# What is the definition of a topological manifold

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 $$\mathbb{R}^{n}$$?

• 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. – Vercassivelaunos Oct 15 '20 at 16:40
• 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. – Paul Frost Oct 15 '20 at 23:39

## 1 Answer

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 $$\mathbb{Q}\subset\mathbb{R}$$.

• Does paracompact + Hausdorff imply second countability? – aneet kumar Oct 16 '20 at 11:44
• @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. – Kajelad Oct 16 '20 at 13:21