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I am reading Differential forms and applications by Manfredo P.do Carmo (springer), and I think that the definition of differentiable manifold given enter image description here

applies to any set that is the image by $f$ (an injective function) of an open subset $U$ of ${\bf R}^n$ with the manifold being the set $f(U)$ and the family $[(U,f)]$, however this includes sets that might not be even connected. Am I misunderstanding the definition?

Alekos Robotis
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Nomix Mar Rue
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    You’re correct. A manifold need not be connected. The disjoint union of any number of manifolds is a manifold. Each connected component is a manifold. – Deane Jun 16 '25 at 19:41
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    Correct, $M$ need not be connected. But regarding terminology, I believe Do Carmo uses smooth and differentiable interchangeably to mean $C^{\infty}$. – peek-a-boo Jun 16 '25 at 19:41
  • @Deane: Hmmm, I would have thought that the majority (but not all) authors don't consider the disjoint union of a $1$-dimensional manifold and a $2$-dimensional manifold to be a manifold. There was some discussion of this issue in the past in the thread. – Joe Jun 16 '25 at 20:34
  • @Joe I’m sure Deane implicitly meant the connected components having the same dimension. But even if not, as long as one is consistent in the terminology, this shouldn’t be an earth-shattering issue. As an aside, Dieudonne defines a chart to be a triple $(U,\phi, n)$, and he calls $n$ the dimension of the chart. If two charts contain a point $p$, then the corresponding $n$’s are the same, and he calls this common value the dimension of the manifold at $p$. If this is the same constant at all points, he calls the manifold pure. – peek-a-boo Jun 16 '25 at 21:57
  • (I don’t like having non-constant dimensions.. heck in practical calculations, I, and probably almost everyone else, treat different connected components as different manifolds, but no doubt it’s occasionally convenient to allow non-connected ones :) – peek-a-boo Jun 16 '25 at 21:57
  • @joe, you’re right. My preference is to assume all of the connected components have the same dimension. It is possible however that in some specialized situations, you don’t want to assume that. You would say so explicitly, to warn the reader. – Deane Jun 17 '25 at 00:36

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You understood correctly: any open subset of $\bf R^n$ (connected or not) is an $n$-dimensional manifold, and it is smooth since in the case of only one index $\alpha$, condition 2) is vacuously true.

Anne Bauval
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