definition of differential does not depend on the choice of the curve, differential is a linear map

Question

Let $\varphi: M \to N$ be a differentiable map. How do I show that the definition of the differential $d\varphi_p: T_pM \to T_{\varphi(p)}N$ of $\varphi$ at $p$ does not depend on the choice of the curve and that $d\varphi_p$ is a linear map? Any help would be appreciated, thanks.

Answer 1

We may pass to local coordinates and assume that $M$, $N$ are open subsets of Euclidean spaces for which $p$ and $\varphi(p)$ correspond to $0$. Let $\gamma_1, \gamma_2: (-\epsilon, \epsilon) \to M$ be two curves with $\gamma_1(0) = \gamma_2(0)$ and $\gamma_1'(0) = \gamma_2'(0)$; then for any smooth function $f: M \to \mathbb{R}$ we get that $f \circ \varphi \circ \gamma_1$ and $f \circ \varphi \circ \gamma_2$ have the same derivative at $0$, by the chain rule. Thus the curves $\varphi \circ \gamma_i$ are in the same equivalence class for $i = 1, 2$.

Now to prove linearity. Use the same charts as above, and introduce two curves $\gamma_3$, $\gamma_4$ with $\gamma_3(0) = \gamma_4(0) = 0$ as well $($in the coordinate of $M$$)$, with the property that $\gamma_3$ and $\gamma_4$ are the restrictions of linear maps with respect to our chart for $M$. Fix $\lambda \in \mathbb{R}$; then for sufficiently small $\epsilon' > 0$, $\gamma_5(t) :=\lambda \gamma_3(t) + \gamma_4(t)$ is defined for $t \in (-\epsilon', \epsilon')$ and is also linear. Now the definition of a differentiable map implies that for the coordinate functions on $N$, $\varphi(\lambda \gamma_3(t) + \gamma_4(t)) = \lambda\varphi(\gamma_3(t) + \gamma_4(t) +$ three error terms for first order approximations of $\varphi$, where the error terms are $o(t)$. It follows then, that $\lambda(\varphi \circ \gamma_3)'(0) + (\varphi \circ \gamma_4'(0) = (\varphi \circ \gamma_5)'(0)$ and linearity follows.