Suppose $\pi:X\to Y$ is a flat projective morphism with connected fibers. Suppose $Y$ is a smooth projective variety ($X$ and $Y$ are over $\mathbb C$). Let $E$ be a locally free sheaf on $X$ such that its restriction to fibers of $\pi$ is trivial. Suppose finally that for each fiber $F$ of $\pi$ the higher cohomology of its structure sheaf vanish $H^i(O\_F)=0$ for $i>0$. Is it true that 


$$\pi^\* \pi\_\*(E)\cong E?$$


If yes, then what would be a reference for this statement? If no, I would like to know a general enough restriction on the morphism so that the statement holds. (In the case that is of interest to me all fibers are irreducible, and a generic fiber is a smooth Fano manifold). 



Locally free sheaves trivial on fibers of a flat projective morphism
One possible approach to this is to divide the problem into two and figure out what conditions are needed for the two parts.


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First assume that $\pi\_\*E$ is locally free of the same rank as $E$ (say $r$). By adjointness of $\pi^\*$ and $\pi\_\*$ we have a natural map
$$
\nu:\pi^\*\pi\_\*E\to E
$$
**If** $E|\_F$ is trivial, then $\nu$ is surjective (this can be checked locally, to do that at $x\in X$, take a set of local generators of $\pi\_\*E$ near $\pi(x)$. Their pre-images in $\pi^\*\pi\_\*E$ will map to a set of generators in $E\_x$). 
But then $\nu$ actually has to be an isomorphism, because $\pi^\*\pi\_\*E$ and $E$ has the same rank, so it is an isomorphism at the generic points of $X$ and hence the kernel of $\nu$ is a torsion sheaf. Since $\pi^\*\pi\_\*E$ is torsion-free this shows that $\nu$ is injective and hence an isomorphism.




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OK, so how can we guarantee that $\pi\_\*E$ is locally free of the same rank as $E$?


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We certainly need that $E|\_F$ is trivial, but probably a little more. 


Let me say, without exploring whether weaker conditions suffice, that the following works:


**If** $Y$ is integral and all fibers of $\pi$ are reduced, then this is OK. This is essentially Grauert's theorem (see Hartshorne, III.12.9). 


(The condition on the fibers is to ensure that $h^0(F, E|\_F)$ is constant. See Mohan's comment to Qing Liu's answer). 




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So we have the following:



> 
> Let $\pi:X\to Y$ be a flat projective morphism and assume that $Y$ is integral and all fibers of $\pi$ are reduced and connected. 
>  Further let $E$ be a locally free sheaf on $X$ such that $E$ restricted to any fiber of $\pi$ is trivial. Then $$\pi^\*\pi\_\*E\simeq E.$$
> 
> 
> 




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**Note** some of the assumptions you made or was willing to make are not necessary. 
You don't need $Y$ to be smooth or the higher cohomology of $\mathscr O\_F$ to vanish.


