I am trying to prove the following. (I posted this on [math.se](https://math.stackexchange.com/questions/4568521/initially-horizontal-geodesic-is-always-horizontal) with no success)



> 
> Let $E,B$ be Riemannian manifolds. Suppose
> $\pi: E\to B$ is a Riemannian submersion.
> 
> 
> For each $x\in E$, define $V\_x E = \ker \pi\_{\*}$ and $H\_xE = (V\_x
>  E)^{\perp}$. $W\in T\_x E$ is said to be *horizontal* if $W\in H\_x E$.
> 
> 
> Let $\gamma:[0,1]\to E$ be a geodesic curve such that $\gamma'(0)$ is
> horizontal. Prove that $\gamma'(t)$ is horizontal for all $t\in
>  [0,1]$.
> 
> 
> 


**My attempt**


I mostly followed the proof [here](https://www.ime.usp.br/%7Egorodski/teaching/mat5771-2021/fabricio-submersions.pdf)(Theorem 3.6).


Firstly, for any smooth curve $\gamma:[0,1]\to B$, there is always a locally defined *horizontal lift* to $E$. That is, for any $p\in \pi^{-1}(\gamma(0))$, there is $\epsilon>0$ and a smooth curve $\gamma\_E:[0,\epsilon) \to E$ such that $\pi\circ \gamma\_E = \gamma$, $\gamma\_E(0)=p$ and $\gamma\_E'$ is always horizontal.


So, we take a geodesic on $B$ with initial condition $\gamma\_B(0)=\pi\circ\gamma(0), \gamma\_B'(0)=\pi\_{\*}\gamma'(0)$. We can do so (at least locally) through the uniqueness and existence of geodesic.


Then, take a horizontal lift $\gamma\_E:I\to E$ with $\gamma\_E(0)=\gamma(0)$. Here, $I$ is an open subset of $[0,1]$ containing $0$.


We can prove that $\gamma\_E$ is a geodesic. (This argument is not the main topic of this question.)


Then, by the uniqueness of geodesic, we have $\gamma\_E(t) = \gamma(t)$ for $t\in I$.


**Here is the question.** It seems that we need to prove that $I$ is a closed(and therefore clopen) set of $[0,1]$, too. Then since $[0,1]$ is connected we have $I=[0,1]$. But I do not know how to do it.


For example, [this book](https://www.google.com/books/edition/Riemannian_Geometry/wEcBBQAAQBAJ?hl=en&gbpv=1&dq=initially+horizontal++submersion&pg=PA101&printsec=frontcover) on the proof of Proposition 2.109, (ii),


(Here $\tilde{c}$ corresponds to our $\gamma$, and $c$ corresponds to $\gamma\_B$)



> 
> ...Hence the set of parameters where the geodesic $\tilde{c}$ is
> horizontal, and where it is a lift of $c$ is an open set containing
> $0$. These two conditions being also closed, they are satisfied on the
> maximal interval of definition of $\tilde{c}$.
> 
> 
> 


Any help?



Initially horizontal geodesic is always horizontal
After choosing local coordinates, by the implicit function theorem (I'm omitting a bunch of technical computations) there is a smooth function $\varphi:TE \to TE$ such that $\varphi(x,-): T\_x E \to T\_x E$ is the projection to $V\_x E$.


Let $I\subseteq [0,1]$ be the set on which $\gamma'$ is horizontal: this is the set on which $\varphi\circ \gamma' = 0$ (the zero section). As a composition of two smooth functions ($\gamma'$ is a smooth function from $[0,1]\to TE$ and $\varphi$ is smooth), the function $\varphi\circ\gamma'$ is continuous, and hence $(\varphi\circ\gamma')^{-1}(0)$ is closed (the zero section is a closed subset of $TE$).


