Transitivity of Equivalence Relation of Tangent Vectors on Manifold I'm just starting to learn differential topology, and came across a notion when defining tangent vectors on a smooth manifold that confused me a bit. Let $M$ be a smooth ($C^\infty$) manifold, with $p \in M$, and define a smooth curve at $p$ as a smooth map $\gamma \colon (-1, 1) \to M$ such that $\gamma(0) = p$. Let $(U, \phi)$ be any coordinate chart on $M$ containing $p$. Two smooth maps at $p:\,$ $\gamma_1, \gamma_2$ are called equivalent if $(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_2)'(0)$; this is denoted $\gamma_1 \sim_p \gamma_2$. A tangent vector is defined as an equivalence class of all smooth curves at $p$ under this relation.
The problem I'm having with this is proving that it is an equivalence relation. Obviously, reflexivity and symmetry follow immediately from the definition, but I'm trying to prove the transitivity of this relation. I wrote out a proof but wasn't sure if it was valid because of the use of the chain rule outside Euclidean space. Here it is:
Let $\gamma_1, \gamma_2, \gamma_3 \colon (- 1, 1) \to M$ be smooth curves at $p$. Assume that $\gamma_1 \sim_p \gamma_2$ and $\gamma_2 \sim_p \gamma_3$. Then there exists coordinate charts $(U, \phi), (V, \psi)$ containing $p$ such that:
$$(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_2)'(0) \,\,\text{ and } \,\, (\psi \circ \gamma_2)'(0) = (\psi \circ \gamma_3)'(0)$$
Using the chain rule, the first equality implies that
$$ \phi'(\gamma_1(0)) \cdot \gamma_1'(0) = \phi'(\gamma_2(0)) \cdot \gamma_2'(0)$$
As $\gamma_1(0) = \gamma_2(0) = p$, this means that $\gamma_1'(0) = \gamma_2'(0)$. Likewise, we have that
$$(\psi'(\gamma_2(0)) \cdot \gamma_2'(0) = (\psi'(\gamma_3(0)) \cdot \gamma_3'(0) \Rightarrow \gamma_2'(0) = \gamma_3'(0)$$
This means that $\gamma_1'(0) = \gamma_3'(0)$, meaning that
$$\phi'(\gamma_1(0))\cdot \gamma_1'(0) = \phi'(\gamma_3(0)) \cdot \gamma_3'(0),$$
so that $$(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_3)'(0) \Rightarrow \gamma_1 \sim_p \gamma_3$$
I feel like it isn't legal to use the chain rule there because the codomain of our smooth maps isn't Euclidean space, but I can't think of any other way to prove the transitivity of this relation. Is my proof correct? If not, could someone give me a hint as to how to prove transitivity? Thanks!
 A: Indeed, at this point the use of the chain rule is still illegal because the domains of charts are open subsets of the manifold and you still "don't know" what a derivative means there. Here's how to fix it.
If $\gamma_1\sim_p\gamma_2$ and $\gamma_2\sim_p \gamma_3$, there are $(U,\varphi)$ and $(V,\psi)$ around $p$ with $(\varphi\circ \gamma_1)'(0) = (\varphi\circ \gamma_2)'(0)$ and $(\psi\circ \gamma_2)'(0)=(\psi\circ \gamma_3)'(0)$, alright. Compute $$\begin{align} (\varphi\circ \gamma_3)'(0) &= (\varphi\circ\psi^{-1}\circ \psi\circ\gamma_3)'(0) \\ &= {\rm d}(\varphi\circ \psi^{-1})_0((\psi\circ\gamma_3)'(0)) \\ &= {\rm d}(\varphi\circ \psi^{-1})_0((\psi\circ\gamma_2)'(0)) \\ &= (\varphi\circ \psi\circ \psi^{-1}\circ\gamma_2)'(0) \\ &= (\varphi \circ \gamma_2)'(0) \\ &= (\varphi\circ \gamma_1)'(0).\end{align}$$My use of the chain rule here was legal because I made sure to apply it only for the transition between charts, which is a map between open subsets of Euclidean spaces, and composing a curve in the manifold with a chart gives a curve in Euclidean space.
