linear isomorphism between $T_p(M)$ and $T_\mathbf{x}(\mathbb{R}^k)$

Question

Assume the following definition:

Given $p\in M$, a tangent vector to $M$ at $p$ is a function $\mathbf{v}$ that assigns, to each coordinate patch $\alpha : U\to V$ in $M$ about $p$, a column matrix of size $k$ by $1$ which we denote $\mathbf{v}(\alpha)$. if $\alpha_0$ and $\alpha_1$ are two coordinate patches about $p$, we require that $$ \mathbf{v}(\alpha_1)=Dg(\mathbf{x}_0) \cdot \mathbf{v}(\alpha_0),$$ where $g=\alpha_1^{-1}\circ\alpha_0$ is the transition function and $\mathbf{x}_0=\alpha^{-1}_0(p)$.

Then it is stated that the map $\mathbf{v}\rightarrow (\mathbf{x};\mathbf{v}(\alpha))$, which carries $T_p(M)$ onto $T_\mathbf{x}(\mathbb{R}^k)$, is a linear isomorphism. but if $(\mathbf{x};\mathbf{a})$ be an arbitrary element of $T_\mathbf{x}(\mathbb{R}^k)$ then what is the action of the inverse of this isomorphism on it?

Answer 1

Fix a coordinate chart $\alpha: U \to V$ where $V \subset \mathbb{R}^n$, now fix $p \in U$ and let $x=\alpha(p) \in \mathbb{R}^n$. Your map could be written as:

$F_\alpha(v_p) = (\alpha(p),v(\alpha))$.

Now given $(x,a) \in T_x(\mathbb{R}^n)$ then $F_\alpha ^{-1}(x,a) \in T_p(M)$ is the tangent vector at $p=\alpha^{-1}(x)$ defined by $[F_\alpha ^{-1}(x,a)] (\alpha) = a$. According to the definition you provided, the tangent vector is uniquely specified only if we know its value for every coordinate chart $\alpha$ not just one, but the value of the tangent vector in all all other coordinate charts can be obtained via the transformation law you provided, hence we have indeed defined a tangent vector.