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Let $f: (X, O_X) \rightarrow (Y, O_Y)$ be a scheme morphism, $F$ - module over $O_X$, $G$ - module over $O_Y$. How to prove, that $$ Hom_{O_X}(f^*G, F) = Hom_{O_Y}(G, f_* F). $$ Please give the most detailed proof.

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What did you try? –  Bruno Joyal Oct 28 '12 at 17:36
    
Are you happy with the analogous statement in the category of abelian sheaves, and just want it generalized to $\mathcal{O}_X$-modules, or are you confused about the analogous statement in the category of abelian sheaves? –  user29743 Oct 28 '12 at 17:45
    
I have proved it for functors $f_*$ and $f^{-1}$ in Hartshorne's notation. But when we construct $f^*$, there is an additional tensor product, and I don't understand it. It is very difficult to prove it directly, because there is 2 sheafifications, and it's impossible for me to deal with. –  user46336 Oct 28 '12 at 17:49
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This is given in detail (more than any other reference I've looked at, though I've not checked EGA) in the Stacks Project chapter called Sheaves on Spaces, near the end. I think that proving that $f^*$ and $f_*$ are adjoint in detail is kind of a mess, but with the intermediate results proved in Stacks, it basically follows from a series of adjunctions, beginning with $f^{-1}$ and $f_*$. Incidentally, this doesn't have anything to do with schemes, and holds for arbitrary ringed spaces, although given the concrete description of pullback for morphisms of affine schemes, it might actually be –  Keenan Kidwell Oct 28 '12 at 18:07
    
easier to prove directly for schemes. –  Keenan Kidwell Oct 28 '12 at 18:09

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