# Injectivity of pullback composed with pushforward

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?

• 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.
– abx
Oct 31 '19 at 5:20
• @abx But $\mathcal{O}_X(E)$ is not a subsheaf of a direct sum of $\mathcal{O}_X$.
– Ron
Oct 31 '19 at 7:17
• Oops, sorry, I overlooked that hypothesis.
– 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 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: $$Tor^1(J_Z(m),O_{X}) = Tor^{2}(\mathcal{O}_{Z}(m),\mathcal{O}_{X}) = \bigwedge^2\Omega_{E/Z}(-2E) \otimes \phi^* \mathcal{O}_{Y}(m).$$