How to define submanifolds of $\mathbb{R}^n$ with boundary without using (explicitly) the notion of diffeomorphism?

I know how to define submanifolds of $\mathbb{R}^n$ (without boundary) without using the notion of diffeomorphism:

Definition: Let $M\subseteq \mathbb{R}^n$ be any subset. We say that $M$ is a $C^\infty$-submanifold of $\mathbb{R}^n$ with dimension $m$ if $m\leq n$ and for all $p\in M$ there're an open neighborhood $U\subseteq \mathbb{R}^n$ of $p$ and a smooth map $\varphi :U\to \mathbb{R}^{n-m}$ such that the differential $(d\varphi)_p$ is surjective and $\varphi ^{-1}(\{0\})=U\cap M$.

At first I don't want to use the notion of diffeomorphism because I don't want initially define differentiability of a map defined in a non-open subset of $\mathbb{R}^n$.

Thank you for your attention!


1 Answer 1


If your only concern is avoiding a differentiability on nonopen sets, then the normal definition is quite adequate, though it does involve diffeomorphisms:

Def: $M\subseteq\mathbb{R}^n$ is an $m$-manifold with boundary if for each $p\in M$ there exist open sets $U,V\subseteq\mathbb{R}^n$ with $p\in U$ and a diffeomorphism $\varphi:U\to V$ such that $M\cap U=\varphi^{-1}\left([0,\infty)\times\mathbb{R}^{m-1}\times\{0\}^{n-m}\right)$.

Note that while we do take the image of a nonopen set, the domain and codomain of $\varphi$ are open, and so there's no issue with defining smoothness only on open sets.

If, however, you want something which generalizes the level set definition, this is also possible:

Def: $M\subseteq\mathbb{R}^n$ is an $m$-manifold with boundary if for each $p\in M$ there exist an open set $U\subseteq\mathbb{R}^n$ with $p\in U$ and a smooth submersion $\varphi:U\to\mathbb{R}^{n-m+1}$ such that $M\cap U=\varphi^{-1}\left([0,\infty)\times\{0\}^{n-m}\right)$.


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