Let $(G,\langle\,,\,\rangle)$ a connected Lie group with bi-invariant semi-Riemannian metric. Let $H$ be a closed subgroup, and denote by $h$ the left-invariant metric on the homogeneous space $N=G/H$ such that $\pi:(G,\langle\,,\,\rangle)\to (N,h)$ is a semi-Riemannian submersion.
Under these conditions we have an orthogonal decomposition of the Lie-algebra $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}$, where $\mathfrak{h} = Lie(H)$, and in particular $N$ is reductive. Further, the geodesics of $(N,h)$ through $o=eH$ are exactly $\gamma(t) = \pi \circ exp(tx)$, with $x\in \mathfrak{m}$.
Is it now true that for $G$ noncompact $\pi\circ \exp|_\mathfrak{m}$ is surjective, i.e. $G/H = \pi(\exp\mathfrak{m})$? How could one prove it?
In the case where $G$ is compact it is easy, because then $N=G/H$ is compact and the geodesics through $o$ hit every other point on the manifold.
