4
$\begingroup$

Let $f:X \to Y$ be an affine morphism; is it true that the counit map

\begin{equation*} f^* f_* \mathcal{F} \to \mathcal{F} \end{equation*} is surjective for every (coherent) sheaf $\mathcal{F}$?

Improve this question
$\endgroup$
2
  • $\begingroup$ The counit of adjunction goes in the opposite direction. $\endgroup$
    – Sasha
    Commented May 28, 2018 at 20:02
  • $\begingroup$ Sorry for the mistake Sasha, I am going to correct right now; thank you! $\endgroup$
    – Lorenzo
    Commented May 29, 2018 at 9:22

1 Answer 1

Reset to default
6
$\begingroup$

Yes. Since $f$ is affine, it is enough to check surjectivity after applying $f_*$. But the composition $$ f_* F \to f_* f^* f_* F \to f_* F $$ of $f_*({\rm counit})$ with the unit of $f_* F$ is the identity (triangle identities).

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .