Proving the well-definedness of $df$: How to place the limit inside the argument of $\psi \circ f \circ \phi^{-1}$?

Question

In the book of Chillingworth, the author defines the tangent space of a point $p$ in the smooth manifold $M$ as the set of all conjugacy classes of smooth paths with $\alpha (o) = p$ s.t $\alpha \sim \beta$ iff $$\lim_{t\to 0} \frac{\phi \circ \alpha(t) - \phi \circ \beta (t)}{t } = 0,$$ where $\phi$ is a local coordinate chart around $p\in M$.

Now, given a smooth map from $f : M \to N$, I'm trying to show that $df: T_p M \to T_{f(p)} N$ given by $[\alpha] \mapsto [f\circ \alpha]$ is a well-defined map.

However, to show that, I need to show that $$\lim_{t\to 0} \frac{\psi \circ f \circ \alpha(t) - \psi \circ f \circ \beta (t)}{t } = 0,$$ where $\psi$ is a local coordinate chart around $f(p)$ and $[\alpha] = [\beta]$.

In $\mathbb{R}^n$, I'm aware of this property, but even if I modify the limit as $$\lim_{t\to 0} \frac{\psi \circ f \circ \phi^{-1} \circ [\phi \circ \alpha(t) - \phi \circ \beta (t)]}{t } = 0,$$ how to put the factor $t$ in the denominator inside the argument of $\psi \circ f \circ \phi^{-1}$ ?

Answer 1

$\psi\circ f\circ \phi$ is differentiable in $\phi(p)$ so you have that

$$\lim_{t\to 0} \frac{\psi \circ f \circ \alpha(t) - \psi \circ f \circ \beta (t)}{t } $$

$$= \lim_{t\to 0}\frac{\psi\circ f\circ \phi^{-1}( \phi\circ \alpha(t))-\psi\circ f\circ \phi^{-1}( \phi\circ \beta(t))}{t}$$

$$=\lim_{t\to 0}\frac{\psi\circ f\circ \phi^{-1}( \phi\circ \alpha(t))- \psi\circ f\circ \phi^{-1}( \phi\circ \alpha(0)) + \psi\circ f\circ \phi^{-1}( \phi\circ \alpha(0)) - \psi\circ f\circ \phi^{-1}( \phi\circ \beta(t))}{t}$$

$$=\lim_{t\to 0}\frac{\psi\circ f\circ \phi^{-1}( \phi\circ \alpha(t))- \psi\circ f\circ \phi^{-1}( \phi\circ \alpha(0)) + \psi\circ f\circ \phi^{-1}( \phi\circ \beta(0)) - \psi\circ f\circ \phi^{-1}( \phi\circ \beta(t))}{t}$$

$$=\frac{d}{dt}_{t=0}(\psi\circ f\circ \phi^{-1}\circ \phi\circ \alpha) (t)- \frac{d}{dt}_{t=0}(\psi\circ f\circ \phi^{-1}\circ \phi\circ \beta) (t) $$

$$=J(\psi\circ f\circ \phi^{-1})(\phi(p)) \frac{d}{dt}_{t=0}(\phi\circ \alpha)(0)-J(\psi\circ f\circ \phi^{-1})(\phi(p)) \frac{d}{dt}_{t=0}(\phi\circ \beta)(0)$$

but by our hypothesis

$$\frac{d}{dt}_{t=0}(\phi\circ \alpha)(0)= \frac{d}{dt}_{t=0}(\phi\circ \beta)(0)$$

so you have that

$$\lim_{t\to 0} \frac{\psi \circ f \circ \alpha(t) - \psi \circ f \circ \beta (t)}{t } =0$$