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"Every point of M has a neighborhood homeomorphic to an open subset of $R^n$."

I would like to understand this definition a bit better. With a homeomorphism, I understand it to be a continuous map with a continuous inverse. But why just continuous why not also differentiable? And why do we map to open subsets of $R^n$, why not map to closed sets?

Higgsino
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1 Answers1

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  1. Homeomorphic is enough to have a topological manifold. If, in addition, it is a diffeomorphism, then you've got a smooth manifold. Both are interesting objects themselves.

  2. By neighborhood it is actually meant an open neighborhood, so you want (small enough) open subsets of the manifold to look like $\mathbb{R}^n$, since the properties studied are usually local (continuity, differentiability and so on).

Javi
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    ok, but why open subsets? I don't understand this. – Higgsino Jul 13 '18 at 10:16
  • Local properties are defined in terms of open subsets, in particular, in open balls of $\mathbb{R}^n$. They can usually be defined in closed subsets as well, but that would need some refinement. – Javi Jul 13 '18 at 10:21
  • Another good thing about open subsets is that with they are also open submanifolds. For closed subsets you need some extra hypothesis, as you can see here In general, open sets have more desirable properties in this field. – Javi Jul 13 '18 at 10:26
  • Would it be possible to define a set with a differentiable structure (which a Differetiable Manifold is) that is not necessarily locally diffeomorphic to R^n? – Higgsino Jul 13 '18 at 10:37
  • I guess it is possible to define it in a more abstract way, just like you can do the formal derivative of a polynomial without needing the usual topology. But I don't know if that would be as powerful and useful. – Javi Jul 13 '18 at 11:17