that $B(\phi(p),r) \subseteq V$.

Now let $W=\phi^{-1}(B(\phi(p),r))$.

As $W$ is homeomorphic to $B(\phi(p),r)$ and $B(\phi(p),r)$ is homeomorphic to $\mathbb R^n$, $W $ is homeomorphic to $\mathbb R^n$.

Mayuresh L
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