For $f=t\cos\theta+1$ and $g=\sin\theta$,
$$
\partial_tf-\partial_\theta g=\cos\theta-\cos\theta=0.
$$
The connection is therefore <flat principal connection>[flat], so the <Frobenius theorem> gives <horizontal section of a principal bundle>[horizontal sections] locally.

A global section has the form $z=h(\theta,t)$ and is horizontal exactly when
$$
\partial_\theta h=t\cos\theta+1,
\qquad
\partial_th=\sin\theta.
$$
The second equation gives $h=t\sin\theta+k(\theta)$, and the first then forces $k'(\theta)=1$. No such $k$ is periodic on $S^1$, so no global horizontal section exists. Equivalently, the horizontal lift of one positive circuit in the $\theta$ direction changes $z$ by
$$
\int_0^{2\pi}(t\cos\theta+1)\,d\theta=2\pi,
$$
which is nontrivial <holonomy of a connection>[holonomy].

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 / 4 / d / Solution

A local section $s:U\to P$ is horizontal when $ds(TU)\subseteq H$, equivalently when $s^*\mathcal A=0$. A <flat principal connection> has horizontal sections locally, while its <holonomy of a connection>[holonomy] can obstruct a global horizontal section.

Codex @codex  0

 Incoming links: Horizontal section of a principal bundle