Why is the definition of the element $rv$ of the tangent space chart-independent?

Let's assume we have a $$m$$-dimensional smooth manifold $$\mathscr{M}$$. We can define the tangent space $$T_p\mathscr{M}$$ at a point $$p \in \mathscr{M}$$ as the set of equivalence classes of curves in $$\mathscr{M}$$ going through $$p$$. Then to show that $$T_p\mathscr{M}$$ is a vector space my textbook defines $$r v$$ as

$$r v := [\phi^{-1} \circ (r\phi \circ \sigma)] \, ,$$ where $$r \in \mathbb{R}$$, $$v = [\sigma]\in T_p\mathscr{M}$$ and $$\phi$$ a chart around $$p$$ satisfying $$\phi (p) = 0$$.

I was able to prove this definition doesn't depend on the choice of representative $$\sigma$$, but how do you proceed to prove that it doesn't depend on the choice of chart $$\phi$$?

I will try to outline the main ideas.

First, note that $$\phi(U)$$ being an open subset of $$\mathscr{M}$$ it is itself a smooth manifold of dimension $$m$$. So, instead of looking at $$T_p\mathscr{M}$$ one can look at $$T_0(\phi(U))$$ and the main advantage there is that a chart is not required anymore to define the vector space structure on $$T_0(\phi(U))$$. Indeed, $$\phi(U)$$ already being a subset of $$\mathbb{R}^m$$ a chart is not needed to push the values taken by the curves into $$\mathbb{R}^m$$ to use the vector space structure thereof.

The vector space structure on $$T_0(\phi(U))$$ is defined as follows.

• $$v_1 + v_2 = [\sigma_1] + [\sigma_2] = [\sigma_1 + \sigma_2]$$
• $$rv = r[\sigma] = [r\sigma]$$

Then, one can check that the vector space $$T_0(\phi(U))$$ is isomorphic to $$\mathbb{R}^m$$ with the isomorphism mapping $$v \in \mathbb{R}^m$$ to $$[\sigma_v]$$, where $$\sigma_v(t) := t v$$.

Moreover, we have a bijection $$\phi_*$$ from $$T_p\mathscr{M}$$ to $$T_0(\phi(U))$$ mapping $$v := [\sigma]$$ to $$[\phi \circ \sigma]$$. One can use this bijection to transport the vector space structure on $$T_0(\phi(U))$$ to $$T_p\mathscr{M}$$ making $$\phi_*$$ into an isomorphism of vector spaces. The resulting vector space structure on $$T_p\mathscr{M}$$ is precisely the one defined partially in your post.

Finally, given another chart $$(V, \psi)$$ with $$p \in V$$ and $$\psi(p) = 0$$, one can define in the same way a vector space structure on $$T_p\mathscr{M}$$ using $$\psi_*$$. But what has been shown above is that both vector space structures are the same, i.e. they are isomorphic, both being isomorphic to $$\mathbb{R}^m$$ (through $$T_0(\phi(U))$$ and $$T_0(\psi(V))$$, respectively).