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.
tensor product of sheaves commutes with inverse image
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 $f_*\mathscr{H}om(f^*\mathcal{N},\mathcal{P}) \cong \mathscr{H}om(\mathcal{N},f_*\mathcal{P})$, and then we see that
\begin{align*} \text{Hom}(f^*\mathcal{M} \otimes_{\mathcal{O}_X} f^*\mathcal{N},\mathcal{P}) &\cong \text{Hom}(\mathcal{M},f_*\mathscr{H}om(f^*\mathcal{N},\mathcal{P}))\\ &\cong \text{Hom}(\mathcal{M},\mathscr{H}om(\mathcal{N},f_*\mathcal{P}))\\ &\cong \text{Hom}(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.
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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3But tensor product is not a colimit, so I think this is not a stronger claim. – 2017-12-14