Smooth structure on open subsets of manifolds

Question

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.

Hello,

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 $U$.

Now:

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.

Answer 1

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 $p$.