Integral curves are horizontal

Suppose we're given an Ehresmann connection on a submersion (or a fiber bundle, but I don't think it's needed) $$\begin{smallmatrix}X\\ \downarrow\\ Y \end{smallmatrix}$$. Given a curve $$\gamma$$ in the base $$Y$$, consider the pullback of the submersion and its connection along the curve. The horizontal bundle of the pulled-back bundle $$\begin{smallmatrix}\gamma ^{\ast}X\\ \downarrow\\ I \end{smallmatrix}$$ is a line subbundle of the tangent bundle $$\begin{smallmatrix}\mathrm T\gamma^{\ast}X\\ \downarrow\\ \gamma^\ast X \end{smallmatrix}$$. Thus we locally have integral curves in $$\gamma^\ast X$$ for the horizontal bundle. These integral curves are also transverse to the fibers of the bundle $$\gamma^\ast X\to I$$.

Write $$c_{(s,x)}(t)\in \gamma^\ast X$$ for the value of at $$t$$ of the integral curve $$c$$ based at $$(s,x)\in \gamma^\ast X$$. How can I prove the first coordinate of $$c_{(s,x)}(t)$$ is always $$s+t$$? I drew a picture, which tells me I should use transversality, but I don't know how to formulate a proof.

(The context: I want to prove that flow along the horizontal bundle defines parallel transport, and for that I need to know the horizontal position of an integral curve at each instance.)

• The horizontal curves are the trajectories of a horizontal vector field which you seem to ignore. This vector field is characterized by two properties. The first is horizontality, as you explain in your question. The second property is that it projects to the vector field $\partial/\partial t$ on $I$. – Amitai Yuval Dec 20 '18 at 20:41
• Dear @AmitaiYuval, thank you for your comment. I posted an answer in an attempt to make sure I understood you well. Would appreciate if you could take a look and point out any errors! – Arrow Dec 20 '18 at 22:21

The horizontal subbundle amounts to a multiple of a global horizontal vector field by a scalar function. Of these it is naturally best to choose a lift of the global vector field $$\partial/\partial t$$ on $$I$$. With respect to this horizontal vector field the integral curves are now defined.
That this field lifts $$\partial/\partial t$$ implies by the chain rule that $$\frac{\partial}{\partial t}(\pi\circ c_{(s,x)})\equiv 1$$. But $$\pi\circ c_{(s,x)}$$ is precisely the map $$f$$ giving the first coordinate of the integral curve $$c$$. By assumption $$f(0)=s$$ and $$f^\prime \equiv 1$$. By uniqueness of solutions to ODE it must then by given by $$f(t)=s+t$$, as desired.