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Suppose $\mathcal{F}$ is a sheaf of module on $X$,$f:X\to Y$,suppose $\mathcal{F}$ is generated by global sections. Is $f^*f_*(\mathcal{F})\to \mathcal{F}$ is surjective ?

To check on stalks, $f^*f_*(\mathcal{F})_x \cong f_*(\mathcal{F})_{f(x)} \to \mathcal{F}_x$, and it became messy..

And is there counterexample or is this condition necessary?

2 Answers2

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Let's call $F \in \mathsf{Mod}(X)$ good if $f^* f_* F \to F$ is an epimorphism.

$1.$ Step: $f^* G$ is good for every $G \in \mathsf{Mod}(Y)$. In fact, then the map is a split epimorphism by the triangle identities of the adjunction between $f_*$ and $f^*$.

$2.$ Step: If $F'$ is a quotient of $F$ and $F'$ is good, then $F$ is good. This follows from the obvious commutative diagram.

$3.$ Step: If $F$ is generated by global sections, then $F$ is good. This is because $F$ is a quotient of a direct sum of copies of $\mathcal{O}_X = f^* \mathcal{O}_Y$, so that we may apply Steps 1 and 2.

Martin Brandenburg
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  • Does step 1 mean $Hom_X(f^G,f^G)\cong Hom_Y(G,f_f^G) \cong Hom_X(f^G,f^f_f^G)$, so it implies $f^G \cong f^f_f^G$?Could you explain your terms a little bit? Thanks, –  Nov 10 '13 at 00:53
  • And can we replace "generated by global sections" just by quasi-coherent in step 3? I think it also holds.. –  Nov 10 '13 at 00:57
  • No, Step 1 doesn't mean this and your isomorphism is wrong. And no, I don't see why Step 3 works for qc sheaves. – Martin Brandenburg Nov 10 '13 at 10:23
  • I'm sorry my second isomorphism is wrong, and the third part you are also right.I confused qc to be globally written as cokernal of direct sum of structure sheaf.. Can you explain the term "triangle identity" and why is it epimorphism? Thanks, –  Nov 10 '13 at 12:26
  • http://ncatlab.org/nlab/show/triangle+identities – Martin Brandenburg Nov 10 '13 at 13:18
  • It is really clear using categorical language, thanks! –  Nov 11 '13 at 06:07
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For a counterexample if $\mathcal{F}$ is not generated by global sections, take $f: X \rightarrow Y=\{pt\}$ and $\mathcal{F}$ a non-zero sheaf of $\mathcal{O}_X$-modules, which has no (non-trivial) global sections. Then the natural map $$ f^*f_*\mathcal{F} \rightarrow \mathcal{F} $$ isn't surjective.

Indeed, let $x \in X$ be such that $\mathcal{F}_x \neq 0$. We have \begin{split} (f^*f_*\mathcal{F})_x \cong (f_*\mathcal{F})_{pt} \otimes_{\mathcal{O}_{Y,pt}} \mathcal{O}_{X,x} = 0 \end{split} since $(f_*\mathcal{F})_{pt} \cong \mathcal{F}(X)$, and the latter is zero by assumption. Hence $f^*f_*\mathcal{F} \rightarrow \mathcal{F}$ cannot be surjective at $x$.

Nils Matthes
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