A Soft Question on Differential Geometry Suppose you have your ambient space $\mathbb{R}^n$ and some subset $M$ of $\mathbb{R}^n$. Now, when I learned differential geometry in college, I learned that the notion of differentiability depends on the coordinate charts and transition functions that you use to map out $M$.
However, something about this just doesn't seem intuitively correct. Let $M$ be the locus of points corresponding to the surface of some weird Weierstrass-like sphere-like shape that's continuous everywhere but too jagged to have a tangent plane anywhere. No matter how hard you try, it seems impossible to define a coherent notion of a derivative for $M$, regardless of how many coordinate charts or exotic transition functions you use. So it seems that the differentiability of a manifold shouldn't depend on the atlas or transition functions you use, but rather is inherent to the specific locus of points and the way those points are connected.
 A: Let $\mathcal{N}$ and $\mathcal{M}$ be two $\mathcal{C}^{r}$-differentiable manifolds of dimensions $n$ and $m$ respectively. The map $f: \mathcal{N} \rightarrow \mathcal{M}$ will be said to be a $\mathbf{\mathcal{C}^{r}}$ -differentiable function at $P \in \mathcal{N}$ if there are local charts $(U_{P}, \phi)$ and $(V_{f(P)}, \psi)$ respectively, such that the real (vector) function 
$$
\psi \circ f \circ \phi^{-1}: \phi \big( f^{-1}(V_{f(P)}) \cap U_{P}\big) \rightarrow \psi(V_{f(P)})
$$
is $\mathcal{C}^{r}$-differentiable at $\phi(P) \in \mathbb{R}$ (in the usual sense of calculus). 
Notice that $\phi \big( f^{-1}(V_{f(P)} \cap U_{P})\big) \subseteq \mathbb{R}^{n}$ and $\psi(V_{f(P)}) \subseteq \mathbb{R}^{m}$. The situation is visually described in the following figure

The function $f: \mathcal{N} \rightarrow \mathcal{M}$ will be said to be a $\mathbf{\mathcal{C}^{r}}$ -differentiable function if it is $\mathcal{C}^{r}$-differentiable for every $P \in \mathcal{N}$.
The following lemma verifies the validity and the usefulness of the previous definition, by proving that the definition is actually independent of the charts used:
Lemma:
The definition of $\mathcal{C}^{r}$-differentiable function given above is independent of the choice of local charts
$(U_{P}, \phi)$ and $(V_{f(P)}, \psi)$.
Proof:
If we had chosen two different local charts $(U^{'}_{P}, \phi^{'})$ and $(V^{'}_{f(P)}, \psi^{'})$ then
$$
\psi^{'} \circ f \circ \phi^{' -1} =  (\psi^{'} \circ \psi^{-1}) \circ (\psi \circ f \circ \phi^{-1}) \circ (\phi \circ \phi^{' -1})
$$
But all the (real) functions in the rhs of the above are $\mathcal{C}^{r}$-differentiable (in the usual sense of calculus) by the definition of the notion of $\mathcal{C}^{r}$-differentiable manifold.
Consequently, we have shown that if $f: \mathcal{N} \rightarrow \mathcal{M}$ is a $\mathcal{C}^{r}$-differentiable function for one choice of the local charts $(U_{P}, \phi)$ and $(V_{f(P)}, \psi)$ then it will be so for any other choice of local charts.
The real (vector) function $\psi \circ f \circ \phi^{-1}(x_{1}, x_{2}, ..., x_{n}) = (y_{1}, y_{2}, ..., y_{n})$ can be written equivalently
$$
y_{i} = y_{i}(x_{1}, x_{2}, ..., x_{n})
$$
for all values $i = 1, 2, ..., n$.
