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?

  • 8
    $\begingroup$ Lee's book (page 114) for the uniqueness of the topology/smooth structures on $S$ that makes $i$ an embedding. $\endgroup$
    – jimbo
    Jan 22, 2015 at 2:45
  • $\begingroup$ @jimbo Your response certainly helped. Can you also see if my working above is correct? $\endgroup$ Jan 22, 2015 at 3:16
  • $\begingroup$ 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. $\endgroup$
    – suncup224
    Nov 29, 2020 at 2:36
  • $\begingroup$ Just to add, this also nicely motivates theorem 5.8 in Lee $\endgroup$
    – suncup224
    Nov 29, 2020 at 3:24

1 Answer 1


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 $S$.


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