Lemma 5.43.4074D8F716632

Proof. Fix a Hermitian metric on $M$. The orthogonal complement of $\ker \mathrm dp$ gives a smooth horizontal lift of tangent vectors on $Y$. Around $y_0\in Y$ choose a coordinate ball $B$ centered at $y_0$, with compact closure inside a larger coordinate ball. For $y\in B$, lift the straight segment from $y_0$ to $y$. The ordinary differential equation for horizontal transport has a solution along the whole segment, since its projection remains in the compact closure of $B$, whose inverse image is compact. Smooth dependence on the starting point and on $y$, together with transport along the reversed segment, gives a smooth trivialization

over $B$.

The compact manifold $p^{-1}(y_0)$ has finitely many connected components. The trivialization partitions $p^{-1}(B)$ into finitely many open and closed pieces, each of which meets every fiber in one connected component. On overlaps, two such local partitions identify the same fiber components, and their labels differ by locally constant permutations. Glue one copy of each base ball for every component, using these identifications. Equivalently, take the quotient of $M$ that identifies points in the same connected component of a fiber. The local partitions make this quotient a Hausdorff covering space: distinct points over one base point lie on different sheets, and points over different base points are separated by their images in $Y$.

Call the quotient $Y'$ and its projection $r$. Coordinates of $Y$ on each sheet give holomorphic charts with holomorphic transition maps, so $r$ is a local biholomorphism. It is a finite covering, and $Y'$ is compact because a finite covering of a compact space is compact. Locally, the quotient map $\pi $ is $p$ restricted to one of the open pieces, so it is a holomorphic submersion with the asserted connected fibers. Surjectivity of $\pi $ and connectedness of $M$ make $Y'$ connected. $\square$