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Let $M$ be a smooth $n$-manifold and let $U\subseteq M$ be any open subset. Define an atlas on $U$ $$\mathcal{A}_{U}=\big\{\text{smooth charts}\;(V,\varphi)\;\text{for}\; M\;\text{such that}\;V\subseteq U\big\}.$$

I must prove that $\mathcal{A}_{U}$ is a smooth atlas for $U$, that is

(1) $U=\bigcup{V}$, where $V$ is the domain of charts such that $V\subseteq U.$

This point is ok.

and

(2) It remains to prove that $\mathcal{A}_U$ is a smooth atlas for $U$.

My attempt. Let $\big(V_1,\varphi_1\big)$, $\big(V_2,\varphi_2\big)\in\mathcal{A}_U$, since they are smooth charts for $M$, that is are charts of maximal atlas of $M$, the maps $$\varphi_2\circ\varphi_1^{-1}\colon\varphi_1\big(V_1\cap V_2\big)\to \varphi_2\big(V_1\cap V_2\big)\quad\text{and}\quad \varphi_1\circ\varphi_2^{-1}\colon \varphi_2\big(V_1\cap V_2\big)\to \varphi_1\big(V_1\cap V_2\big)$$ are $C^{\infty}$.

Since $V_1\subseteq U$, $V_1=U\cap V_1$, then $V_1$ is open in $U$, similary $V_2$ is open in $U$, then $\big(V_1,\varphi_1\big)$ and $\big(V_2,\varphi_2\big)$ are charts of $U$, morever $V_1\cap V_2$ is open in $U$, and then they are $C^{\infty}$ compatible.

Question It's correct?

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  • $\begingroup$ I just don't see why you need to add that $V_1\cap V_2$ is open at the end, but this proof is correct. $\endgroup$ May 5 '19 at 9:53
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Your proof is correct, but I think it is uncessary to prove (2). If $\mathcal A$ denotes the maximal smooth atlas of $M$, then any subset $\mathcal A' \subset \mathcal A$ is automatically a smooth atlas on $M' = \bigcup_{(V,\varphi) \in \mathcal A'} V$ which is an open subset of $M$.

It is perhaps worth to mention that $\mathcal A_U$ is a maximal smooth atlas of $U$. To see this, consider any smooth atlas $\mathcal B$ on $U$ containing $\mathcal A_U$.

Let $(W,\psi)$ be any chart in $\mathcal B$. It is compatible with all charts in $\mathcal A_U$. Now let $(\varphi,V) \in \mathcal A$. Its restriction $(\varphi' = \varphi \mid_{V \cap U}, V' = V \cap U)$ also belongs to $\mathcal A$, and since $V \cap U \subset U$, it belongs to $\mathcal A_U$. Since $W \cap V' = W \cap V \cap U = W \cap V$, the charts $(W,\psi)$ and $(\varphi,V)$ have the same transition function as the charts $(W,\psi)$ and $(\varphi',V')$. The latter is smooth, which shows that $(W,\psi)$ is compatible with $(\varphi,V)$.

Hence $(\psi,W)$ is compatible with $\mathcal A$ and we conclude $(\psi,W) \in \mathcal A$. But this shows $(\psi,W) \in \mathcal A_U$ because $W \subset U$.

Therefore $\mathcal B = \mathcal A_U$.

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  • $\begingroup$ @PaulFrostThanks for your answer. But a priori we cannot say that $\mathcal{A}_U$ is a maximal smooth atlas for $U$, but only that it is contained in a unique maximal smooth atlas or not? $\endgroup$
    – Jack J.
    May 5 '19 at 14:41
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    $\begingroup$ It is a maximal smooth atlas, but this requires a short proof. If you add a chart $(V,\psi)$ with $V \subset U$ which does not belong to $\mathcal A_U$, then certainly $(V,\psi) \notin \mathcal A$ which means that you find a non-smooth transition function with a chart in $\mathcal A$. You have to show that you can also find a chart in $\mathcal A_U$ with the same "defect". $\endgroup$
    – Paul Frost
    May 5 '19 at 14:52
  • $\begingroup$ @PaulFrostI believe I have found a way to show that $\mathcal{A}_U$ is a smooth maximal atlas: let's suppose it is absurd that $\mathcal{A}_U$ is not a smooth maximal atlas, then exists a smooth atlas $\mathcal{B}$ such that $\mathcal{A}_U\subset \mathcal{B}$, therefore exists $(W,\psi)\in\mathcal{B}\setminus \mathcal{A}_U$ with $W\cap U\ne W.$ A chart $(V,\varphi)\in\mathcal{A}_U$ must be compatible with $(W,\psi)$, but $V\subseteq U$, then $V\cap W\subseteq U\cap W\ne W$, absurd. It's correct? $\endgroup$
    – Jack J.
    May 6 '19 at 14:17
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    $\begingroup$ Not quite correct. $\mathcal B$ is a smooth atlas for $U$, thus for each $(W,\psi) \in \mathcal B$ you have $W \subset U$ and therefore $W \cap U = W$. I shall edit my answer. $\endgroup$
    – Paul Frost
    May 6 '19 at 15:13

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