I Smooth Manifold Chart Lemma

cianfa72
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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.
 
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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.
 
pasmith said:
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.
 
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).
 
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!
 
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