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Let $M^n$ be a smooth, connected $n$-manifold. By definition, this means that every point $p\in M$ has a neighborhood $U$ that is homeomorphic to some open subset of $\mathbb{R}^n$. Now let us assume we only know that for some point $p\in M$ we know that it is in a neighborhood homeomorphic to $\mathbb R^n$ and for the other points we just know that they have a neighborhood that is homeomorphic to $\mathbb R^k$ for some $k\in \mathbb N$. I think that we can show using path-connectedness that $k$ is always equal to $n$ but I don't know how exactly.

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  • $\begingroup$ Would that be guaranteed by the transition maps? (they are bijective homeomorphisms after all.) $\endgroup$ – Justin Benfield Mar 14 '16 at 16:29
  • $\begingroup$ I think you should be clearer about your assumptions on $M$. $\endgroup$ – Friedrich Philipp Mar 14 '16 at 16:31
  • $\begingroup$ How much do you know about the rigorous definition of a smooth manifold? The transitions maps are the composites given by $\alpha (\beta^{-1}(x))$ where $x$ is an element of $\mathbb{R}^k$. They lift such an $x$ into the manifold, then project it back down to $\mathbb{R}^k$ via a different coordinate chart (and they are defined on their overlap in $M$) $\endgroup$ – Justin Benfield Mar 14 '16 at 16:37
  • $\begingroup$ @JustinBenfield I have understood your point just a few seconds before your second comment and therefore deleted my comment, I'm sorry for that. Your argument shows that dimension is well defined on chain connected charts. However all connected spaces are chain connected w.r.t. open coverings, hence this is sufficient. Thanks! $\endgroup$ – Sebastian Bechtel Mar 14 '16 at 16:41
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Let $\sim$ be the equivalence relation which states that $x \sim y$ if there exists charts around them which go to the same dimension (they are not necessarily on a same chart. This only states that each one has a chart around them that both have the same dimension). This is clearly an equivalence relation, and each class is clearly open. By connectedness, we must have only one class. By invariance of domain (or by looking at the derivatives if you are assuming smoothness), there can be no two charts around the same point which go to different dimensions. Hence, the charts only go to a fixed dimension.

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