Differentiability of a map between manifolds I have a question about this answer.
Let $M$ be a a manifold and let $(U, (x^1, \dots, x^n))$ be a chart around a fixed point $m \in M$. The goal is to define a vector field $F$ such that $F(m) = v$ for a fixed $v \in T_mM$. 
One can define the map $\Phi(w) = (q, [w])$ from $TM|_U$ to $U\times\mathbb{R}^n$, where $q = \pi(w)$ and $[w]$ is the coordinate vector of $w$ with respect to the basis $\{\frac{\partial}{\partial x^1}|_q, \dots, \frac{\partial}{\partial x^n}|_q\}$ of $T_qM$. Define $F: U \to TM|_U$ as $F(p) = \Phi^{-1}((p, g(p)[v]))$ where $g$ is a certain plateau function around $m$. 
My question is: how can I prove that $F$ is differentiable?
 A: Let $\alpha = (x^1, \dots, x^n) : U \to \mathbb{R}^n$ and $\beta := (\alpha, \operatorname{id}_{\mathbb{R}^n})\circ\Phi : TM|_U \to \mathbb{R}^n\times\mathbb{R}^n$. Note that $\beta$ is the chart on $TM|_U$ induced by the chart $\alpha$ on $U$.
Let's consider the map $\hat{F} := \beta\circ F\circ\alpha^{-1} : \mathbb{R}^n \to \mathbb{R}^n\times\mathbb{R}^n$; this is the coordinate expression of the map $F$. The map $F$ is smooth if and only if $\hat{F}$ is smooth (this is the definition of smoothness of $F$). Let $\alpha^{-1}(p) = (y^1, \dots, y^n)$ and $[v] = (v^1, \dots, v^n)$, then we have
\begin{align*}
\hat{F}(y^1, \dots, y^n) &= (\beta\circ F\circ\alpha^{-1})(y^1, \dots, y^n)\\
&= (\beta\circ F)(\alpha^{-1}(y^1, \dots, y^n))\\
&= (\beta\circ F)(p)\\
&= \beta(F(p))\\
&= \beta(\Phi^{-1}((p, g(p)[v]))\\
&= ((\alpha, \operatorname{id}_{\mathbb{R}^n})\circ\Phi)(\Phi^{-1}((p, g(p)[v])))\\
&= ((\alpha, \operatorname{id}_{\mathbb{R}^n})\circ\Phi\circ\Phi^{-1})((p, g(p)[v]))\\
&= (\alpha, \operatorname{id}_{\mathbb{R}^n})((p, g(p)[v]))\\
&= (\alpha(p), g(p)[v])\\
&= (y^1, \dots, y^n, g(\alpha^{-1}(y^1, \dots, y^n))[v])\\
&= (y^1, \dots, y^n, (g\circ\alpha^{-1})(y^1, \dots, y^n)v^1, \dots, (g\circ\alpha^{-1})(y^1, \dots, y^n)v^n).
\end{align*}
Note that $g$ and $\alpha^{-1}$ are smooth by assumption, so $g\circ\alpha^{-1}$ is smooth. As each of the component functions of $\hat{F}$ is smooth, $\hat{F}$ itself is smooth and therefore so is $F$.
