# tensor product of sheaves commutes with inverse image

Let $f : X \to Y$ be a morphism of ringed spaces and $\mathcal{M}$, $\mathcal{N}$ sheaves of $\mathcal{O}_Y$-modules. Then one has a canonical isomorphism $f^*(\mathcal{M} \otimes_{\mathcal{O}_Y} \mathcal{N}) \cong f^*\mathcal{M} \otimes_{\mathcal{O}_X} f^*\mathcal{N}$, but I cannot find a proof in any of the standard references. The problem is that the definitions of the functors $f^*$ and $\otimes$ are so cumbersome that I cannot even write down a map between these two sheaves. Surely there is a nice way to do this: to give you an idea of what I mean by "nice," I am the type of person who likes to define such functors as adjoints to some less complicated functor, prove that they exist, and then forget the construction.

## 2 Answers

I figured it out: let $$\mathcal{P}$$ be a sheaf of $$\mathcal{O}_X$$-modules. It is easy to check from the definition of $$\mathscr{H}om$$ and the adjointness of $$f^*$$ and $$f_*$$ that $$f_*\mathscr{H}om_{\mathcal{O}_X}(f^*\mathcal{N},\mathcal{P}) \cong \mathscr{H}om_{\mathcal{O}_Y}(\mathcal{N},f_*\mathcal{P})$$ as $$\mathcal{O}_Y$$-modules, and then we see that

\begin{align*} \text{Hom}_{\mathcal{O}_X}(f^*\mathcal{M} \otimes_{\mathcal{O}_X} f^*\mathcal{N},\mathcal{P}) &\cong \text{Hom}_{\mathcal{O}_Y}(\mathcal{M},f_*\mathscr{H}om_{\mathcal{O}_X}(f^*\mathcal{N},\mathcal{P}))\\ &\cong \text{Hom}_{\mathcal{O}_Y}(\mathcal{M},\mathscr{H}om_{\mathcal{O}_Y}(\mathcal{N},f_*\mathcal{P}))\\ &\cong \text{Hom}_{\mathcal{O}_X}(f^*(\mathcal{M} \otimes_{\mathcal{O}_Y} \mathcal{N}),\mathcal{P}). \end{align*}

So $$f^*\mathcal{M} \otimes_{\mathcal{O}_X} f^*\mathcal{N}$$ and $$f^*(\mathcal{M} \otimes_{\mathcal{O}_Y} \mathcal{N})$$ represent the same functor, whence they are canonically isomorphic.

• But don't you need to require that at least $\mathcal M$ be locally free of finite rank for these equalities to work? See Hartshorne Exercise II.5.1c and here Commented Jul 25, 2020 at 14:04
• @quantum, no, this is not necessary. This is the tensor-hom adjunction for sheaves, see here on MSE, or tag 01CN on StacksProject. Commented Sep 24, 2020 at 21:37
• @KReiser But in the Stacks link they don't switch $\operatorname{Hom}$ for $Hom$. From what I understand they use the $Hom$ sheaf so why can we use this here ? Commented Sep 30, 2023 at 16:03
• @raisinsec do you know the relation between the hom sheaf and the sheaf homs? The linked results are actually stronger. Commented Sep 30, 2023 at 19:53
• @KReiser You mean that Hom is the global section of $Hom$? So the result linked is indeed for the Hom sheaf but we consider the global section. Commented Sep 30, 2023 at 20:31

It's worth noting that the same proof idea more or less proves the stronger claim:

Let $F:\mathcal{C}\rightarrow\mathcal{D}$ be a left adjoint functor and let $C^\bullet$ be a diagram in $\mathcal{C}$. Then there is a natural isomorphism

$\text{colim } FC^{\bullet}\cong F \text{ colim } C^{\bullet}$.

Proof: For any object $D\in \mathcal{D}$, we have

\begin{align*} \text{Hom}(F\text{ colim }C^{\bullet},D)&\cong \text{Hom}(\text{ colim }C^{\bullet}, F^{\perp} D)\\ &\cong \text{ lim }\text{Hom}(C^{\bullet},F^{\perp} D)\\ &\cong \text{ lim }\text{Hom}(FC^{\bullet},D)\\ &\cong \text{Hom}(\text{ colim }FC^{\bullet},D). \end{align*}

• Shouldn't that be a $\lim$ rather than $\operatorname{colim}$ after you pull it out of the Hom?
– ಠ_ಠ
Commented Aug 31, 2016 at 8:00
• But tensor product is not a colimit, so I think this is not a stronger claim. Commented Dec 14, 2017 at 21:09