Let $Y$ be a rationally connected variety over an algebraically closed field, and let
$$\phi:X\dashrightarrow Y$$
be a rational fibration such that the general fiber of $\phi$ is rationally chain connected. **Is it true that $X$ is rationally chain connected?**


**If we assume that the general fiber of $\phi$ is smooth and rationally connected can we conclude that $X$ is rationally connected?**



Rationally connected varieties and rational fibrations
Over $\mathbb{C}$, the answer to the second question is *yes*.


In fact, since rational connectedness is a birational property, one can solve the indeterminacy of the rational map $\phi$ in order to obtain a dominant morphism $f \colon Z \to Y$ (whose general fiber is birational to the general fiber of $\phi$) and then apply the following result, due to Graber, Harris and Starr:



> 
> **Theorem.** Let $f \colon Z \to Y$ be any dominant morphism of complex varieties. If $Y$ and the general fiber of $f$ are rationally connected, then $Z$ is rationally connected.
> 
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See the paper [Families of rationally connected varieties](http://www.ams.org/journals/jams/2003-16-01/S0894-0347-02-00402-2/), *J. Amer. Math. Soc.* **16** (2003), no. 1, 57–67, in particular Corollary 1.3. 


