Let $\phi:X \to Y$ be a projective/proper, birational morphism between complex algebraic varieties, with connected fibers and $\phi_*\mathcal{O}_X \cong \mathcal{O}_Y$. Suppose further that $X$ is a nonsingular. Let $F$ be a subsheaf of a free $\mathcal{O}_X$module (not just locallyfree) such that the quotient is torsionfree. Is the natural morphism $$\phi^*\phi_*F \to F$$ injective or surjective?

$\begingroup$ It is certainly not surjective: take for $\phi$ the blowing up of $Y$ at a smooth point, let $E\subset X$ be the exceptional divisor, and let $F=\mathcal{O}_X(E)$. It is easy to see that $\phi_*F\cong \mathcal{O}_Y$, so the homomorphism $\phi^*\phi_*F\rightarrow F$ is not surjective. $\endgroup$ – abx Oct 31 '19 at 5:20

1$\begingroup$ @abx But $\mathcal{O}_X(E)$ is not a subsheaf of a direct sum of $\mathcal{O}_X$. $\endgroup$ – Ron Oct 31 '19 at 7:17

$\begingroup$ Oops, sorry, I overlooked that hypothesis. $\endgroup$ – abx Oct 31 '19 at 7:26
The answer to both questions (injectivity ans surjectivity) is no without further hypothesis.
Surjectivity : Let $Y$ be a smooth projective variety and let $\phi : X \longrightarrow Y$ be the blowup of $Y$ along a smooth subvariety. Denote by $L = \mathcal{O}_{X}(E) \otimes \phi^*\mathcal{O}_{Y}(1)$, where $E$ is the exceptional divisor and $\mathcal{O}_{Y}(1)$ is an ample line bundle on $Y$. The line bundle $L$ is ample so a sufficiently high power of $L$ is globally generated. That is, we have an exact sequence: $$0 \longrightarrow F \longrightarrow \mathcal{O}_{X}^{\oplus r} \longrightarrow L^{\otimes m} \longrightarrow 0,$$ for some $m,r \in \mathbb{N}$, when $m$ is big enough. Dualizing this sequence, we get:
$$0 \longrightarrow L^{\otimes m} \longrightarrow \mathcal{O}_{X}^{\oplus r} \longrightarrow F^* \longrightarrow 0,$$
with $F^*$ being torsionfree as the dual of a coherent sheaf. We have: $$\phi_* L^{m} = \mathcal{O}_{Y}(m) \otimes \phi_*(\mathcal{O}(mE)) = \mathcal{O}_{Y}(m),$$ by the projection formula and the fact that $\phi_* \mathcal{O}_{X}(mE) = \mathcal{O}_{Y}$ for $m \geq 0$. As a consequence, the map $\phi^* \phi_* L^{m} \longrightarrow L^{m}$ has cokernel equal to $\phi^*\mathcal{O}_{Y}(m) \otimes \mathcal{O}_{mE}(mE)$, it is not surjective.
injectivity : Let $Y$ be a smooth projective variety and $Z$ be a smooth codimension $3$ subvariety of $Y$. Let $J_{Z}$ be the ideal sheaf of $Z$ and let $\mathcal{O}_{Y}(1)$ be an ample line bundle on $Y$. For $m$ big enough, the sheaf $\mathcal{J}_{Z}(m)$ is globally generated, so that we have an exact sequence:
$$0 \longrightarrow F \longrightarrow \mathcal{O}_{Y}^{\oplus r} \longrightarrow \mathcal{J}_{Z}(m) \longrightarrow 0,$$ for some $m,r \in \mathbb{N}$, when $m$ is big enough. Let $\phi : X \longrightarrow Y$ be the blowup along $Z$. Pulling back the above exact sequence along $\phi$, we get a long exact sequence:
$$ 0 \longrightarrow Tor^1(\mathcal{J}_Z(m),\mathcal{O}_{X}) \longrightarrow \phi^*F \longrightarrow \mathcal{O}_{X}^{\oplus r} \longrightarrow \phi^*\mathcal{J}_Z(m) \longrightarrow 0.$$ There are no higher $Tor$ because $Z$ has codimension $3$ in $Y$. Denote by $G$ the cokernel of: $$0 \longrightarrow Tor^1(\mathcal{J}_Z(m),\mathcal{O}_{X}) \longrightarrow \phi^*F.$$ We then have an exact sequence:
$$0 \longrightarrow G \longrightarrow \mathcal{O}_{X}^{\oplus r} \longrightarrow \phi^* \mathcal{J}_{Z}(m) \longrightarrow 0.$$
Let's push forward this exact sequence by $\phi_*$, we have a long exact sequence:
$$0 \longrightarrow \phi_* G \longrightarrow \mathcal{O}_{Y}^{\oplus r} \longrightarrow \phi_* \phi^* \mathcal{J}_{Z}(m) \longrightarrow R^1 \phi_* G \longrightarrow 0.$$
Let's show that $\phi_* \phi^* \mathcal{J}_{Z}(m) = \mathcal{J}_{Z}(m)$. We have an exact sequence:
$$ 0 \longrightarrow Tor^1(\mathcal{O}_{Z}, \mathcal{O}_X) \longrightarrow \phi^* \mathcal{J}_Z \longrightarrow \mathcal{O}_{X}(E) \longrightarrow 0.$$ Since $Tor^1(\mathcal{O}_{Z}, \mathcal{O}_X) = \Omega_{E/Z}(E)$, (where $\Omega_{E/Z}$ is the bundle of relative differentials for the map $E \longrightarrow Z$), The relative version of Bott's vanishing Theorem implies that: $$\phi_* Tor^1(\mathcal{O}_{Z}, \mathcal{O}_X) = R^1 \phi_* Tor^1(\mathcal{O}_{Z}, \mathcal{O}_X) = 0.$$ We deduce that $\phi_* \phi^* \mathcal{J}_Z(m) = \mathcal{J}_Z(m)$.
But the map $\mathcal{O}_{Y}^{\oplus r} \longrightarrow \mathcal{J}_{Z}(m)$ is the one we started with and it is surjective. We thus get $R^1 \phi_* G = 0$ and $\phi_* G = F$. Finally, we get that the map: $$ \phi^* \phi_* G \longrightarrow G$$ is not injective as its kernal is: $$Tor^1(J_Z(m),O_{X}) = Tor^{2}(\mathcal{O}_{Z}(m),\mathcal{O}_{X}) = \bigwedge^2\Omega_{E/Z}(2E) \otimes \phi^* \mathcal{O}_{Y}(m).$$