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Let $M$ be a manifold with a $C^k$ atlas, $k\geq 1$. I was asked to show that the dimension $\dim M$ is well-defined on each connected component of $M$. I'm not sure if I misunderstood the meaning of being well-defined in question. To my knowledge, what I need to do is to prove any two coordinate domains must be homeomorphic to opens set in the same Euclidean space. And for connected components, it occurs to me that each point of $M$ belongs to a uniquely determined connected component of $M$, that is, $M$ is a disjoint union of connected components. Now I'm in trouble associating connectedness with homeomorphisms. Is there any useful topological property that I haven't considered, please? Thank you.

Arctic Char
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Boar
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    What is your definition of dimension of manifold? – Arctic Char Sep 21 '20 at 06:07
  • The dimension of a manifold is defined to be the common value, if any, of the dimensions of the Euclidean spaces that all coordinate charts are mapped into. – Boar Sep 21 '20 at 06:19
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    Within each connected component there is a sequence of charts connecting any two charts such that any two consecutive ones have a point in common. Any two charts that have a point in common must have the same dimension. – Conifold Sep 21 '20 at 06:32
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    Another related post here – Arctic Char Sep 21 '20 at 06:38
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    This issue only arises if you (or your instructor) are following Tu's textbook on manifolds. Tu explicitly allows different connected components of a manifold to have different dimension. If you are following Tu's book, then the linked question is not a duplicate. A hint for a solution in this case is: Given $n$, define a subset $U_n\subset M$ consisting of points with local dimension $n$. Then argue that each $U_n$ is open and $U_n\cap U_m=\emptyset$ if $n\ne m$. Lastly, use the connectedness assumption. – Moishe Kohan Sep 22 '20 at 17:19

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