i need to prove that the two following definitions of differentiability on regular surfaces are equivalent:
Let $S$ be a regular surface. A function $f: S \rightarrow \mathbb{R^n}$ is differentiable if:
i) for any local chart of $S$ like $X: U \rightarrow S$ we have $f \circ X : U \rightarrow \mathbb{R^n}$ differentiable.
or
ii) for every point $p$ in $S$ there exists a local chart $X_p : U_p \rightarrow S$ around $p$ such that $f \circ X_p: U_p \rightarrow \mathbb{R^n}$ is differentiable.
Clearly i implies ii but i'm not seeing how ii implies i. Can anyone help me?
Thx!
The key point is the compatibility requirement for two overlapping charts in $S$. Suppose $X\colon U\to S$ is a chart and $p\in X(U)$. Then there is, by assumption, a chart $X_p\colon U_p\to S$ with $p\in X_p(U_p)$. By definition of a surface, $$X_p^{-1}\circ X\big|_{X^{-1}(X_p(U_p)\cap X(U))}\colon X^{-1}(X_p(U_p)\cap X(U))\to X_p^{-1}(X_p(U_p)\cap X(U))$$ is a differentiable map from an open subset of $\Bbb R^2$ to another. But then $f\circ X = \big(f\circ X_p\big)\circ \big(X_p^{-1}\circ X)$ is the composition of differentiable functions on a neighborhood of $X^{-1}(p)$ and is differentiable.