Let $M$ be a smooth manifold of dimension $n$, and $x\in M$. I want to double check that I understand the definition of the tangent vector space $T_xM$ using velocity vectors correctly.
- We define $T_xM$ as a set of tangent vectors at $x$ where each tangent vector is the equivalence class of curves $\gamma$ that goes through point $x$ and share the same tangent vector. In other words, $$T_xM=\{\gamma^{\prime}(0)|\gamma:I\to M\text{ is a smooth path}\}$$ where $\gamma^{\prime}(0)$ is the equivalence class of curves $\gamma:I\to M$, and we say that $\gamma_1\sim\gamma_2$ iff $(\varphi\circ \gamma_1)^{\prime}(0)=(\varphi\circ \gamma_2)^{\prime}(0)$ for some choice of open chart $(U,\varphi)$ of $M$.
So, I have the following question: at this step, we need to show that our equivalence class does not depend on the choice of the chart $(U,\varphi)$, correct? In other words, let's take two charts $(U,\varphi)$ and $(V,\psi)$, then we want to show that $$\text{if }(\varphi\circ \gamma_1)^{\prime}(0)=(\varphi\circ \gamma_2)^{\prime}(0)\text{, then }(\psi\circ \gamma_1)^{\prime}(0)=(\psi\circ \gamma_2)^{\prime}(0)$$ i.e. if the curves are equivalent via one chart, then they are equivalent in another chart.
emphasized textTo show this, we need just to observe that $$(\psi\circ \gamma_1)^{\prime}(0)=(\psi\circ\varphi^{-1}\circ\varphi\circ\gamma_1)^{\prime}(0)=d(\psi\circ\varphi^{-1})_0((\varphi\circ\gamma_1)^{\prime}(0))=d(\psi\circ\varphi^{-1})_0((\varphi\circ\gamma_2)^{\prime}(0))=(\psi\circ \gamma_2)^{\prime}(0)$$ where $\psi\circ\varphi^{-1}$ is a diffeomorphism since $\varphi$ and $\psi$ are diffeomorphisms. Finally, as a step $(1)$, we defined a tangent space $T_xM$ as a set of equivalence class and we showed that those equivalence classes does not depend on the choice of the chart. So, for the step $(2)$, we need to show that $T_xM$ is actually a vector space.
