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Given a Morphism f of Schemes X->Y and two sheaves F, G of modules on Y, is it right that the tensor product of F and G as modules commutes with the inverse image (not the module pullback but only the inverse image f^{-1}) construction? Here I mean one time tensor product over O_Y and the other time over f{-1}O_Y.


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You mean tensor product over $f^{-1} \mathcal{O}_Y$? – Karl Schwede Aug 1 '11 at 16:55
Yes, that's what I mean. – Descartes Aug 1 '11 at 17:23

2 Answers 2

up vote 6 down vote accepted

$\newcommand{mc}{\mathcal}$ I guess that the problem lies in all the sheafifications so let me explain how to get rid of them in some small independent steps which are of some interest on their own. I denote by $\cdot^\#$ the sheafification. The following two statements follow without difficulty by checking that the object in question has the right universal property (in particular there is no need to adhere to any construction of the sheafification):

i) Let $X$ be a topological space, let $\mc O$ be a presheaf of rings on $X$ and let $\mc M$ and $\mc N$ be presheaves of $\mc O$-modules. Then $(M\otimes^p_{\mc O}\mc N)^\#=M^\#\otimes^s_{\mc O^\#}\mc N^\#$, where the superscript of the tensor product indicates whether I mean the tensor product of sheaves or the one of presheaves.

ii) Let $f:X\to Y$ be a continuous map of topological spaces and let $\mc M$ be a presheaf of abelian groups on $Y$. Then $f^{-1,s}(M^\#)=(f^{-1,p}(M))^\#$, where the superscript to the pullback indicates whether I mean the one for sheaves or for presheaves.

Concerning your question (and using your notation): By ii) we have $f^{-1,s}(\mc F\otimes^s_{\mc O_Y}\mc G)=(f^{-1,p}(\mc F\otimes^p_{\mc O_Y}\mc G))^\#$. By i) we have $f^{-1,s}\mc F\otimes^s_{f^{-1,s}\mc O_Y}f^{-1,s}\mc G=(f^{-1,p}\mc F\otimes^p_{f^{-1,p}\mc O_Y}f^{-1,p}\mc G)^\#$. Thus it suffices to show the corresponding statement for presheaves. By the definition of $f^{-1,p}$ it follows from the following fact: Let $(A_i)_{i\in I}$ be a direct system of rings and let $(M_i)_{i\in I}$ and $(N_i)_{i\in I}$ be direct systems of $A_i$-modules. Then $\varinjlim M_i\otimes_{\varinjlim A_i}\varinjlim N_i=\varinjlim (M_i\otimes_{A_i}N_i)$.

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The tensor product $F \otimes G$ has the following universal property:

$\hom(F \otimes G,H) = \text{Bilin}(F \times G,H)$

Here, the right hand side is the set of bilinear sheaf homomorphisms $F \times G \to H$. You should keep this in mind instead of the explicit construction involving sheavifications! The idea behind tensor products is not their construction, but rather that they classify bilinear maps. This idea is familiar from commutative algebra and should not be lost in algebraic geometry.

Now we also have a canonical identification

$\text{Bilin}(F \times G,H) = \hom(F,\underline{\hom}(G,H))$.

The pullback $f^\*$ has also a universal property, it is left adjoint to the pushforward $f_\*$. Thus, we get:

$\hom(f^\* F \otimes f^\* G,-) \cong \hom(f^\* F,\underline{\hom}(f^\* G,-)) \cong \hom(F,f_* \underline{\hom}(f^\* G,-))$ $\cong \hom(F,\underline{\hom}(G,f_\* -)) \cong \hom(F \otimes G,f_\* -) \cong \hom(f^\* (F \otimes G),-)$

and the Yoneda lemma gives us $f^\* F \otimes f^\* G \cong f^\* (F \otimes G)$.

PS: Try to prove $F \otimes (G \otimes H) \cong (F \otimes G) \otimes H$ using sheavifications and compare the proof with the one using trilinear maps.

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Is your spelling "sheavification" instead of sheafification intentional ? – Zoran Skoda Feb 19 '14 at 15:58

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