Horizontal curves on a submersion $\pi: M \to N$ pass through unique elements of the fibers Let $M$ and $N$ be smooth Riemannian manifolds, and consider a smooth surjective submersion $\pi: M \to N$. We define the vertical tangent space at $x$ by $V_x := \ker d\pi_x$ and the horizontal tangent space at $x$ by $H_x := V_x^\perp$. I've seen that for any curve $\gamma: [a, b] \to N$ and $x \in \pi^{-1}(\{\gamma(a)\})$, there exists a unique curve $\tilde{\gamma}: [a, b] \to M$, called the horizontal lift of $\gamma$, such that $\tilde{\gamma}(a) = x$, $\ \pi \circ \tilde{\gamma} = \gamma$, and $\dot{\gamma}$ is horizontal.
Suppose that $\tilde{\gamma}(b) = y$. Intuitively, it seems that any curve $c: [a, b] \to M$ with $c(a) = x$, $c(b) \in \pi^{-1}(\{\gamma(b)\})$, and $\dot{c}$ horizontal should in fact have $c(b) = y$, as (roughly speaking) a horizontal curve cannot move vertically along the fibers. Is this true? How would I prove it?
It is clear that it holds in the case that $M = \mathbb{R}^m$, $N = \mathbb{R}^{n}$ (with $m > n$ and the standard metrics), and $\pi$ is the canonical projection. Since for any $p \in M$, there exists charts around $p \in M$ and $\pi(p) \in N$ such that the chart representation of $\pi$ takes the form $\pi(x^1, \dots, x^m) = (x^1, \dots, x^n)$, I imagine that this result can be proven locally, at which points curve segments could be glued together somehow to prove the result. Would this strategy work? Is there a coordinate-free way of doing it? I thought an alternative strategy could be to show that closed loops in $N$ are horizontally lifted to closed loops in $M$ (at which point a contradiction could be derived in assuming the proposition doesn't hold), but I'm not sure if this is any more tractable.
 A: This is not true. Here's a counterexample.
Let $M=\mathbb R^3$ with the Riemannian metric for which $(X,Y,Z)$ is an orthonormal basis, where
$$
X = \frac{\partial}{\partial x}+ y \frac{\partial}{\partial z},
\quad
Y = \frac{\partial}{\partial y},
\quad
Z = \frac{\partial}{\partial z},
$$
and let $N = \mathbb R^2$ with its Euclidean metric.
Let $\pi\colon M\to N$ be the projection $\pi(x,y,z)=(x,y)$.
Then the vertical space is spanned by $Z$ everywhere, so the horizontal space is spanned by $X$ and $Y$.
Now consider the curve $\gamma\colon [0,1]\to N$ given by $\gamma(t) = (t,0)$.
Its horizontal lift starting at the origin is $\widetilde \gamma(t) = (t,0,0)$, which ends at $(1,0,0)$.
Another curve $\sigma\colon [0,1]\to N$ connecting the same two points in $N$ is given by $\sigma(t) = (t, t-t^2)$. Its horizontal lift starting at the origin is $\widetilde \sigma(t) = (t,t-t^2, \frac 1 2 t^2 - \frac 1 3 t^3)$, as you can check. (This is just a matter of verifying that $\widetilde \gamma'(t)$ lies in the span of $X$ and $Y$ at each point.) But this lift ends at $(1,0,\frac 1 6)$.
What's going on here is that the horizontal tangent spaces form a distribution on $M$. If that distribution is integrable, then every horizontal curve has to stay in one single leaf, so horizontal curves that start in the same leaf must end in the same leaf. This justifies your intuition that a horizontal curve "can't move vertically." But if it's not integrable there's no way to make sense of "staying on the same level." My example is cooked up so that the horizontal distribution is the simplest non-integrable distribution to write down.
A: Here's another counter example. Let both $M$ and $N$ be the circle $S^1=[0,1]/\sim$. Consider the Riemannian submersion $\pi \colon M \to N$ by $\pi(\theta)=2\theta$. This defines a connected double cover of the circle $S^1 \to S^1$.
Define $\gamma\colon [0,\varepsilon] \to N= [0,1]/\sim$ by $\gamma(\theta)=\theta$. Pick the point $0 \in \pi^{-1}(0)\subset M$.
The unique lift of $\gamma$ is $\tilde{\gamma}\colon[0, \varepsilon] \to M = [0,1]/\sim$ defined by $\tilde{\gamma}(\theta) = \theta/2$. Note that $\pi(\tilde{\gamma}(\theta)) = \gamma(\theta)$ and $\tilde{\gamma}(0)=0$.
Now define the curve $\sigma\colon [0, 1 + \varepsilon] \to M$ by $\sigma(\theta) = \theta/2$. Observe that $\sigma(0)=0$ and $\pi(\sigma(1 + \varepsilon)) =  \pi(\frac{1 + \varepsilon}{2}) = 1 + \varepsilon\in N = [0,1]/\sim$. But $1 + \varepsilon \sim \varepsilon$ so $\pi(\sigma(1 + \varepsilon)) = \gamma(\varepsilon)$.
However we can clearly see that $\sigma(1+\varepsilon) = \frac{1}{2} + \frac{\varepsilon}{2} \not= \frac{\varepsilon}{2} = \tilde{\gamma}(\varepsilon)$.
