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 non-singular. Let $F$ be a subsheaf of a *free* $\mathcal{O}\_X$-module (not just locally-free) such that the quotient is torsion-free. Is the natural morphism $$\phi^\*\phi\_\*F \to F$$ injective or surjective?



Injectivity of pullback composed with pushforward
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 blow-up 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 torsion-free 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 blow-up 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:
$$Tor^1(J\_Z(m),O\_{X}) = Tor^{2}(\mathcal{O}\_{Z}(m),\mathcal{O}\_{X}) = \bigwedge^2\Omega\_{E/Z}(-2E) \otimes \phi^\* \mathcal{O}\_{Y}(m).$$ 


