Maybe this is a stupid question, but I got irritated by it: Suppose $f: X \rightarrow Y$ is a morphism of schemes. That comes with a map of sheaves $f^\#: \mathcal{O}_Y \rightarrow f_* \mathcal{O}_X$. Because $f_*$ and $f^*$ are adjoint to each other, this map corresponds to a homomorphism $f^*\mathcal{O}_Y \rightarrow \mathcal{O}_X$ of $\mathcal{O}_X$-modules. But as far as I understand, $f^*\mathcal{O}_Y = f^{-1}\mathcal{O}_Y \otimes_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X = \mathcal{O}_X$. So the map $f^\#$ really is the same as an $\mathcal{O}_X$-module homomorphism $\mathcal{O}_X \rightarrow \mathcal{O}_X$, which is the same as giving a global section $s \in \Gamma(X, \mathcal{O}_X)$, because the map is fully determined by the value of the global section $1$.

Is this reasoning correct, or did I make a mistake?

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    $\begingroup$ Yes, the canonical map of sheaves on $X$: $f^*\mathcal{O}_Y = f^{-1}\mathcal{O}_Y \otimes_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X \stackrel {\cong} {\to} \mathcal{O}_X$ is an isomorphism. $\endgroup$ Jan 18, 2019 at 13:04

1 Answer 1


This is a community-wiki post recording the answer from the comments so that this question may be marked as answered.

Yes, the canonical map of sheaves on $X$: $f^*\mathcal{O}_Y=f^{-1}\mathcal{O}_Y\otimes_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X\stackrel{\cong}{\to}\mathcal{O}_X$ is an isomorphism. - Georges Elencwajg


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