Let $f:X\rightarrow Y$ be a birational morphism between smooth varieties and $F$ a torsion free sheaf. Is $f_{\ast} F$ torsion free? If not, are there conditions on either $F$, $f$, $X$ or $Y$ that could ensure the torsion freeness of $F$? For example if $f$ is surj. and $F$ is the canonical bundle on $X$ a theorem of Kollar ensure the torsion freeness of $f_*(\omega_X)$.

5$\begingroup$ If $F$ is torsion free on $X$, then $F\to F\otimes K(X)$ is injective where $K(X)$ is the constant sheaf associated to the function field. The injectivity persists for $f_*F$ since $f_*$ is left exact. So yes, and you don't need anything as deep as Kollar's theorem. $\endgroup$ – Donu Arapura May 13 '11 at 11:17
Actually much more is true. In fact there is the following result, whose proof can be found in
[Grothendieck  Dieudonné: EGA 1 (Elements de Géométrie Algebrique), Proposition 8.4.5 page 351].
Proposition. Let $X$, $Y$ be two integral schemes and $f \colon X \to Y$ be a dominant morphism. Then for any torsionfree $\mathcal{O}_X$module $\mathcal{F}$, the pushforward $f_* \mathcal{F}$ is a torsionfree $\mathcal{O}_Y$module.
Kollar's result is much deeper since he proves the torsionfreeness of the higher direct images $R^if_* \omega_X$.

$\begingroup$ Thank you very much!!! I believed it was true but I could not find the right reference!!! best wishes! $\endgroup$ – Rurik May 13 '11 at 11:54

$\begingroup$ Maybe we don't have the same edition (?), but in mine (french version) the result is Proposition 7.4.5 page 163 (still in EGA I). $\endgroup$ – Henri Nov 10 '16 at 19:54
As already mentioned by Donu and Francesco, you don't need such big guns as Kollár's theorem. Also, the Proposition Francesco cites might seem more serious than it is by virtue of being in EGA...
The point is this: For an open set $V\subseteq Y$, the module $f_*\mathscr F(V)$ is the same as the module $\mathscr F(f^{1}V)$. Being torsion on an integral scheme is equivalent to having a nonempty support that is strictly smaller than the ambient scheme. If $f$ is surjective this already gives you what you want, and if it is only dominant you need to think a little more to see that the statement is true.

$\begingroup$ Thank you very much for your time! this was really helpfull. $\endgroup$ – Rurik May 18 '11 at 8:59