Abstract Manifolds - Is $\mathbb R^m$ only a $C^1$ manifold? I have a question that came to my mind. First, this is the definition of an abstract manifold we once used in a lecture:

Definition. Let $M$ be metric space and $\left(V_{i}\right)_{i\in I}$ be an open cover of $M$ with open sets $U_i\subset\mathbb R^m$ and homeomorphisms $F_i: U_i\rightarrow V_i$. Then we say $M$ is an (abstract) $m$-dimensional $C^k$ manifold if for all two open sets $V_1$, $V_2\subset M$ with maps $F_1$ and $F_2$ the transition map $$F_{2}^{-1}\circ F_1: \quad F_1^{-1}(V_1 \cap V_2) \rightarrow F_{2}^{-1}(V_1\cap V_2)$$ is a $C^k$ diffeomorphism.

(I am aware that one can use an even more general definition by requiring only the spaces $M$ to be only a topological one instead of a metric space, as is done in [1], but let's stick to this for now.)
Question: I was asking myself whether $M = \mathbb R^m$ itself is an abstract $m$-dimensional $C^k$ manifold and to what degree $k$ it is. I myself found until now that $M = \mathbb R^m$ is only a $m$-dimensional $C^1$ manifold, is this really correct?
Explanation: According to [2], we can cover $\mathbb R^m$ by "[t]he collection of all open discs with rational radii and rational center coordinates". As the open sets $U_i$, I would choose the same sets, thus $U_i = V_i$. Now choose as the homeomorphisms $F_i$ simply the identity, and we have for the transition map $F_{2}^{-1}\circ F_1 = \text{Id} \circ \text{Id} = \text{Id}$, which is obviously a $C^1$ diffeomorphism. However, this is not a $C^2$, $\dots$, $C^{\infty}$ diffeomorphism as we do not have an inverse function for the first, second, etc. derivate of the identity mapping.
[1] http://www.math.lsa.umich.edu/~jchw/WOMPtalk-Manifolds.pdf
[2] https://www.quora.com/How-does-the-plane-R-2-with-the-usual-topology-satisfy-the-second-axiom-of-countability?share=1
 A: It think you’re confused what $C^n$ means. It does not mean that the inverse is continuous or anything. It only means that the function is differentiable $n$ times and the $n$-th derivative is continuous.
This means that $\operatorname{Id}$ is $C^\infty$ because it is differentiable infinitely many times.
Also, $\mathbb{R}^m$ is a manifold with the single chart $U=\mathbb{R}^m$.
A: An important part of the definition of a $C^k$ manifold is that the homeomorphisms $\{F_i\}$ are part of the package.  The $C^k$ structure consists of the topological manifold along with an open cover and specific homeomorphisms.
This means the question "is $\mathbb{R}^n$ a $C^k$ manifold?" doesn't make sense on its own.  What is certainly true is that $\mathbb{R}^n$ along with the open cover consisting of the single set $V_1=\mathbb{R}^n$ and the homeomorphism $F_1:V_1\to \mathbb{R}^n$ given by the identity defines an abstract $C^k$ manifold, for all $k\geq 0$.  The open disk cover you mention defines an equivalent $C^\infty$ abstract manifold, for a suitable notion of equivalence (that the transition maps for homeomorphisms from one structure to homeomorphisms in the other are all $C^\infty$).
It turns out that $\mathbb{R}^n$ has other $C^k$ structures that are inequivalent, and which aren't quite so smooth.  For example, consider $\mathbb{R}=\mathbb{R}^1$ and the open cover by $V_1=V_2=\mathbb{R}$.  Define $F_1:V_1\to\mathbb{R}$ by $F_1=\operatorname{id}$ and $F_2:V_2\to\mathbb{R}$ by $F_2(x)=x^3$.  Both of these are homeomorphisms, but the transition map between these is only $C^0$, so while this defines an abstract $C^0$ manifold, it's not a $C^1$ manifold or higher.
