Question: Prove that a vector field $X$ is smooth if and only if its coordinates (or components) $X_i$ are smooth for all chart of manifold $M$.
I have the solution of the forward direction, if vector field $X$ is smooth then its coordinates (or components) $X_i$ are smooth for all chart of manifold $M$ by taking the charts $(U,\varphi)$ of $M$ and $(\pi^{-1}(U),\Phi)$ of $TM$, then the corresponding local representation
$$\Phi\circ X\circ \varphi^{-1}: \varphi(U)\rightarrow \Phi(\pi^{-1}(U))=\varphi(U)\times\mathbb R $$ is smooth, where
$$\pi:TM\rightarrow M,\quad \pi(v)=x\quad\text{if }v\in T_xM$$ $$\Phi:\pi^{-1}(U)\rightarrow \varphi(U)\times\mathbb R^m$$ defined by $\Phi(v):=(\varphi(\pi(v)),\bar\varphi(v))$ where, $$\bar\varphi:T_xM\rightarrow\mathbb R^m,\quad \bar\varphi(v):=(\varphi\circ\alpha)'(0)$$
And looking by $\left(\Phi \circ X \circ \varphi^{-1}\right)(a)$ we have the local representation is smooth.
But I couldn't understand how to solve the reverse direction. Any hint or solution will be appreciated.
A map $f \colon M^m \to N^n$ between two manifolds is smooth if and only if for any $p\in M$, there exists a chart $(U,\varphi)$ around $x\in U \subset M$ and a chart $(V,\psi)$ around $f(p) \in V\subset N$ such that the map $$ \psi\circ f \circ \varphi^{-1}\colon \varphi(U) \subset \Bbb R^m \to \psi(V)\subset \Bbb R^n $$ is smooth in the usual sense. Note that by the usual compatibility of charts, the "there exists" can be substitute with "for all" in the above definition.
Let $X\colon M^m \to TM$ be a vector field. Let $\pi \colon TM \to M$ be the canonical projection. Let $(U,\varphi)$ be a chart on $U$, and $(\pi^{-1}(U),\Phi)$ be the associated chart of $TM$. In these charts, $X$ has the expression $$ \begin{array}{r|ccc} \Phi\circ X \circ \varphi^{-1} \colon & \varphi(U) & \longrightarrow & \Phi(\pi^{-1}(U)) = \varphi(U)\times \Bbb R^m \\ & (x^1,\ldots,x^m) & \longmapsto & \left(x^m,\ldots,x^m,X^1(x^1,\ldots,x^m),\ldots,X^m(x^1,\ldots,x^m)\right). \end{array} $$ Note that these $\{X^j\}$ are the same as in the usual expression $X = \sum_{j=1}^m X^j \frac{\partial}{\partial x^j}$.
Now, if $X$, is smooth, then for all chart of $M$ and $TM$, the composition should be smooth. Applying to the pair of charts $(U,\varphi)$ and $(\pi^{-1}(U),\Phi)$ shows that the components $\{X^j\}$ of $X$ are smooth.
Reciprocally, assume that $\{X^j\}$ are smooth. Then around any point, we have found a specific pair of charts $(U,\varphi)$ and $(\pi^{-1}(U),\Phi)$ in which $\Phi\circ X \circ \varphi^{-1}$ is smooth. It follows that $X$ is smooth.