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Consider the following definitions (correct me if I'm wrong):

Definition (type 1) An n-manifold is a Hausdorff space that is locally homeomorphic to $\mathbb R^n$.

Definition (type 2) An n-manifold is a set together with a set of charts $(U, \kappa)$ where $\kappa: U\rightarrow \mathbb R^n$ is injective with open image. The set of charts shall form an atlas, meaning that all sets $U$ cover $M$ and all changes $\kappa_1\circ\kappa_2^{-1}$ must be continuous.

$M$ of type 2 can be equipped with a topology by taking $\{S\subseteq M | \kappa(S\cap U) \text{ open }\forall (U, \kappa) \text{ atlas}\}$, which then becomes a MF of type 1 (because every open subset of $\mathbb R^n$ is again locally homeomorphic to $\mathbb R^n$), and conversely, for every MF of type 1 we can choose the local homeomorphisms as an atlas. I will skip the proof of compatability of the resulting topologies/atlases.

I now want to prove:

Lemma The topology of a MF is second countable if and only if the corresponding atlas is countable.

If we have a countable atlas, I have a pretty clear adea (just transport the countable base of $\mathbb R^n$ through the $\kappa_i$ onto $M$ and show it's actually a base). However, given a countable base, how can I construct a countable Atlas? I know it's locally homeomorphic to $\mathbb R^n$, but in general, there are/might be uncountably many homeomorphisms, so we cannot just take the set of all those.

Lukas Juhrich
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2 Answers2

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Let $\mathcal{V} = \lbrace V_n \rbrace_{n \in \mathbb{N}}$ be a countable base. Take any atlas $(\kappa_\alpha, U_\alpha)_{\alpha\in A}$ of $M$. Let $K = \lbrace n \in \mathbb{N} \mid V_n \text{ is contained in some } U_\alpha \rbrace$; this is a countable set. For $k \in K$ choose $\alpha_k$ such that $V_k \subset U_{\alpha_k}$.

Let $x \in M$. There exists $\alpha \in A$ such that $x \in U_\alpha$. Since $\mathcal{V}$ is a base, there exists $k \in \mathbb{N}$ such that $x \in V_k \subset U_\alpha$. But then $x \in U_{\alpha_k}$. Therefore the $(U_{\alpha_k})_{k \in K}$ cover $M$ which gives you a countable atlas.

Paul Frost
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$M$ is second-countable implies $M$ is Lindelöf . Thus, you can take the cover $\mathcal{U}$ of $X$ given by $(U,\varphi)$ and take a countable sub-cover. This gives a countable atlas.

David
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