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Instead of defining a smooth manifold to be a manifold whose gluing functions are smooth, what would happen if we defined it as an $n$-manifold $M$ which has an embedding into $\mathbb{R}^{n +1}$?

A smooth map between manifolds $e_M : M \hookrightarrow \mathbb{R}^{n+1}$ and $e_N : N \hookrightarrow \mathbb{R}^{n+1}$ would be a continuous function $f : M \to N$ along with a smooth function $g : \mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$ such that $g \circ e_M = e_N \circ f$.

Would defining them this way be equivalent?

user1172394
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    No. The real projective plane wouldn’t be a smooth manifold, because there are no embeddings into $\mathbb{R}^3$. – Eduardo Longa Jul 08 '20 at 19:19
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    Definitely not equivalent, because not all topological manifolds admit a smooth structure, but all topological manifolds embed in $\Bbb R^n$ for large enough $n$. – pancini Jul 08 '20 at 19:20
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    The definition presented here is certainly not equivalent, though it's worth noting there is a way to equivalently define a smooth manifold as a "sufficiently nice" subset of $\mathbb{R}^n$, due to the Whitney embedding theorem. – Kajelad Jul 08 '20 at 19:24

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For a start, the Klein bottle would no longer be a smooth manifold, as it has no embedding in $\Bbb R^3$. Nor would any non-orientable closed $2$-manifold.

Angina Seng
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