Let $M$ be a differentiable $n$-manifold. I want to prove that $M$ has a countable atlas $\mathcal{A}=\{(U_j,\varphi_j)\}_{j\in J}$, where $J\subseteq\mathbb{N}$, such that $\varphi_j(U_j)=\mathbb{R}^n,\forall\,j\in J$.

First, I have proven that, if $p\in M$, for every open subset $\Omega\subset M$ such that $p\in\Omega$, exists a chart $(U,\varphi)$ such that $p\in U\subseteq\Omega$, $\varphi(U)=\mathbb{R}^n$ and $\varphi(p)=0$.

Now, since $M$ is a differentiable manifold, it has the Lindelof property. So, for every open covery, there exist a countable subcovery of $M$. With that said, let $M=\bigcup_{j\in J} G_j$, where $J\subseteq\mathbb{N}$ and $G_j\subseteq M$ are open sets, $\forall\,j\in J$. Then, by the previous property we can construct a family of charts $\mathcal{A}=\{(U_j,\varphi_j)\}_{j\in J}$ verifying $U_j\subseteq G_j$ and $\varphi_j(U_j)=\mathbb{R}^n$.

Now, we have to prove that $\mathcal{A}$ is an atlas for $M$, that is, we have to prove that $M=\bigcup_{j\in J} U_j$ and that $(U_i,\varphi_i)$ and $(U_j,\varphi_j)$ are compatible charts, if $i\neq j$.

Let's first see that $M=\bigcup_{j\in J} U_j$. Of course, $\bigcup_{j\in J} U_j\subseteq M$. Now, let $p\in M$ and take an arbitrary chart $(U_j,\varphi_j)\in\mathcal{A}$. Then, $\varphi_j(p)\in\mathbb{R}^n$. So, $p\in\varphi_j^{-1}(\mathbb{R}^n)$ and we conclude that $p\in U_j$, so $M=\bigcup_{j\in J} U_j$.

Now, suppose that for two charts $(U_i,\varphi_i)$ and $(U_j,\varphi_j)$, we have $U_i\cap U_j\neq \varnothing$ (if $U_i\cap U_j = \varnothing$, the charts are compatible by definition).

Then, $\varphi_i(U_i \cap U_j)$ and $\varphi_j(U_i \cap U_j)$ are both open sets in $\mathbb{R}^n$, because both $U_i$ and $U_j$ are open sets in $M$ and by definition of the natural topology of $M$.

To finish this proof I have to show that this $\varphi_i\circ\varphi_j^{-1}$ is a diffeomorphism. I've tried to prove it creating a conmutative diagram, but I haven't been able. I would really appreciate some help. Thank you!!

P.D.: Sorry for the long post, but I did want to give the full context. Thank you!

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    $\begingroup$ You have a smooth manifold $M$, so you have a maximal atlas $A$. Now you have found for each $p$ in $M$ a chart $phi_p$ that you like for some reason and whose domain contains $p$ The sep $B=\{\phi_p:p\in M\}$ is an atlas simply because it is contained in $A$ and and the domains of its elements cover $M$. If you keep a countable subset of $B$ that still has this last property you still have a chart. There is no need to check anything, really. $\endgroup$ Sep 29, 2022 at 16:02
  • $\begingroup$ So, the fact that $\mathcal{A}\subseteq\mathcal{D}$, where $\mathcal{D}$ is the differentiable structure (maximal atlas) of $M$, and that $M=\cup_{(U,\phi)\in\mathcal{A}} U$, makes $\mathcal{A}$ an atlas of $M$? $\endgroup$
    – JaviLark01
    Sep 29, 2022 at 16:38

1 Answer 1


There is nothing to prove. By a chart you mean any member of the differentiable structure (maximal atlas) $\mathfrak D$ of $M$. You construct a countable subset $\mathcal A = \{(U_j,\varphi_j)\}_{j\in J} \subset \mathfrak D$ such that $\bigcup_{j\in J} U_j = M$ and $\varphi_j(U_j) = \mathbb R^n$ for all $j \in J$. The set $\mathcal A$ is nothing else than a special atlas for $M$ . But now trivially all transition functions $\varphi_i \circ \varphi_j^{-1}$ are diffeomorphisms because that is true for any two charts in $\mathfrak D$.

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    $\begingroup$ Note that there is another condition that is required, which is $\varphi_j(U_j) = \Bbb R^n$. In other words, the charts are defined on subsets diffeomorphic to $\Bbb R^n$. Your answer does not take this into account. The right answer has been given in the comment section of the question $\endgroup$
    – Didier
    Sep 29, 2022 at 16:30
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    $\begingroup$ @Didier You are right, I did not mention the requirement $\varphi_j(U_j) = \mathbb R^n$ because this is irrelevant for the OP's question to prove that the transition functions $\varphi_i \circ \varphi_j^{-1}$ in $\mathcal A$ are diffeomorphisms. Or do you think it is relevant? Also the comment which you declare as the right answer does not mention it. $\endgroup$ Sep 29, 2022 at 23:56
  • $\begingroup$ You're right, it is not explicit in the comment, but it is hidden under the words that you like for some reason. I think, fir a complete answer, that saying it is relevant yes! $\endgroup$
    – Didier
    Sep 30, 2022 at 7:35

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