As discussed in this other question, given a manifold $M$ and a point $p\in M$, we can define its tangent vectors in $T_p M$ as the set of equivalence classes $[\gamma'(0)]$ defined so that $\gamma_1,\gamma_2\in[\gamma'(0)]$ iff $(\phi\circ\gamma_1)'(0)=(\phi\circ\gamma_2)'(0)$ for all coordinate charts $\phi:U\to\mathbb R^n$, where $p\in U\subset M$.
In this definition, is it sufficient to ask for the curves to have same derivative with respect to one coordinate chart defined around $p$? In other words, given two charts $\phi,\tilde\phi:U\to\mathbb R^n$ defined on some neighbourhood of $p$, suppose $$(\phi\circ\gamma_1)'(0)=(\phi\circ\gamma_2)'(0).$$
Does this imply that $(\tilde\phi\circ\gamma_1)'(0)=(\tilde\phi\circ\gamma_2)'(0)$?
From the definition of a smooth manifold, I know that $\tilde\phi\circ\phi^{-1}$ is a homeomorphism between $\phi(U)$ and $\tilde\phi(U)$. I would therefore expect that if
$$\phi(\gamma_1(\epsilon))-\phi(\gamma_2(\epsilon)) = o(\epsilon),$$
then the same should hold replacing $\phi\to\tilde\phi$. However, I'm not sure what properties of $\tilde\phi\circ\phi^{-1}$ I could use to show this.
Best Answer
To (hopefully) clarify notation, for a map $F:\mathbb{R}^k\to\mathbb{R}^l$, let $D_xF$ denote the derivative of $F$ at $x$, in the sense of multivariable calculus.
Let $\gamma_1,\gamma_2$ be smooth curves with $\gamma_1(0)=\gamma_2(0)=p$, suppose that $D_0(\varphi\circ\gamma_1)=D_0(\varphi\circ\gamma_2)$ for some chart $\varphi$, and let $\psi$ be any other chart containing $p$. Since $M$ is a smooth manifold, the transition function $\tau=\psi\circ\varphi^{-1}$ is a deffeomorphism between open subsets of $\mathbb{R}^n$. We can compute the coordinate derivative of $\gamma_1$ in the new chart. \begin{align*} D_0(\psi\circ\gamma_1)=&D_0(\psi\circ\varphi^{-1}\circ\varphi\circ\gamma_1) \\ =&D_0(\tau\circ(\varphi\circ\gamma_1)) \end{align*} Applying the chain rule: \begin{align*} =&D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_1) \end{align*} We see that $D_0(\psi\circ\gamma_1)$ is related to $D_0(\varphi\circ\gamma_1)$ by the linear map $D_{\varphi(0)}\tau$. The same is true of $\gamma_2$. As a result, of the derivatives are equal in one chart, they are equal in any other: \begin{align*} D_0(\varphi\circ\gamma_1)=&D_0(\varphi\circ\gamma_2) \\ \implies D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_1)=&D_{\varphi(p)}\tau\cdot D_0(\varphi\circ\gamma_2) \\ \implies D_0(\psi\circ\gamma_1)=&D_0(\psi\circ\gamma_2) \\ \end{align*}