I'm studying Newman's Semi-Riemannian Geometry book and covered the proof of this theorem (using Einstein summation convention throughout):
A vector field $X$ on a smooth manifold $M$ is smooth iff its components are smooth in any coordinate chart.
The $\implies$ proof there is a bit contrived for me.
Let $X=X^i(\partial/\partial x^i)$ be the local coordinate expression of $X$ w.r.t. chart $(U,x^i)$, and let $p\in U$. Since $x^i$ is a function in $C^{\infty}(U)$, (by some previous theorem) there is a function $\tilde x^i$ in $C^{\infty}(M)$ and a neighborhood $\tilde U\subseteq U$ of $p$ in $M$ such that $x^i$ and $\tilde x^i$ agree on $\tilde U$. Thus $$X(\tilde x^i)=\alpha^j\frac{\partial}{\partial x^j}(\tilde x^i)=\alpha^i$$ on $\tilde U$. Since $X$ is smooth and $\tilde x^i\in C^{\infty}(M)$, then $X(\tilde x^i)\in C^{\infty}(M)$. So $\tilde\alpha^i$ is smooth on $\tilde U$, and since $p$ was arbitrary, $\alpha^i$ is smooth on $U$.
Now my question is what's the purpose behind this elaborate construction of $\tilde U$? Can't we just do it like this:
Let $X$ be smooth and $(U,x^i)$ be some chart. Let $\{\partial/\partial x^i\}$ ($i=1,\ldots,n$) be a local frame in that chart, and let $\alpha^i$ be components of $X$ w.r.t. that chart. Act $X$ on the $x^j$ coordinate function to get: $$X(x^j)=\alpha^i\frac{\partial}{\partial x^i}(x^j)=\alpha^j$$
Since the LHS is a smooth vector field acting on a $C^{\infty}(U)$ function, the RHS is also a $C^{\infty}(U)$ function.
Isn't this a much simpler proof of the $\implies$ statement? Anything I'm missing?