Let $M$ be a smooth manifold.
An embedded submanifold of $M$ is a subset $S$ of $M$ such that $S$ is a topological manifold under the subspace topology induced by $M$, endowed with a smooth structure such that the inclusion map $i:S\to M$ is a smooth embedding.
Is it true that if $M$ is a smooth manifold and $S$ is a subset $M$, then there is a unique smooth structure, if one exists, which makes $S$ into an embedded submanifold?
Suppose there is a smooth structure on S which makes the inclusion map $i:S\to M$ a smooth embedding. Let $\mathscr A =\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in J}$ be a smooth atlas on $M$. Then I think the set
$$\mathscr B=\{(U_\alpha\cap S,\varphi_\alpha|_{U_\alpha\cap S}):\alpha\in J\}$$
is a smooth atlas on $S$, since any two elments in $\mathscr B$ are smoothly compatible (borrowing from the smooth compatibility of $(U_\alpha,\varphi_\alpha)$ and $(U_\beta,\varphi_\beta)$).
The inclusion map can also be shown to be smooth under this smooth structure.
Since any smooth atlas is contained in a unique maximal smooth atlas, there is a unique smooth strucure on $S$.
Is this reasoning correct?
Best Answer
Let $p\in{S}$, $(\phi, U)$ such $\phi(S\cap{U})=\phi(U)\cap(\mathbb{R}^n\times{0})$
$$\phi\circ{i_S}\circ(\phi|_{S\cap{U}})^{-1}:\phi(U)\cap(\mathbb{R}^n\times{0})\to\phi(U)\subset\mathbb{R}^n$$ is the inclusion map, hence smooth.
In fact it´s a local representation of $i_S$. Then induced structure por $M$ is the same as the original structure of $S$.