Intuition of a smooth vector field I read the following statement in pg180, of John Lee's smooth manifold. I am not having much intuition. 

Let $M$ be a smooth manifold, $X:M \rightarrow TM$ be a rough vector field (i.e. it does not have to be smooth map). Then the following are equivalent:
  
  
*
  
*$X$ is smooth. 
  
*For every $f \in C^\infty(M)$ the map $Xf$ is also smooth on $M$. 
  

I understand the proof. But what is the geometric meaning/ significance of this statement? Is there an analogue somewhere else? 
 A: Let us work locally around a point $p$, which is enough since the notion of differentiability is local. Assume you have local coordinates $(x^{\mu}), \mu =1,\dots,\dim M=n$. The vector field $X$ can be written as $X = \sum_{\mu=1}^nX^{\mu}\partial_{\mu}$
and, by definition, it is smooth if all the functions $X^{\mu} \colon M \to \mathbb{R}$ are smooth. Thinking geometrically, this tells you that the vector $X_p = \sum_{\mu=1}^n X^{\mu}(p)\partial_{\mu|p}$ does not change much when you move around $p$. 
En passant, observe that if $q \in M$ is a point 'close' to $p$, there is no making sense of the difference $X_q-X_p$, because the two vectors belong to different vector spaces, i.e. $T_qM$ and $T_pM$ respectively. You should think topologically here, considering that the topology of $TM$ is the natural one obtained from the base manifold $M$. 
Now, taking a smooth function $f$ defined around $p$, the function $Xf$ is nothing but the derivative of $f$ in the direction of $X$. Since $X$ and $f$ are both smooth objects, one would expect that $Xf$ is also a smooth function, and that is what the proposition is telling you. In other words, if you move around $p$ and $X$ does not change much, also the derivative of $f$ in the direction of $X$ will not change too much. I hope this is what you were looking for.
