Smooth Manifold Chart Lemma

[cianfa72]

TL;DR

About the proof of Lemma 1.35 - Smooth Manifold Chart Lemma in J. Lee book "Introduction to Smooth Manifolds"

I've a doubt regarding Lemma 1.35 (Smooth Manifold Chart Lemma) from J. Lee "Introduction to Smooth Manifolds"

Lemma 1.35 (Smooth Manifold Chart Lemma)

Let $M$ be a set, and suppose we are given a collection $\{U_{\alpha} \}$ of subsets of $M$ together with maps $\varphi_{\alpha}: U_{\alpha} \to \mathbb R^n$ such that the following properties are satisfied:
(i) For each $\alpha, \varphi_{\alpha}$ is a bijection between $U_{\alpha}$ and an open subset $\varphi_{\alpha}(U_{\alpha}) \subseteq \mathbb R^n$.
(ii) For each $\alpha$ and $\beta$, the sets $\varphi_{\alpha}(U_{\alpha} \cap U_{\beta})$ and are open in $\mathbb R^n$.
(iii) Whenever $U_{\alpha} \cap U_{\beta} \neq 0$, the map $\varphi_{\beta} \circ \varphi_{\alpha}^{-1}: \varphi_{\alpha}(U_{\alpha} \cap U_{\beta}) \to \varphi_{\beta}(U_{\alpha} \cap U_{\beta})$ is smoot.
(iv) Countably many of the sets $U_{\alpha}$ cover $M$.
(v) Whenever $p,q$ are distinct points in $M$, either there exists some $U_{\alpha}$ containing both $p$ and $q$ or there exist disjoint sets $U_{\alpha},U_{\beta}$ with $p \in U_{\alpha}$ and $q \in U_{\beta}$.

Then $M$ has a unique smooth manifold structure such that each $(U_{\alpha}, \varphi_{\alpha})$ is a smooth chart.

The proof claims that Hausdorff property follows from v). However v) includes the case where both $p$ and $q$ are included in the same $U_{\alpha}$, i.e. their neighborhoods are not disjoint though.

 

[pasmith]

My edition of Lee's Introduction to Smooth Manifolds (Springer, 2003) does not contain a Lemma 1.35; the lemma you cite appears as 1.23 (Smooth Manifold Construction Lemma).

$\{U_\alpha\}$ is neither the topology on $M$ nor a basis for it. The topology on $M$ is defined by the basis $$\bigcup_{\alpha}\{ \phi_\alpha^{-1}(V) : V\mbox{ open in \mathbb{R}^n}\}.$$ Note that this requires us to prove that each $U_\alpha$ is open in this topology.

Let $p \in M$ and $q \in M$ be distinct. If there does not exist $\alpha$ such that $\{p, q\}\subset U_\alpha$ then by (v) there exist disjoint open $U_\alpha \ni p$ and $U_\beta \ni q$ and we are done.
Otherwise, let $(U, \phi)$ be the chart whose domain contains both $p$ and $q$. If $\phi(p) = \phi(q)$ then $p = q$ by bijectivity of $\phi$, and if $\phi(p) \neq \phi(q)$ then (by Hausdorffness of $\mathbb{R}^n$) there exist disjoint open $V_p \subset \phi(U)$ and $V_q \subset \phi(U)$ such that $\phi(p) \in V_p$ and $\phi(q) \in V_q$. The preimages $\phi^{-1}(V_p) \ni p$ and $\phi{-1}(V_q) \ni q$ are then open and disjoint by bijectivity of $\phi$.

Note that if we did not have (v) then if there was no $U_\alpha$ containing both $p$ and $q$, we would not be able to conclude that $p$ and $q$ have disjoint open neighbourhoods.

 

[cianfa72]

the lemma you cite appears as 1.23 (Smooth Manifold Construction Lemma).

$\{U_\alpha\}$ is neither the topology on $M$ nor a basis for it. The topology on $M$ is defined by the basis $$\bigcup_{\alpha}\{ \phi_\alpha^{-1}(V) : V\mbox{ open in \mathbb{R}^n}\}.$$ Note that this requires us to prove that each $U_\alpha$ is open in this topology.

Yes, from (i) $\varphi_{\alpha}(U_{\alpha})$ is open in $\mathbb R^n$. Since $\varphi$ is bijective $U_{\alpha} = \varphi_{\alpha}^{-1} \circ \varphi_{\alpha}(U_{\alpha})$ hence it is open in the given topology by its definition.

 

[fresh_42]

I suspect that (v) is a failed attempt to describe Hausdorff, namely that M is T_1 and T_0. (v) is phrased here as T_0 or T_1 whereas it should be only T_1 (implying T_0).

 

[martinbn]

Isn't all this obvious! If two points belong to the same $U_\alpha$, since it is bijective to an open set of some $\mathbb R^n$ and the topology comes from that, they are separable!