**EDIT:** Let $f\colon X\to C$ be a flat proper morphism of complex algebraic (or analytic) varieties. Assume the special fiber over a point $p\in C$ is smooth.


**Is it true that there exists a neighborhood $U$ of $p$ (in Zariski or analytic topology) such that all fibers of $f$ over $U$ are smooth? Or more precisely, $U$ should be such that the morphism $f\colon f^{-1}(U)\to U$ is smooth.**


In my case one may assume that $C$ is a smooth curve or an infinitesimal neighborhood of a point $p$ inside such a curve, and the fibers of $f$ are one dimensional. The complex analytic situation is particularly important for me, but the algebraic situation is also of interest.


**UPDATE:** As was mentioned by abx in his comment to his answer below, flatness of $f$ and smoothness of the special fiber $f^{-1}(p)$ imply that $f$ is smooth along $f^{-1}(p)$, i.e. $f$ is smooth in a neighborhood of the special fiber. This is the key fact for my question. I was shown a reference to it: Lemma 130.16 here <http://stacks.math.columbia.edu/download/algebra.pdf>



When flatness of a morphism implies smoothness?
No if you don't assume that $f$ is proper. For a typical example, let $S$ be a curve with one singular point $s$; take $X=(S\times C) \smallsetminus \{(s,p)\} $.


On the other hand if you assume $f$ proper, this is (trivially) true: the locus of points $x\in X$ such that $T\_x(f)$ is not surjective is a closed subset $F$ of $X$, $f(F)$ is closed and does not contain $p$, just take $U=C \smallsetminus f(F)$.


