Is a subset of $\mathbb{R}^n$ with a $C^q$ atlas always a $C^q$ submanifold of $\mathbb{R}^n$?

Question

Let $q$ belongs to $\mathbb{N}^{\times}\cup$ {$\infty$}.

A subset $M$ of $\mathbb{R}^n$ is said to be an $m$-dimensional $C^q$ submanifold of $\mathbb{R}^n$ if, for every $x_0\in M$, there is in $\mathbb{R}^n$ an open neighborhood $U$ of $x_0$, an open set $V$ in $\mathbb{R}^n$, and a $\varphi\in \textrm{Diff}^q(U,V)$ such that $\varphi(U\cap M)=V\cap (\mathbb{R}^m\times${$0$}$)$. $\textrm{Diff}^q(U,V)$ is the set of all $C^q$ diffeomorphisms from $U$ to $V$.

Suppose $M$ is a subset of $\mathbb{R}^n$ and $p\in M$. We denote by $$i_{M}: M \rightarrow \mathbb{R}^{n}, \quad x \mapsto x$$ the canonical injection of $M$ into $\mathbb{R}^n$. The map $\varphi$ is called an $m$-dimensional (local) $C^q$ chart of $M$ around $p$ if

• $U:=\operatorname{dom}(\varphi)$ is an open neighborhood of $p$ in $M$;
• $\varphi$ is a homeomorphism of $U$ onto the open set $V := \varphi(U)$ of $\mathbb{R}^m$;
• $g:=i_{M} \circ \varphi^{-1}$ is a $C^q$ immersion.

An $m$-dimensional $C^q$ atlas for $M$ is a family $\{\varphi_\alpha ; \alpha \in A\}$ of $m$-dimensional $C^q$ charts of $M$ whose charted territories $U_\alpha:=\operatorname{dom}(\varphi_\alpha)$ cover the set $M$.

It is easy to show that an $m$-dimensional $C^q$ submanifold of $\mathbb{R}^n$ always has an $m$-dimensional $C^q$ atlas. What about the converse?

Answer 1

1. Suppose $X$ is open in $\mathbb{R}^m$ and $f\in C^q(X,\mathbb{R}^n)$ is an immersion. Then there is for every $x_0\in X$ an open neighborhood $X_0$ in $X$ such that $f(X_0)$ is an $m$-dimensional $C^q$ submanifold of $\mathbb{R}^n$. (This can be shown by the inverse function theorem.)
2. If $M$ has a $C^q$ atlas, we can use the theorem above for each $g_\alpha$ and get a cover of $M$: $$\{g_\alpha(X_0); \alpha\in A \text{ and } X_0 \text{ corresponding to each } x_0\in V_\alpha\}$$ Then each $g_\alpha(X_0)$ is an $m$-dimensional $C^q$ submanifold of $\mathbb{R}^n$ and open in $M$. (Note that each $\varphi_\alpha$ is a homeomorphism.)
3. For each $p\in M$, suppose $p\in g_\alpha(X_0)$. Since $g_\alpha(X_0)$ is a submanifold of $\mathbb{R}^n$, there is in $\mathbb{R}^n$ an open neighborhood $U$ of $p$, an open set $V$ in $\mathbb{R}^n$, and a $\phi\in \textrm{Diff}^q(U,V)$ such that $\phi(U\cap g_\alpha(X_0))=V\cap (\mathbb{R}^m\times\{0\})$. Suppose that $g_\alpha(X_0)=W\cap M$. $W$ is open in $\mathbb{R}^n$. Then $U\cap W$ is an open neighborhood of $p$ in $\mathbb{R}^n$. $\phi(U\cap W)$ is open in $\mathbb{R}^n$ and $\phi|_{U\cap W}\in \textrm{Diff}^q(U\cap W,\phi(U\cap W))$. And we have $$\phi|_{U\cap W}((U\cap W)\cap M)=\phi(U\cap g_\alpha(X_0))=V\cap (\mathbb{R}^m\times\{0\})=\phi(U\cap W)\cap (\mathbb{R}^m\times\{0\}).$$ The final equation holds because $\phi(g_\alpha(X_0))\subset\phi(W)$. Hence $M$ is a submanifold.