Derivative between two smooth manifolds is well defined I am reading a (non-English) book in Differential Geometry. The author has introduce this mapping: $T_x f : T_x M \to T_{f(x)} N$ by $T_x f ([\alpha]_x) = [f \circ \alpha]_{f(x)}$ where $[\alpha]_x$ is the equivalence class of smooth curves $\alpha : (-\epsilon, \epsilon) \to M$ passing through x. And $f : M \to N$ is a smooth map between the two smooth manifolds.
My question is how this can be a well defined map? That is how $\alpha \sim \beta$ (which by definition equals $\lim_t \dfrac{\| \phi \circ \alpha (t) - \phi \circ \beta (t)\|}{t} =0$ for some $(U, \phi)$) implies $f \circ \alpha \sim f \circ \beta$?
I believe such things require detailed proof but the book considers it a trivial/immediate fact!
My first thought is to show that $\lim_{t \to 0} \dfrac{\| \phi \circ \alpha (t) - \phi \circ \beta (t)\|}{t} =0$ implies that $\lim_{t \to 0} \dfrac{\| \phi \circ f \circ \alpha (t) - \phi \circ f \circ \beta (t)\|}{t} =0$. Though it is based on the definition, but a comment below says that it is incorrect!
 A: Here is a suggestion: At first, forget about charts and assume that $M$ and $N$ are open subsets in ${\mathbb R}^m$ and ${\mathbb R}^n$ respectively. Then the notion of equivalence between smooth curves $\alpha, \beta$ (in $M$) at $x=\alpha(0)=\beta(0)$ that your book uses reads:
$$
\lim_{t\to 0} \frac{\alpha(t)-\beta(t)}{t}=0 \iff 
 \lim_{t\to 0} \frac{\alpha(t)-x - \beta(t)+x}{t}=0
$$
Using the definition of the derivative (and the limit rule that informally reads "limit of the difference is the difference of limits") , this condition is equivalent to
$$
\alpha'(0)-\beta'(0)=0 \iff \alpha'(0)=\beta'(0).  
$$
The same interpretation, of course, applies to the paths $\tilde\alpha=f\circ \alpha, \tilde\beta=f\circ \beta$. Now, your task is to show:
$$
\alpha'(0)=\beta'(0) \Rightarrow \tilde\alpha'(0)= \tilde\beta'(0). 
$$
If you know the Chain Rule in multivariable calculus, you should be able to establish this implication easily. This would mean that for $x=\alpha(0)=\beta(0)$, $y=\tilde\alpha(0)=\tilde\beta(0)$,
$$
[\alpha]_x= [\beta]_x \Rightarrow [\tilde\alpha]_y= [\tilde\beta]_y,
$$
as required. Once you are done with this part, you should think about the charts on general manifolds and how the above observation implies the desired conclusion in general.
