Analytically isomorphic fibers. Suppose that $S$ is a non-singular complex projective surface with a fibration $f$ over $\mathbb P^1(\mathbb C)$. Suppose also that:


*

*There are only finitely many points $y_1,\ldots,y_n\in\mathbb P^1(\mathbb C)$ such that $f^{-1}(y_i)$ is a singular curve.

*The non-singular fibers of $f$ are analytically isomorphic.



I would like to prove that all the fibers of $f$ are analytically
  isomorphic.

Do you have any idea?
 A: Is this true?  Two examples:


*

*Let $X = \mathbb P^1 \times \mathbb P^1$ blown up at a point, with projection to $\mathbb P^1$.  All the fibers except one are just $\mathbb P^1$, and the singular fiber is a union of two $\mathbb P^1$ meeting at a point.  It's not analytically isomorphic to the others. OK, this isn't flat, but...

*Consider the family $y^2 = x^3 + t$ (inside $\mathbb A^2 \times \mathbb P^1$).  The smooth fibers ($t \neq 0$) are all isomorphic to $y^2= x^3 +1$ by an easy change of variables.  But over $t=0$ you get a cuspidal cubic.
A: Let $f:X\to C$ be a relatively minimal projective family of curves with $X$ non-singular such that all (but a finite number of) fibres are isomorphic (analytically or algebraically...). Assume the fiber genus of $f$ is at least two. (In genus one you have to be a bit more careful.)
In this case, $f$ is "isotrivial". So that $f$ becomes a trivial family after a finite base-change.
It is a non-trivial fact that the following statements are equivalent (for $f$ a family of curves of genus at least two over a curve)
1) $f$ is isotrivial (i.e.., $f$ becomes a trivial family after a finite base-change)
2) $f$ becomes trivial after a finite \'etale base-change
3) All the fibres of $f$ are isomorphic (and $f$ is smooth).
This is what you were asking for. In genus zero this is not true as mark points out, and in genus one you have to work with split elliptic surfaces (so that $f$ has a section).
