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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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    @Zhen Lin Nice to notice such properties of $f^{-1}$, nor do I know whether tensor can can be constructed from colimits. But I guess no, since in tensor product, the objects are in the same position, but in colimits, the morphism(it exists) always has direction.2011-08-14

2 Answers 2

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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!2011-08-14
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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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    Yes, it's the internal hom, in this case known as the sheaf hom.2019-02-15