How to establish the chart-independence of this equivalence relation on curves on a manifold Let $\mathscr{M}$ be a differentiable manifold of dim m. We define the following equivalence relation on curves on $\mathscr{M}$.
Two curves $\sigma_1$ and $\sigma_2$ are tangent at a point $p$ in $\mathscr{M}$ iff

*

*$\sigma_1(0) = \sigma_2(0) = p$

*In some local coordinate system $(x^1, \dots, x^m)$ around the point $p$ the two curves satisfy
$$\frac{dx^i}{dt}(\sigma_1(t))\Bigr|_{t=0} = \frac{dx^i}{dt}(\sigma_2(t))\Bigr|_{t=0}$$
for $i=1, \dots, m$.

How would check that if $\sigma_1$ and $\sigma_2$ are tangent in one coordinate system, then they're tangent    in any other coordinate system that covers the point $p$?
 A: Assume one has a chart $(U, \phi)$ with $p \in U$ and $x^i$ is $u^i \circ \phi$, where $u^i$ is the $i$-th projection from $\mathbb{R}^m$ to $\mathbb{R}$. Your assumption $2.$ means that
$$ D(\phi \circ \sigma_1)(0) = D(\phi \circ \sigma_2)(0) \, , $$
where $D(\phi \circ \sigma_1)(0)$ (resp. $D(\phi \circ \sigma_2)(0)$) denotes the differential of $\phi \circ \sigma_1$ (resp. $\phi \circ \sigma_2$) at $0$. Now, assume that $(V, \psi)$ is another chart with $p \in V$ and $y^i$ is $u^i \circ \psi$.
One has to prove
$$\frac{dy^i}{dt}(\sigma_1(t))\Bigr|_{t=0} = \frac{dy^i}{dt}(\sigma_2(t))\Bigr|_{t=0}$$
for $i=1, \dots, m$.
One has the following sequence of equalities:
\begin{equation*}
\begin{split}
D (y^i \circ \sigma_1)(0) & = D(u^i \circ \psi \circ \sigma_1)(0) \\
& = D(u^i \circ \psi \circ \phi^{-1} \circ \phi \circ \sigma_1)(0) \\
& = D(u^i \circ \psi \circ \phi^{-1})(\phi (p)) \circ D(\phi \circ \sigma_1 )(0) \\
& = D(u^i \circ \psi \circ \phi^{-1})(\phi (p)) \circ D(\phi \circ \sigma_2 )(0) \\
& = D(u^i \circ \psi \circ \sigma_2)(0) \\
& = D(y^i \circ \sigma_2)(0) \, ,
\end{split}
\end{equation*}
where to go from line 2 to line 3 one uses the chain rule. This is the required equality.
