Given a morphism $f$ of Schemes $X \to Y$ and two sheaves $\mathcal F$, $\mathcal G$ of modules on $Y$,
is it right that the tensor product of $\mathcal F$ and $\mathcal G$ 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 $\mathcal O\_Y$ and the other time over $f^{-1}\mathcal O\_Y$.


Regards!



Inverse Image and Tensor Product
$\newcommand{mc}{\mathcal}$
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 $\varinjlim
M\_i\otimes\_{\varinjlim A\_i}\varinjlim N\_i=\varinjlim (M\_i\otimes\_{A\_i}N\_i)$.


