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EDIT (after the comment by Jason Starr): Let $X$ be a complex algebraic (or, more generally, analytic) variety, possibly singular and non-compact. Let $f\colon X\to D^*$ be a proper algebraic morphism to a complex smooth curve $D$ with a removed point $p\in D$, i.e. $D^*=D\backslash \{p\}$. (In the complex analytic situation $D$ is a complex disk, and one has to assume in addition that there is a proper holomorphic morphism $\bar f\colon \bar X\to D$ such that $X=f^{-1}(D^*)$ and $\bar f|_X=f$.)

Is it true that for all $q\in D^*$ close enough to $p$ the fibers $f^{-1}(q)$ are homeomorphic to each other?

Remark. If $X$ is smooth then the answer is positive.

A reference would be helpful, at least in the algebraic case.

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    $\begingroup$ That is not true for holomorphic morphisms of complex analytic varieties. For $X=\mathbb{C}\times D^*\setminus\{(0,1/n) : n\in \mathbb{Z}_{>0} \}$ with projection to $D^*$, that fails. $\endgroup$
    – Jason Starr
    Commented Nov 28, 2015 at 11:47
  • $\begingroup$ @JasonStarr: Thanks, you are right. In the complex analytic case I have to edit my question. $\endgroup$
    – asv
    Commented Nov 28, 2015 at 11:50

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