Let $X$ be a smooth manifold and $U\subseteq X$ open. Define a canonical smooth structure on $U$, for which the embedding $U\to X$ is smooth.


I want to solve this task. My try was as follows:

$X$ is a smooth manifold, hence a second-countable Hausdorff space, with a smooth structure, therefore a maximal smooth atlas $\mathcal{A}$. So every "transition between maps" (I hope this is the right translation) is smooth.

I need a maximal atlas on $U$.


Since $X$ is a smooth manifold, we have a maximal atlas $\mathcal{A}$. Because $X$ is second-countable, we have a countable set $B=\{U_i,i\in I\}$, with $U_i\subseteq X$ open for every $i\in I$, and every open subset of $X$ can be written as an union of these sets. Let $u$ be an element of $U$. Since $X$ is a Hausdorff space, for every $u\in U$ exists an open set in $B$ with $u\in U_{j_u}$ and $U=\bigcup_{u\in U} U_{j_u}$.

Let $V\subset X$ be open and $\varphi$ a homeomorphism. It is $\bigcup_{(V,\varphi)\in\mathcal{A}} V=X$.

Choose now an open set $V_i$ with homeomorphism $\varphi_i$, with $V_i\supseteq U_i$ and observe $(U_i, \varphi_{i|U_i})$.

Then is $\bigcup_{\displaystyle (U_{j_u},\varphi_{j_{u}|U_{j_u}})} U_{j_u}=U$

and $\iota: U\to X$, $\iota(x)=x$ is smooth.

I would be thankfull for your thoughts about my proof. Am I to sloppy at some points, or is it even wrong?

Thanks in advance.

  • $\begingroup$ It seems correct $\endgroup$ – Spotty Jun 30 '16 at 8:56
  • $\begingroup$ I am unsure, if you really get a maximal atlas on $U$. Why is the embedding smooth? Is this just trivial, since you can differentiate this function infinite times? $\endgroup$ – MrTopology Jun 30 '16 at 8:59
  • $\begingroup$ I don't understand why you did get a downvote. Everything looks correct to me. The embedding is smooth because in some well chosen charts, your embedding is simply inclusion $\mathbb R^k \to \mathbb R^n$ which is of course smooth. For maximality of the atlas, you can simply cover $U$ by charts and then take the bigger atlas which contains your charts. $\endgroup$ – user171326 Jun 30 '16 at 9:06
  • $\begingroup$ I think i got the downvote, because I asked this question again, but I deleted the first post, before reposting it, to draw more attention to my question. Thanks for your comment. Do you mind writing it as an answer? $\endgroup$ – MrTopology Jun 30 '16 at 9:10
  • 1
    $\begingroup$ @MrTopology : Sure ! I did add a formal proof that $\iota$ is smooth, I think this is important to be able to prove rigourously propositions, especially because the theory of manifolds is not easy at the beginning, and one can be easily lost by the formalism of charts, tangent spaces, ... $\endgroup$ – user171326 Jun 30 '16 at 9:24

Yes this is correct. For getting a maximal atlas on a manifold $X$, one usually take a familly of charts $\phi_i$ which covers $U$, and then take all the $\psi : V \to \mathbb R^n$ (where $V \subset U$ is open) which are compatible with $\phi_i$. You get a maximal atlas like this.

For a map $M \to N$ between manifolds, it's enough to check it is smooth in some charts. Then, take $p \in U$, $(\phi, V)$ a chart on $X$ which contains $p$ and the chart $(\psi := \phi_{|V \cap U}, V \cap U)$. Then $ \phi \circ \iota \circ \psi^{-1} = id_{\mathbb R^k} $ so $\iota$ is smooth at $p$.

  • $\begingroup$ Thanks. I have one final question. Why is $\phi\circ\iota\circ\psi^{-1}=id_{\mathbb{R}^k}$. Is it because for manifolds it is always $U\subseteq\mathbb{R}^k$ $\endgroup$ – MrTopology Jun 30 '16 at 9:29
  • $\begingroup$ Here I took $U \subset X$ as an abstract manifold. $\psi : V \cap U \to \mathbb R^k$ so $\phi \circ \iota \circ \psi^{-1} : \mathbb R^k \to \mathbb R^n$. Is it more clear now ? $\endgroup$ – user171326 Jun 30 '16 at 10:23
  • $\begingroup$ Yes, I think it is clear. $\endgroup$ – MrTopology Jun 30 '16 at 10:30

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