# Is the natural map $f^* f_* \mathcal{F} \to \mathcal{F}$ surjective?

I'm trying to solve Exercise III 12.4 from Hartshorne's Algebraic geometry. There is a flat projective morphism $$f: X \to Y$$ of schemes of finite type over an algebraically closed field $$k$$. Also $$Y$$ is assumed to be integral, and all fibers are integral schemes.

Now suppose $$\mathcal{F}$$ is an invertible sheaf on $$X$$, that is trivial on each fiber $$X_y$$. I was able to show that $$f_*\mathcal{F}$$ is an invertible sheaf on $$Y$$ (this is essentially Cor 12.9), and now I would like to show that the natural map $$f^*f_* \mathcal{F} \to \mathcal{F}$$ is an isomorphism, for which it is enough to show that it is surjective, because both sheaves are locally free.

• @KReiser Is it sufficient to choose $U \subset Y$ such that $f_* \mathcal{F}|_U \cong \mathcal{O}_Y|_U$? If so, why? May 5, 2020 at 16:19
• My previous comment was not as helpful as it should have been, and I am sorry. I have posted a solution, do let me know if it resolves your issues. May 6, 2020 at 6:03

Since whether $$f^*f_*\mathcal{F}\to\mathcal{F}$$ is surjective is a local condition, we can check it on a cover $$X$$ by open subsets of the form $$f^{-1}(U)$$ where $$U\subset Y$$ is open and $$f_*\mathcal{F}|_U$$ is free. So it suffices to treat the case where $$Y=\operatorname{Spec} R$$ is affine and $$f_*\mathcal{F}=\mathcal{O}_Y$$.
Now let's recall some facts about the natural map $$f^*f_*\mathcal{F}\to\mathcal{F}$$. First, the natural map is the image of $$id_{f_*\mathcal{F}}\in \operatorname{Hom}_{\mathcal{O}_Y}(f_*\mathcal{F},f_*\mathcal{F})$$ under the isomorphism of $$\mathcal{O}_Y(Y)=R$$-modules given by the adjunction $$\operatorname{Hom}_{\mathcal{O}_Y}(f_*\mathcal{F},f_*\mathcal{F})\cong \operatorname{Hom}_{\mathcal{O}_X}(f^*f_*\mathcal{F},\mathcal{F}).$$
But since $$f_*\mathcal{F}=\mathcal{O}_Y$$, we have $$\operatorname{Hom}_{\mathcal{O}_Y}(f_*\mathcal{F},f_*\mathcal{F})=\operatorname{Hom}_{\mathcal{O}_Y}(\mathcal{O}_Y,\mathcal{O}_Y)=\mathcal{O}_Y(Y)=R$$, and $$id_{f_*\mathcal{F}}=id_{\mathcal{O}_Y}=1\in R$$. On the other hand, the pullback of the structure sheaf is the structure sheaf, so $$\operatorname{Hom}_{\mathcal{O}_X}(f^*f_*\mathcal{F},\mathcal{F})=\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{F})=\mathcal{F}(X)$$, so $$\mathcal{F}(X)=R$$. As the $$R$$-linear endomorphisms of $$R$$ are exactly given by multiplication by an element of $$R$$, the endomorphisms that are isomorphisms are exactly multiplication by a unit. So we see that $$id_{f_*\mathcal{F}}$$ must be sent to the map $$\mathcal{O}_X\to\mathcal{F}$$ which picks out an invertible element of $$\mathcal{F}(X)$$, that is, a non-vanishing global section. So $$\mathcal{F}$$ is trivial and the natural map $$f^*f_*\mathcal{F}\to\mathcal{F}$$ is an isomorphism.
• I don't see why the global section $s \in \mathcal{F}(X)$ should be non-vanishing. Your reasoning appears a bit circular, we don't know yet that $\mathcal{O}_X \to \mathcal{F}$ is an isomorphism. May 6, 2020 at 7:12
• Also, this would prove the general implication $f_* \mathcal{F} \cong \mathcal{O}_Y \implies \mathcal{F} \cong \mathcal{O}_X$, without any conditions on $Y$, $X$ or $f$. That makes me a bit suspicious, although I would like to be convinced if it's true. May 6, 2020 at 7:15
• @red_trumpet $s\in \mathcal{F}(X)=R$ is the image of the global section $1\in \mathcal{O}_X(X)=R$ under an $R$-linear isomorphism of $R$. Since $\operatorname{Hom}_R(R,R)=R$ for commutative rings, with composition of maps corresponding to multiplication of the corresponding elements of $R$, we see that all the isomorphisms are given by multiplication by units. So $s$ is a unit, which means it's not in any maximal ideal, which is equivalent to saying it doesn't vanish anywhere. May 6, 2020 at 8:04
• We did also use the fact that $\mathcal{F}$ was a line bundle in here - the conclusion that $\mathcal{F}$ is trivial if one can produce a nowhere vanishing global section is essential and is only true for line bundles. May 6, 2020 at 8:07
• No, $\mathcal{O}_X(X)=R$. By Stein factorization combined with the fact that $f:X\to Y$ is projective with geometrically irreducible fibers, we have $f_*\mathcal{O}_X=\mathcal{O}_Y$. May 6, 2020 at 8:12