Let $k,d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$, $U\subseteq\mathbb R^k$ and $\phi:\Omega\to U$ be a homeomorphism and $\psi:=\phi^{-1}$. Assume $$\phi=\left.\tilde\phi\right|_\Omega\tag1$$ for some $\tilde\phi\in C^1(O,\mathbb R^k$ for some $\mathbb R^d$-open neighborhood $O$ of $\Omega$ and $$\psi=\left.\tilde\psi\right|_V\tag2$$ for some $\tilde\psi\in C^1(V,\mathbb R^d$ for some $\mathbb R^k$-open neighborhood $V$ of $U$.

Let $x\in\Omega$ and $u:=\phi(x)$. Are we able to show that ${\rm D}\phi(x)$ and ${\rm D}\psi(u)$ are bijective? If not, are we at least able to show that ${\rm D}\phi(x)$ is surjective and ${\rm D}\psi(u)$ is injective?

My problem with this task is that I think we would need, for example, that $\tilde\psi\circ\tilde\phi$ is the identity map when restricted to an open subset of $\mathbb R^d$.

Clearly, since $\Omega\subseteq O$ and $U\subseteq V$, $\tilde\phi$ is differentiable at $x$ and $\tilde\psi$ is differentiable at $u:=\tilde\phi(x)=\phi(x)\in U$. Thus, $\tilde\psi\circ\left.\tilde\phi\right|_{\tilde\phi^{-1}(V)}$ is differentiable at $x$ and $${\rm D}\left(\tilde\psi\circ\tilde\phi\right)(x)={\rm D}\tilde\psi(u){\rm D}\tilde\phi(x)\tag3.$$ We clearly would like the left-hand side to be $\operatorname{id}_{\mathbb R^d}$, but I think we need what I wrote before for that.

EDIT: I'm using the following definition for differentiability on arbitrary sets, which can be found in this book:

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  • $\begingroup$ The differential of the identity map is the identity map on the respective tangent space, pretty much by definition. $\endgroup$ – Thorgott Aug 4 at 15:19
  • $\begingroup$ @Thorgott What do you mean by "tangent space" here? $\endgroup$ – 0xbadf00d Aug 4 at 15:42
  • $\begingroup$ What do you mean by $D\phi(x)$ if there's no tangent spaces involved? $\endgroup$ – Thorgott Aug 4 at 15:51
  • $\begingroup$ @Thorgott By $(1)$, ${\rm D}\phi(x)$ is equal to ${\rm D}\tilde\phi(x)$. As I said before, my problem is that $U\xrightarrow\psi\Omega\subseteq O\xrightarrow{\tilde\phi}\mathbb R^k$ and $\left.\tilde\psi\right|_U=\psi$, but since $U$ is not open it seems like I cannot conclude ... $\endgroup$ – 0xbadf00d Aug 4 at 15:54
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    $\begingroup$ Try taking a look at the beginning of Milnor's Topology from the differentiable viewpoint. $\endgroup$ – Thorgott Aug 4 at 17:13

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