TITLE: Basic functions and vertical tangent vector
QUESTION [1 upvotes]: If $\pi:E\rightarrow M$ is a fiber bundle with connected fibers and $f$ is a smooth function on $E$, I want to prove that if $v(f)=0$ for all vertical tangent vector $v$ then $f$ is a basic function i.e. $f=g\circ\pi$ for some smooth function $g$ on $M$.
I think that in fibre coordinates on $E$, $v=v(x,y)\partial_y$ where $y$ are the coordinates on the fiber $F$. So, $v(f)=0$ implies that $f$ is (locally) costant on the fiber. Say that, if $\phi^{-1}(U\times F)$ where $\phi$ is a local trivialization and $U$ an open subset of $M$, we can say that $f=f|_{\phi^{-1}(U)}\circ\pi$.
What do you think? Make sense?
Thanks for all help me.

REPLY [0 votes]: In fibre coordinates on $E$, $v=v(x,y)\partial_y$ where $y$ are the coordinates on the fiber $F$. So, $v(f)=0$ implies that $f$ is (locally) costant on the fiber. Say that, if $\phi^{-1}(U\times F)$ where $\phi$ is a local trivialization and $U$ an open subset of $M$, we can say that $f=f|_{\phi^{-1}(U)}\circ\pi$.
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