Let $f:(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ be a map of ringed spaces, and let $\mathscr{F}$ be an $\mathcal{O}_Y$-module, then we can easily define $f^{-1}\mathscr{F}$ as a sheaf of Abelian groups. But how do we define the morphism $$f^{-1}\mathcal{O}_Y\times f^{-1}\mathscr{F}\to f^{-1}\mathscr{F}$$ so as to make $f^{-1}\mathscr{F}$ into an $f^{-1}\mathcal{O}_Y$-module?
Let $f:(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$, $\mathscr{F}$ an $\mathcal{O}_Y$-module. How is $f^{-1}\mathscr{F}$ an $f^{-1}\mathcal{O}_Y$ module?
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algebraic-geometry
commutative-algebra
sheaf-theory
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0You just take the morphism $\mathcal O_Y \times \mathcal F \to \mathcal F$ and apply $f^{-1}$. Of course one has to use, that $f^{-1}$ commutes with finite products. – 2017-01-27
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0How do we show that that map induces $f^{-1}\mathcal{O}_Y(U)$-module structure on $f^{-1}\mathscr{F}$? – 2017-01-27
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0I do not understand what you are asking. A module structure is nothing else but a map $R \times M \to M$. And thats precisely what you have there. – 2017-01-29
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0If $M$ is an Abelian group, then an $R$-module structure on $M$ is precisely a $\mathbb{Z}$-bilinear map $R\times M\to M$ that satisfies associativity and identity properties. That is to say, it is a map $$\varphi:R\otimes_\mathbb{Z} M\to M$$ such that $\varphi(r\otimes\varphi(s\otimes m)) = \varphi(rs\otimes m)$. – 2017-01-29
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0I am totally aware of this. – 2017-01-29
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0Then how do we know that the map $$f^{-1}\mathcal{O}_Y\times f^{-1}\mathscr{F}\to f^{-1}\mathscr{F}$$ actually induces $f^{-1}\mathcal{O}_Y(U)$-module structure on $f^{-1}\mathscr{F}(U)$? – 2017-01-29