# Is the pull-back of the structure sheaf the structure sheaf?

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?

• 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. Jan 18, 2019 at 13:04

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