Let $S \subseteq \mathbb{R}^3$ be a surface and $x:U \to V$ be a chart of $S$, i.e., $U$ is an open subset of $S$, $V$ is an open subset of $\mathbb{R}^2$, $x$ is onto and one-to-one and $x^{-1}:V \to U$ is $C^{\infty}$.
Let $\gamma$ and $\tau$ be curves from an open interval $(-\epsilon,\epsilon)$ to $U$ with $\gamma(0)=\tau(0)$ and $\gamma'(0)=\tau'(0)$. Is it true that $(x \circ \gamma)'(0)=(x \circ \tau)'(0)$? If so, what is the proof?
In my studies of basic differential geometry, I was confronted with the above question on several occasions but have no idea how to answer it. The main problem I have is that we cannot apply the chain rule to $(x \circ \gamma)'$ since it is not allowed to differentiate the map $x:U\to V$.
For people who know do Carmo's Differential geometry of curves and surfaces for example, Do Carmo seems to implicitly assume that the claim is true in his proof of Proposition 2, Section 2-4, Chapter 2. The proposition is about the differential of a smooth map between two surfaces. He defines the differnetial via curves representing a tangential vector and among other things, he shows that his definition is independent of the curve. The above claim seems to be one step in his proof but he doesn't explain why it holds.
Thanks for help!
It drives me nuts that DoCarmo writes his undergraduate geometry text treating surfaces in $\Bbb R^3$ as abstract differentiable manifolds. It is more maddening yet that he uses the same letter $\mathbf x$ both for a chart on $S$ and for a parametrization of $S$. The derivative $\alpha'(0)$ makes sense only because $S\subset\Bbb R^3$ and you differentiate $\alpha$ as a map to $\Bbb R^3$. In fact, the proof you refer to is working with parametrized surfaces, with the $\mathbf x$ mapping going in the other direction. He shows that the tangent plane, defined as the image of the derivative map of the parametrization, consists of all tangent vectors of curves. In this direction, the chain rule works fine. And, yes, DoCarmo uses the Inverse Function Theorem liberally when he needs it.
(For your amusement, see this post for an instance of DoCarmo's using the "usual" sloppy conventions that people do with abstract manifolds. This is in exactly the portion of the text you're reading, as it happens.)
So, can you edit your question to clarify where the situation you have set up actually shows up in DoCarmo?