# Embedded Submanifolds Have a Unique Smooth Structure

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?

• Lee's book (page 114) for the uniqueness of the topology/smooth structures on $S$ that makes $i$ an embedding. Jan 22, 2015 at 2:45
• @jimbo Your response certainly helped. Can you also see if my working above is correct? Jan 22, 2015 at 3:16
• I think the issue (with what OP suggested) is that $\phi_{\alpha} | _{U_{\alpha}\cap S}$ maps into a "$\dim S$ surface" in $\mathbb{R}^{\dim M}$ but this surface is not necessarily inside a $\mathbb{R}^{\dim S}$ copy in $\mathbb{R}^{\dim M}$. Then you would have to homeomorphically identify the surface with $\mathbb{R}^{\dim S}$ and that gets messy/ambiguous/etc. TLDR, $\mathscr{B}$ doesn't consist of charts, rightly speaking. Nov 29, 2020 at 2:36
• Just to add, this also nicely motivates theorem 5.8 in Lee Nov 29, 2020 at 3:24

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 for $$M$$ is the same as the original structure of $$S$$.