When is a chart of a submanifold not only a homeomorphism, but a diffeomorphism?

Question

I've got trouble to understand the concept of a "smooth structure" associated to a submanifold.

Let $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$. Say $M\subseteq\mathbb R^d$ is a $k$-dimensional embedded submanifold

1. without boundary if $M$ is locally homeomorphic to $\mathbb R^k$;
2. with boundary if $M$ is locally homeomorphic to $\mathbb H^k$.

If I didn't make a mistake, (1.) should imply (2.): If $x\in M$, then (since $\mathbb R^d$ is locally compact) there is a homeomorphism $\varphi$ from a compact neighborhood $\Omega$ of $x$ onto an open subset $U$ of $\mathbb R^k$. Now $\varphi-\inf_\Omega\varphi_k$ is a homeomorphism from $\Omega$ onto $U-\inf_\omega\varphi_k\subseteq\mathbb H^k$.

Now $(\Omega,\phi)$ is called a $k$-dimensional chart of $M$ if $\Omega$ is an open subset of $M$ (equipped with the subspace topology) and $\phi$ is a homeomorphism from $\Omega$ onto an open subset of $\mathbb R^k$ or $\mathbb H^k$. In the first case, it is called an interior chart and if in the second case it additionally holds $\phi(\Omega)\cap\partial\mathbb H^k=\emptyset$, then it is called a boudary chart.

If $(\Omega_i,\phi_i)$ is a $k$-dimensional chart of $M$, then $(\Omega_1,\phi_1)$ and $(\Omega_2,\phi_2)$ are called $C^\alpha$-compatible, if $\phi_2\circ\phi_1^{-1}:\phi_1(\Omega_1\cap\Omega_2)\to\phi_2(\Omega_1\cap\Omega_2)$ is a $C^\alpha$-diffeomorphism.$^1$ Now an atlas $\mathcal A$ for $M$ is a collection of charts whose domain cover $M$ and $\mathcal A$ is called $C^\alpha$-atlas if any two of its charts are $C^\alpha$-compatible.

Now my question is: If I got such an atlas $\mathcal A$, is it somehow possible to show that the charts itself are $C^\alpha$-diffeomorphism?

What I also want to know: If $x\in M$, then a function $f$ from $M$ into a Banach space is called $C^\alpha$-differentiable at $x$, if there is a chart $(\Omega,\phi)$ of $M$ with $x\in\Omega$ and $f\circ\phi^{-1}$ is $C^\alpha$-differentiable at $\phi(x)$. How strongly does this notion depend on the particular chart? I guess we can show that it also follows that $f\circ\psi^{-1}$ is $C^\alpha$-differentiable at $\psi(x)$ for any other chart $\psi$ which is $C^\alpha$-compatible to $\phi$; but can we show more?

EDIT: It seems like I've found the claim in this book, but I cannot really follow the argumentation given there:

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$^1$ Is this notion well-defined even when one of the $(\Omega_i,\phi_i)$ is a boundary chart? Note that I say that a function on an arbitrary subset of $\mathbb R^d$ is differentiable, if it is the restriction of a differentiable map on an open subset of $\mathbb R^d$.

Answer 1

I think the answer in ''no'' since the topological manifold has no intrinsic differentiable structure. Thus you can not define ''differentiability'' both ways. Hence charts can not be diffeomorphisms themselves.