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Let $f:X\rightarrow Y$ be a proper morphism of notherian schemes over a DVR $R$, $Y$ being a smooth connected scheme. Assuming the generic fiber of $f_K:X_K\rightarrow Y_K$ to be connected (even smooth connected), where $K=Frac(R)$, is it true that the generic fiber of $f_k:X_k\rightarrow Y_k$ is also connected?
It seems that applying Stein factorization gives this result but I am not sure of my interpretation.

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    $\begingroup$ Is $f$ flat? The result is not true if you allow $f$ to be non-flat. $\endgroup$ – Jason Starr Nov 14 '16 at 16:55
  • $\begingroup$ Thank you for your answer. What if $Y$ is proper and $f_K$ and $f_k$ generically smooth ? $\endgroup$ – pi_1 Nov 14 '16 at 17:26
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    $\begingroup$ A counterexample in the absence of flatness is when f is a disjoint union of two maps, one of which lies over the special fiber. (For instance, $f= Id_Y \cup Y_k \hookrightarrow Y$.) Then both $f_K$ and $f_k$ are smooth, while the special fiber is obviously disconnected. $\endgroup$ – user84144 Nov 14 '16 at 18:19
  • $\begingroup$ Thank you. What if $X$ is connected? $\endgroup$ – pi_1 Nov 14 '16 at 18:22

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