The purpose of the present note is to show that holomorphic fiber bundles with hyperbolic fiber are almost trivial : They are trivial if the base is simply connected, and have a simple structure otherwise. Kobayashi [2],[3] has introduced the notion of a hyperbolic complex manifold.We need not be concerned with the definition here, but only with the following consequence [3], [4]: If M is a hyperbolic complex manifold, p a point of M, and U a neighborhood of /, then there is a <5>0 such that any holomorphic map <f> of the unit disk into M with (f,(0)=p maps the disk {z:|z|<} into U. From this, it follows that, if iv1, , wn are coordinates at/, and / an integer, then there are constants Al depending only on / and the choice of coordinates such that, if-p is any map of the unit disk into M with cf>(0)=p, then \dlwidzl\<Al at z=0.The crux of our argument is the following proposition, which generalizes a result of Kobayashi's that a complex Lie group (of dimension greater than 0) cannot act holomorphically on a hyperbolic manifold. Proposition.Let M and N be complex manifolds with M hyperbolic.Let F=F(w, z) be a holomorphic map of MxN into M. Suppose that, for some z0 e N, the restriction of F to Mx{z0} is a biholomorphic map onto M. Then F is independent of z; i.e., F(w, z)=F(w, z0). Proof.It suffices to prove the proposition for the case that N is a ball, and hence for the case that Nis a disk A, which we take to be the unit disk with z0=0.Without loss of generality, we may assume F(w, z0)=w.Near a point (w0, z0), the map F has a power series expansion of the form F\w, z) = w' + a\w)zl + 0(zl+1).-
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