# If $\phi$ is a homeomorphism and both $\phi$ and $\phi^{-1}$ extend to differentiable functions, is ${\rm D}\phi(x)$ invertible?

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:

• The differential of the identity map is the identity map on the respective tangent space, pretty much by definition. – Thorgott Aug 4 at 15:19
• @Thorgott What do you mean by "tangent space" here? – 0xbadf00d Aug 4 at 15:42
• What do you mean by $D\phi(x)$ if there's no tangent spaces involved? – Thorgott Aug 4 at 15:51
• @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 ... – 0xbadf00d Aug 4 at 15:54
• Try taking a look at the beginning of Milnor's Topology from the differentiable viewpoint. – Thorgott Aug 4 at 17:13