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Let $f:X\rightarrow C$ be a fibration of complex manifold over a smooth curve $C$. Let $D$ be an irreducible component of singular fibers. How can one prove that $X$ does not admit any fibration with general fiber class $D$? Or any counterexample?

I think this is intuitively true because $D$ is an irreducible component of singular fibers, so it cannot $move$. But I don't know how to prove it.

I would appreciate it if you can provide me a proof or reference.

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    $\begingroup$ Without further hypothesis, I don't see how you would prove something like this. Consider the following example. Suppose $g:Y\to C$ is a surjective map with general fibre $D$, blow up $Y$ along a point on $D$ to get $X$. Now $D$ appears as a component of singular fibre of $X$, yet all the other smooth fibres are deformation equivalent to $D$. $\endgroup$
    – Donu Arapura
    Commented Mar 17, 2013 at 15:24
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    $\begingroup$ Correction: I guess I was thinking of $Y$ as a surface. In general, you would need to blow up along a smooth codim 2 subset of $X$ supported on $D$. $\endgroup$
    – Donu Arapura
    Commented Mar 17, 2013 at 15:26

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