I have come across two different defintions of a topological manifold -

Def 1: A topological manifold of dimension n is a second-countable Hausdorff space M such that for all $p\in $M, there exists open neighborhood $U$ at $p$ and a homeomorphism $x:U\to x(U)\subseteq \mathbb{R}^{n}$

Def 2: A topological manifold M of dim. n is a Hausdorff topological space with an open cover $C$ with countable elements $U_i\in C$ and a collection of homeomorphism $\phi_i:U_i\to \phi_i(U)\subseteq\mathbb{R}^{n}$ where $\phi_i(U)$ is an open subset in $\mathbb{R}^{n}$.

  1. Are these two equivalent? If not, which one of them is the correct one (if any of them is)?

  2. Is second-countable same as to have an open cover $C$ with countable elements?

  3. Does the target of chart map ($x/\phi$) need to be an open subset in $\mathbb{R}^{n}$?

  • $\begingroup$ All the neighborhoods in definition 1 are the open cover from definition 2, and an open cover like in definition 2 contains neighborhoods of all points, so it also fulfills definition 1. $\endgroup$ – Vercassivelaunos Oct 15 '20 at 16:40
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    $\begingroup$ If we do not require $x(U)$ to be open in $\mathbb R^n$, then all subsets $M \subset \mathbb R^n$ would be manifolds (take the identity on $M$ as a universal chart around all $p$). This doesn't make much sense. $\endgroup$ – Paul Frost Oct 15 '20 at 23:39
  1. Definition 1 is missing (or assuming) the requirement that $x(U)$ be open. With that addition, both definitions are equivalent.

  2. Yes. This is because $\mathbb{R}^n$ is itself second countable. To show a countable cover implies a second countable manifold, choose a countable basis $\mathcal{B}$ for $\mathbb{R}^n$ (e.g. balls of rational center/radii), and let $\mathcal{B}'=\{\varphi_i^{-1}(B):B\in\mathcal{B},i\in\mathbb{N}\}$. I claim this is a countable basis for $M$.

  3. Yes, removing that requirement allows for objects which are not conventionally thought of as manifolds, such as graphs or $\mathbb{Q}\subset\mathbb{R}$.

  • $\begingroup$ Does paracompact + Hausdorff imply second countability? $\endgroup$ – aneet kumar Oct 16 '20 at 11:44
  • $\begingroup$ @aneetkumar Not quite. Paracompact + Hausdorff + locally Euclidean + countably many connected components $\implies$ second countable, but all four are necessary if I'm not mistaken. Also see here. $\endgroup$ – Kajelad Oct 16 '20 at 13:21

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