Let $X$ be a quasiprojective variety, $Y$ a projective variety, and $f:X \rightarrow Y$ be an open immersion. If $\mathcal{F}$ is a locally free coherent sheave, what can be said about $f_\ast \mathcal{F}$? Is it coherent? Is it torsion free? Is it reflexive?

Dear Yemon, about your new question: Let $X$ be a projective variety and $Y=X\setminus S$ an open subset (with inclusion denoted $f:Y\to X$). Let $\mathcal F$ be an algebraic coherent sheaf without torsion on $Y$. Theorem (SerreGrothendieck) Suppose that $X$ is normal and that $S$ has codimension $\geq 2$. Then the sheaf $f_\ast \mathcal F$ is coherent. Serre, Prolongement de faisceaux analytiques cohérents, Ann.Inst.Fourier 16 (1966), 363374 


Dear Yemon, a)The sheaf $f_\ast \mathcal{F}$ is not coherent in general since its stalk will not be finitely generated over the local ring of a point of $Y\setminus X$. For example take $P$ a point of $\mathbb P^1=Y$ and $X= \mathbb P^1 \setminus P=\mathbb A^1$. Then for $\mathcal F =\mathcal O_X$, you get $(f_\ast \mathcal{F})_P= Rat(Y)$ b) The direct image $f_\ast \mathcal{F}$ will be torsion free because an inductive limit of torsion free modules over a domain is torsion free ( I assume that variety means in particular integral scheme.) c) I'm not sure reflexive is a reasonable concept for a noncoherent sheaf. 


Thank you for your answer. My question was motivated by the fact that I would like to construct a reflexive coherent sheave $\mathcal{G}$ on $Y$ such that $\mathcal{G}_X = \mathcal{F}$. Is it possible? What if I suppose that $Y$ is normal and codim($Y\setminus X) \geq 2$? 


By the way, assuming by varieties, you mean irreducible varieties, then for the second question, the answer is yes. For the torsion freeness, suppose that $r \in H^0(U, O_X)$ kills some nonzero element $z \in H^0(U, f_* \mathcal{F}) = H^0(U \cap X, \mathcal{F})$. By restriction, $r$ is a nonzero element of $H^0(X \cap U, \mathcal{O}_Z)$. We still have $rz = 0$ even in this setting, and so by restricting to an affine cover of $X$, it still happens. This will contradict the torsionfreeness (and thus in particular the locallyfreeness) of $\mathcal{F}$. 

