Rationally connected varieties and rational fibrations 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?
 A: What are you assuming about $X$?  Consider the closed subscheme $X$ of $\mathbb{P}^2 \times \mathbb{P^2}$ consisting of those points
$([s,t],[X,Y,Z])$ satisfying the equation $$s^3(X+uY + u^2Z)(X+\omega uY+\omega^2 u^2 Z)(X+\omega^2 u Y +\omega^2 u^2 Z) = 0,$$
where $u$ is a coordinate on some irrational cyclic $3$-sheeted cover of $\mathbb{P}^1$, e.g., $u$ is a cube root of $(t/s)^3 + 1$.  Of course this $X$ is very much not normal.  But it fibers over the base $\mathbb{P}^1$, the fibers are rationally chain connected (just triangles of lines), yet every rational curve in $X$ is contained in a fiber of the projection to $\mathbb{P}^1$.
A: 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.

See the paper Families of rationally connected varieties, J. Amer. Math. Soc. 16 (2003), no. 1, 57–67, in particular Corollary 1.3. 
