Holomorphic fiber bundle

Remarks

The underlying smooth map is a fiber bundle in the usual sense. On overlaps the trivializations differ by maps $(b, f) \mapsto (b, g_{ij}(b)(f))$, where each $g_{ij}(b)$ is a biholomorphism of $F$ that depends holomorphically on $b$. When $F = \C^r$ and each $g_{ij}(b)$ is linear, the bundle is a holomorphic vector bundle.

Example: the map $\pi \colon \C^2 \setminus \set{0} \to \mathbb{P}^1$, $(z_0, z_1) \mapsto [z_0 : z_1]$, with fibre $F = \C^*$. Over $U_0 = \set{z_0 \ne 0}$ the map $(z_0, z_1) \mapsto ([z_0 : z_1], z_0)$ is a biholomorphism $\pi^{-1}(U_0) \to U_0 \times \C^*$, with inverse $([1 : w], c) \mapsto (c, c w)$, and similarly over $U_1$.

Non-example: $\pi \colon \C^2 \to \C$, $(z, w) \mapsto zw$. Each fibre over $t \ne 0$ is biholomorphic to $\C^*$, but the fibre over $0$ is the union of the two axes, which is not a manifold. So no neighbourhood of $0$ has a trivialization.

By a theorem of Fischer and Grauert, a proper holomorphic submersion between connected complex manifolds whose fibres are all biholomorphic to one another is a holomorphic fiber bundle.