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.
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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](https://math.stackexchange.com/questions/3229725/pullbackoflocallyfreesheavesislocallyfree) – quantum Jul 25 '20 at 14:04

1@quantum, no, this is not necessary. This is the tensorhom adjunction for sheaves, see [here on MSE](https://math.stackexchange.com/questions/1592338/homtensoradjunctionforoxmodules/1592440), or [tag 01CN on StacksProject](https://stacks.math.columbia.edu/tag/01CN). – KReiser Sep 24 '20 at 21:37
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*}
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Shouldn't that be a $\lim$ rather than $\operatorname{colim}$ after you pull it out of the Hom? – ಠ_ಠ Aug 31 '16 at 08:00

8But tensor product is not a colimit, so I think this is not a stronger claim. – Akatsuki Dec 14 '17 at 21:09