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In about any introductory book on manifolds it is shown that if $M$ is a smooth manifold and $X$ a vector field on $M$, then $X$ is smooth $\Leftrightarrow$ the coefficient functions $X^i$ (where $X=X^i\partial_i$) are smooth in some chart induced frame around each point $\Leftrightarrow$ the $X^i$ are smooth in any chart induced frame.

I'm wondering if one can prove something similar for vector fields along curves. Let $\gamma:I\to M$ be a smooth curve, with $I\subset\mathbb R$ open, and let $V:I\to TM$ be a vector field along $\gamma$, i.e, $V(t)\in T_{\gamma(t)}M$ for all $t\in I$. $V$ is called smooth if it is smooth as a map between manifolds. Now it would be very convenient if the following statement were true:

Hypothesis: V is smooth $\Leftrightarrow$ the coefficient functions $V^i:I\to\mathbb R$ (where $V(t)=V^i(t)\partial_i$) are smooth in some chart induced frame around each point $\Leftrightarrow$ the $V^i$ are smooth in any chart induced frame.

Is this indeed a true statement? If so, I would very much like to see a proof of it.

Some remarks: 1) The reason that I want to prove the hypothesis is that I need the result for proving that every smooth vector field along a (regular) curve has a local extension to a smooth vector field on $M$. So please don't use the latter fact in the proof of the hypothesis. 2) Preferably also do not use the machinery of pull-back bundles in the proof.

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I'll answer my own question. The hypothesis needs to be adjusted a little bit to make things precise, but the general idea is correct. A precise statement would be the proposition below, which I'll prove in this answer using the following lemma.

Lemma. Let $V:I\to TM$ be a vector field along a smooth curve $\gamma:I\to M$, with $I\subset\mathbb R$ open such that $\gamma(I)\subset U$, where $(U,\phi)$ is a chart on $M$. Write $V(t)=V^i(t)\partial_i|_{\gamma(t)}$ for $t\in U$. Then $V$ is smooth if and only if $V^i\in C^\infty(I)$ for all $i$.

Proof. The chart $(U,\phi)$ induces a chart $\widetilde\phi:TU\to \phi(U)\times \mathbb R^n$ on $TM$, namely \begin{align*} \widetilde\phi(v^i\partial_i|_p) = (x^1(p),\dots,x^m(p), v_1,\dots,v_m) \end{align*} for all $p\in U$, $v^i\in\mathbb R$, where $x^i$ are the component functions of $\phi$. Since $\widetilde\phi$ is a diffeomorphism, smoothness of $V$ is equivalent to that of the map $\widetilde\phi\circ V:I\to \phi(U)\times\mathbb R^n$. If we now write $V(t)=V^i(t)\partial_i|_t$, then $ \widetilde\phi \circ V (t) = \widetilde\phi(V^i(t)\partial_i|_{\gamma(t)})=\left(x^1(\gamma(t)),\dots,x^m(\gamma(t)), V^1(t),\dots,V^m(t)\right)$. The first $m$ functions are all smooth functions of $t$, so $\widetilde\phi\circ V$ is smooth if and only if the $V^i$ are all smooth functions of $t$, giving the result. QED

Proposition. Let $V:I\to TM$ be a vector field along a smooth curve $\gamma:I\to M$, with $I\subset\mathbb R$ open. The following are equivalent.

(a) $V$ is smooth.

(b) There exists an open cover $\mathcal U$ of $\gamma(I)$, comprised of charts on $M$, such that for every chart $(U,\phi)\in\mathcal U$ there exists an open subset $J\subset I$ with $\gamma(J)\subset U$ and the coefficient functions $V_i:J\to\mathbb R$ of $V$ with respect to that chart are smooth.

(c) For any chart $(U,\phi)$ on $M$ with $U\cap \gamma(I)\neq 0$ there exists an open subset $J\subset I$ with $\gamma(J)\subset U$, and for any such $J$ the coefficient functions $V_i:J\to\mathbb R$ of $V$ with respect to that chart are smooth.

Proof. (a $\Rightarrow$ c): Let $(U,\phi)$ have nonempty intersection with $\gamma(I)$. Then there is $t\in I$ such that $\gamma(t)\in U$. By continuity of $\gamma$ there exists an open neighborhood $J\subset I$ of $t_0$ such that $\gamma(J)\subset U$. For any such $J$ the lemma implies that the coefficient functions $V^i:J\to\mathbb R$ in the chart are all smooth. (c $\Rightarrow$ b): Trivial. (b $\Rightarrow$ a): For $t_0\in I$, pick a chart $(U,\phi)$ around $\gamma(t_0)$ and a subset $J\subset I$ such that $\gamma(J)\subset U$ and the coefficient functions $V^i:J\to \mathbb R$ are all smooth. Then by the lemma, $V$ is smooth on $J$ and in particular at $t_0$. Since $t_0$ was arbitrary, $V$ is smooth. QED

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