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In general, when someone defines a manifold he defines the concept of topological manifold and then the concept of compatible smooth charts, atlases,...

If I understand it correctly the Whitney embedding theorem sais that every manifold can be embedded in $\mathbb{R^n}$ (I guess with the canonical smooth structure?) for some natural number $n$, so we could instead of defining manifolds just define embedded submanifolds of $\mathbb{R^n}$. Also, this is the takeaway I get from this wikipedia section. But then What would be this precise definition of an embedded submanifold of $\mathbb{R}^n$? If I consider the definition of an embedded submanifold in the same wiki article I think I get that a subset $S\subset \mathbb{R}^n$ is an embedded submanifold if for every $x\in S$ we have a diffeomorphism $U\xrightarrow{\phi}\mathbb{R}^n$ (here $U\subset\mathbb{R}^n$ is open) such that $\phi(U\cap S)=\phi(U)\cap(\mathbb{R}^{(n-k)}\times\{0_{\mathbb{R}^k}\})$. Is that how it is defined? Because in some notes I see the (except for open submanifolds) :

definition : Let $M$ be a subset of a linear space $E$. We say $M$ is a smooth embedded submanifold of $E$ if : For a fixed integer $k\ge 1$ and for each $x\in M$ there exists a neighborhood $U$ of $x$ in $E$ and a smooth function $h:U\to \mathbb{R}^k$ such that

(a) If $y$ is in $U$, then h(y) = 0 if and only if $y\in M$;and

(b) rank Dh(x) = k

Are these two equivalent?

roi_saumon
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  • I'm voting to close as duplicate of the target @PaulFrost proposed. It may not fully supply the information you seek, in part because the target is a more detailed treatment and because you've seemingly overlooked the importance of the dimension (rank) of any embedded manifold in $\mathbb R^n$. So have a look and edit this Question if you would like further clarification. – hardmath Dec 16 '20 at 19:27
  • @PaulFrost, I think it does. Nice answer by the way. +1 – roi_saumon Dec 16 '20 at 21:07

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