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

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 linear 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 $$f(p)$$.

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 $$\psi \circ f \circ \phi^{-1}$$ ?

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