Let
- $d\in\mathbb N$
- $O\subseteq\mathbb R^d$ be open
- $k\in\{1,\ldots,d\}$
- $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary
- $f:O\to M$ be a $C^1$-diffeomorphism
How can we conclude that $M$ is an open subset of $\mathbb R^d$?
I think it's easy to make a stupid mistake here, since the road from the assumption to the conclusion is subtle.
First of all, if $M_i$ is any embedded $C^1$-submanifold of $\mathbb R^d$, $g:M_1\to M_2$ is bijective and $C^1$-differentiable at $x_1\in M_1$ and $g^{-1}$ is $C^1$-differentiable at $g(x_1)$, then
- $g^{-1}\circ g$ is $C^1$-differentiable at $x_1$ and $$\operatorname{id}_{T_{x_1}\:M_1}=T_{g(x_1)}(g^{-1})\circ T_{x_1}(g).$$
- $g\circ g^{-1}$ is $C^1$-differentiable at $g(x_1)$ and $$\operatorname{id}_{T_{g(x_1)}\:M_2}=T_{x_1}(g)\circ T_{g(x_1)}(g^{-1}).$$
- $T_{x_1}(g)$ is an isomorphism from $T_{x_1}\:M_1$ onto $T_{g(x_1)}\:M_2$ with $${T_{x_1}(g)}^{-1}=T_{g(x_1)}(g^{-1}).$$
- $T_{g(x_1)}(g^{-1})$ is an isomorphism from $T_{g(x_1)}\:M_2$ onto $T_{x_1}\:M_1$ with $${T_{g(x_1)}(g^{-1})}^{-1}=T_{x_1}(g).$$
Now I wonder whether we really need to argue with the implicit function theorem. Maybe this is too complicated, but we could do the following (and hopefully I'm not missing a subtlety): Let $x\in O$. Since $T_x(f)$ (which is equal to the Fréchet derivative ${\rm D}f(x)$, since $f$ is defined on an open subset of $\mathbb R^d$) is an isomorphism, it is - in particular - injective and hence there is an $\mathbb R^d$-open neighborhood $N_1\subseteq O$ of $x$ and an $\mathbb R^d$-open neighborhood $N_2\subseteq M$ of $f(x)$ such that $\left.f\right|_{N_1}$ is a $C^1$-diffeomorphism from $N_1$ onto $N_2$. So, $f(x)$ is an interior point of $M$ and hence we're done.
Am I missing something in my argumentation? And, more interestingly, do we really need to argue with the inverse function theorem, if we already know what I wrote about $g$ in general?