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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.

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up vote 16 down vote accepted

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

$\qquad \qquad \qquad \qquad \qquad \text{Hom}(f^*\mathcal{M} \otimes_{\mathcal{O}_X} f^*\mathcal{N},\mathcal{P}) \cong \text{Hom}(\mathcal{M},f_*\mathscr{H}om(f^*\mathcal{N},\mathcal{P}))$ $\qquad \qquad \qquad \qquad \qquad \qquad \cong \text{Hom}(\mathcal{M},\mathscr{H}om(\mathcal{N},f_*\mathcal{P})) \cong \text{Hom}(f^*(\mathcal{M} \otimes_{\mathcal{O}_Y} \mathcal{N}),\mathcal{P})$.

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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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 $C^\cdot$ be a diagram in $\mathcal{C}$. Then there is a natural isomorphism

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

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

$\text{Hom}(F\text{ colim }C^\cdot,D)\cong \text{Hom}(\text{ colim }C^\cdot,F^\perp D)\cong \text{ colim }\text{Hom}(C^\cdot,F^\perp D)\cong \text{ colim }\text{Hom}(FC^\cdot,D)\cong \text{Hom}(\text{ colim }FC^\cdot,D).$

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