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Let $f: X \to Y$ be a morphism of ringed spaces, $\mathcal{G}$ a $\mathcal{O}_Y$-module, $\mathcal{F}$ a $\mathcal{O}_X$-module.

It is well known that the fuctors $f^*, f_*$ are adjunct via the adjunction formula

$$Hom_{\mathcal{O}_X}(f^*\mathcal{G},\mathcal{F})= Hom_{\mathcal{O}_Y}(\mathcal{G}, f_*\mathcal{F})$$

I have following question:

If we take $\mathcal{G}:= \mathcal{O}_Y$ and $\mathcal{F}:= f^*(\mathcal{O}_Y)$ then the adjunction relation becomes

$$Hom_{\mathcal{O}_X}(f^*(\mathcal{O}_Y),f^*(\mathcal{O}_Y))= Hom_{\mathcal{O}_Y}(\mathcal{O}_Y, f_*f^*(\mathcal{O}_Y))$$

And I often read that with this formula the identity $id_{f^*\mathcal{G}}$ on the left side corresponds to the sheaf morphism $f^{\#}: \mathcal{O}_Y \to f_*f^*(\mathcal{O}_Y)=f^* \mathcal{O}_X$ on the right hand side which coinsides with sheaf morphism from the morphism of ringed spaces $f = (f, f^{\#}):(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$.

What I don't understand is that via the adjunction formula $f^{\#}$ has to be a morphism of $\mathcal{O}_Y$-modules, so for every open $U \subset Y$ the induced morphism $f^{\#}_U: \mathcal{O}_Y(U) \to f^* \mathcal{O}_X(U) =\mathcal{O}_X(f^{-1}(U))$ has to be a morphism of $\mathcal{O}_Y(U)$-modules

but on the other hand it is well known that $f^{\#}$ as a morphism arrising from morphism of ringed spaces $f$ is a morphism of sheaves of rings on $Y$, so it gives for each open $U \subset Y$ a ring morphism $f^{\#}_U: \mathcal{O}_Y(U) \to \mathcal{O}_X(f^{-1}(U))$

and a ring morphism $\phi: R \to A$ is in general never a $R$-module morphism.

So I don't understand why we can say that $f^{\#}$ is a morphism of $\mathcal{O}_Y$-modules (this gives the adjunction formula) where each $f^{\#}_U$ is a ring morphism since on the other hand $f^{\#}$ is a morphism of sheaves of rings on $Y$.

References for this construction: e. g.

  • Bosch "Commutative Algebra and Algebraic Geometry" (page 269)

  • Görtz, Wedhorn "Algebraic Geometry" (page 181)

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The way you view $f_\ast \mathcal{F}$ as an $\mathcal{O}_Y$-module is via the morphism $f^\sharp\colon \mathcal{O}_Y\to f_\ast \mathcal{O}_X$. So it all boils down to observing that, given an $R$-algebra $A$, the structure morphism $R\to A$ is an $R$-module morphism.

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  • $\begingroup$ ok so the $\mathcal{O}_Y$-module structure for every morphism $m: \mathcal{G} \to f_*\mathcal{F}$ is settled locally for every open $U \subset Y$ on $m_U:\mathcal{G}(U) \to f_*\mathcal{F}(U)$ via $rn \mapsto m_U(rn) := f^{\#}(r) \cdot m_U(n)$ where $n \in \mathcal{G}(U)$ and $r \in \mathcal{O}_Y(U)$, right? $\endgroup$ – KarlPeter Feb 25 at 14:07
  • $\begingroup$ And in case of $m= f^{\#}$ locally the ring morphism is "swapped" to the $\mathcal{O}_Y(U)$ module stricure by $f^{\#}_U(r) = f^{\#}_U(r \cdot 1) = f^{\#}_U(r) \cdot f^{\#}_U(1) $? $\endgroup$ – KarlPeter Feb 25 at 14:13
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    $\begingroup$ @KarlPeter I don't quiet understand what you want to say. The $\mathcal{O}_Y$-module structure on $f_\ast \mathcal{F}$ is given as follows: For an open $U$ of $Y$, $(f_\ast \mathcal{F})(U)=\mathcal{F}(f^{-1}(U))$ is an $\mathcal{O}_X(f^{-1}(U))$-module (as $\mathcal{F}$ is by assumption an $\mathcal{O}_X$-module). Now you have $f^\sharp_U\colon \mathcal{O}_Y(U)\to \mathcal{O}_X(f^{-1}(U))$ making $\mathcal{O}_X(f^{-1}(U))$ into an $\mathcal{O}_Y(U)$-algebra. In particular, via this algebra structure any module over $\mathcal{O}_X(f^{-1}(U))$ may be viewed as an $\mathcal{O}_Y(U)$-module. $\endgroup$ – user363120 Feb 25 at 16:10

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