# Proof that a differentiable manifold has a countable atlas verifying certain property

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!

• 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. Sep 29, 2022 at 16:02
• 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 $M$? Sep 29, 2022 at 16:38

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 $$\mathfrak D$$.
• 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 Sep 29, 2022 at 16:30
• @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. Sep 29, 2022 at 23:56