I was wondering if someone could help me with the following problem, any help would be greatly appreciated.
Let $f:M \rightarrow N$ be a $C^{\infty}$ map between smooth manifolds. Given $x \in M$, let $c_{0}, c_{1}$ be $C^{\infty}$ curves defined on open intervals $(a_{0},b_{0}), (a_{1},b_{1})$ with $0 \in (a_{0},b_{0})$ , $0 \in (a_{1},b_{1})$ and $c_{0}(0)=c_{1}(0)=x$. Define an equivalence relation $\sim$ by $c_{0} \sim c_{1}$ $\iff$ $\frac{d}{dt}(\psi_{U}\circ c_{0}(t))\mid_{t=0} = \frac{d}{dt}(\psi_{U}\circ c_{1}(t))\mid_{t=0}$ , where $\psi_{U}:U \rightarrow \psi_{U}(U)\subseteq \mathbb{R}^{m}$ is a chart. Define the tangent space $TM_{x}$ of $M$ at $x$ as the set of all equivalence classes $[c_{0}]$ and define $df_{x}:TM_{x} \rightarrow TN_{f(x)}$ by $[c_{0}] \mapsto [f \circ c_{0}]$. Prove this is a well-defined map, namely if $c_{0} \sim c_{1}$ then $f \circ c_{0} \sim f \circ c_{1}$.
Choose a chart $\phi_V \colon V \rightarrow \mathbb{R}^n$ around $f(x)$ and a chart $\psi_U \colon U \rightarrow \mathbb{R}^m$ around $x$ such that $f(U) \subseteq V$. Then $f \circ c_0 \sim f \circ c_1$ if $$ \frac{d}{dt} \left( \phi_V \circ f \circ c_0 \right)|_{t = 0} = \frac{d}{dt} \left( \phi_V \circ f \circ c_1 \right)|_{t = 0}. $$
We have
$$\frac{d}{dt} \left( \phi_V \circ f \circ c_i \right)|_{t = 0} = \frac{d}{dt} \left( \phi_V \circ f \circ \psi_U^{-1} \circ \psi_U \circ c_i \right)|_{t = 0}. $$
Use the chain rule and the fact that $c_0 \sim c_1$ to prove that $f \circ c_0 \sim f \circ c_1$.