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Let $f$ be a morphism of schemes $f: (X,\mathcal{O}_X)\to (Y,\mathcal{O}_Y)$, and $\mathcal{F},\mathcal{G}$ be sheaves of $\mathcal{O}_Y$-modules. I am trying to prove (I do NOT claim this to be true):

$f^{\ast}\mathcal{F}\otimes_{\mathcal{O}_X}f^{\ast}\mathcal{G}\cong f^{\ast}(\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{G})$

By the definition of $f^{*}$, and the property of the tensor product, one can check that this boils down to proving: $\quad f^{-1} \mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y} f^{-1}\mathcal{G} \cong f^{-1}(\mathcal{F} \otimes_{\mathcal{O}_Y}\mathcal{G})$. However, I cannot continue this bare hand computation at the present stage. For one thing $f^{-1}$ and $\otimes$ both require sheafification, and thus I get a compostion of two sheafification objects; for another, I know nothing about good properties of stalks on $f^{-1}$.

I guess the computation may be dirty, but I appreciate any insight on handling the problem.

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not a clue what the question is, but the title is the best i have ever seen. –  Arjang Aug 13 '11 at 8:56
    
I'm not sure whether this helps, but $f^{-1}$ preserves all small colimits (because it is a left adjoint) and all finite limits. But I'm not sure whether the tensor product can be constructed using just those... –  Zhen Lin Aug 13 '11 at 9:54
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In this case, it may be better to write down an explicit sheafification (eg the one in Hartshorne) rather than working with the universal property, since the stalks are pretty easy to write down. –  Soarer Aug 13 '11 at 10:16
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As for stalk of $f^{-1}$, I'm pretty sure that this holds $f^{-1}\mathcal{F}_p = \mathcal{F}_{f(p)}$, because stalk can be treated as pullback to a point. –  Soarer Aug 13 '11 at 10:18
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I don't really know what this means but I love the notation! –  Jonas Teuwen Aug 13 '11 at 10:18
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2 Answers 2

up vote 13 down vote accepted

Yes, we have $f^{\ast}\mathcal{F}\otimes_{\mathcal{O}_X}f^{\ast}\mathcal{G}\cong f^{\ast}(\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{G})\quad$

And, yes, this results from the isomorphism $\alpha: f^{-1} \mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y} f^{-1}\mathcal{G} \overset {\sim}{\longrightarrow} f^{-1}(\mathcal{F} \otimes_{\mathcal{O}_Y}\mathcal{G})$
And, no, the computation is not dirty!

To prove that the natural map $\alpha$ is an isomorphism, it is enough to look at the stalks. The morphism $\alpha_x: (f^{-1} \mathcal{F} \otimes_{f^{-1}\mathcal{O}_Y} f^{-1}\mathcal{G})_x \to (f^{-1}(\mathcal{F} \otimes_{\mathcal{O}_Y}\mathcal{G}))_x$ is indeed an isomorphism because or the following two general results (which do not involve schemes):

Fact 1: For any continuous map $f:X\to Y$ of topological spaces and any sheaf $\mathcal E$ on $Y$, we have for every $x\in X$ a canonical isomorphism $(f^{-1} \mathcal E)_x=\mathcal E _{f(x)}$

Fact 2: Given a sheaf of rings $\mathcal A$ and sheaves $\mathcal C, \mathcal D$ of $\mathcal A$-Modules on the topological space $X$, we have for every $x\in X$ a natural isomorphism $(\mathcal C \otimes _{\mathcal A}\mathcal D)_x=\mathcal C_x \otimes _{\mathcal A_x}\mathcal D_x$.
[Of course, in the discussion at hand $\mathcal A$ is $f^{-1}\mathcal{O}_Y$]

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I cannot agree more, Thank you very much! –  Li Zhan Aug 14 '11 at 2:55
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Alternative proof, using only adjunctions.

First, notice that there is an isomorphism in $\mathsf{Mod}(Y)$

$$f_* \underline{\hom}_X(f^* G,H) = \underline{\hom}_Y(G,f^* H)$$

for $G \in \mathsf{Mod}(Y)$ and $H \in \mathsf{Mod}(X)$. In fact, on an open subset $V \subseteq Y$, we have

$\Gamma(V,f_* \underline{\hom}_X(f^* G,H)) = \hom_{f^{-1}(V)}(f^* G |_{f^{-1}(V)},H|_{f^{-1}(V)})$

$ = \hom_{f^{-1}(V)}(f_V^* G|_V,H|_{f^{-1}(V)}) = \hom_V(G|_V,(f_V)_* H|_{f^{-1}(V)})$

$ = \hom_V(G|_V,(f_* H)|_V) = \Gamma(V,\underline{\hom}_Y(G,f^* H)).$

The rest is purely formal:

$\hom_X(f^* F \otimes f^* G , H) = \hom_X(f^* F , \underline{\hom}_X(f^* G,H)) = \hom_Y(F,f_* \underline{\hom}_X(f^* G,H))$ $ = \hom_Y(F,\underline{\hom}_Y(G,f_* H)) = \hom_Y(F \otimes G,f_* H) = \hom_X(f^* (F \otimes G),H).$

Hence $f^* F \otimes f^* G \cong f^* (F \otimes G)$ by Yoneda. This proof also works in quite general contexts (for example where no stalks are available).

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What does the hom mean by underline? Does it mean the sheaf $Hom(\mathcal{F}(U),\mathcal{G}(U))$ on each U? –  mqx Oct 2 '13 at 2:54
    
Yes, it's the internal hom, in this case known as the sheaf hom. –  Martin Brandenburg Oct 2 '13 at 9:17
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