tensor product of presheaves of modules

Question

Let $\mathscr{O}$ be a presheaf of rings on $X$ and $\mathscr{F}$, $\mathscr{G}$ be presheaves of $\mathscr{O}$-modules on $X$. Let $\mathscr{O}^{\#}$,$\mathscr{F}^{\#}$ and $\mathscr{G}^{\#}$ be respectively the sheafification. Then is the $\mathscr{O}^{\#}$-module $\mathscr{F}^{\#}\otimes_{\mathscr{O}^{\#}}\mathscr{G}^{\#}$ just the sheafification of the presheaf $\mathscr{O}$-module $\mathscr{F}\otimes_{p,\mathscr{O}}\mathscr{G}$ ?

Answer 1

The answer is yes. It follows (as always with these basic isomorphisms) formally form the universal properties; in particular adjunctions. That being said, I don't think that stalks are very enlightening, and in fact they cannot be used when $X$ is a general site.

Let $F,G$ be presheaves of $\mathcal{O}$-modules. If $H$ is a sheaf of $\mathcal{O}^\#$-modules, then there are natural bijections $$\hom(F^\# \otimes_{\mathcal{O}^\#} G^\#,H) \cong \hom_{\mathcal{O}^\#}(F^\#,\underline{\hom}_{\mathcal{O}^\#}(G^\#,H)) \cong \hom_{\mathcal{O}}(F,\underline{ \hom}_{\mathcal{O}}(G,H))$$ $$ \cong \hom_{\mathcal{O}}(F \otimes_{p,\mathcal{O}} G,H).$$ This means that $F^\# \otimes_{\mathcal{O}^\#} G^\#$ is a model of $(F \otimes_{p,\mathcal{O}} G)^\#$.

Answer 2

I think it's important to show that the isomorphism fits in the relevant commutative diagram. Also, stalks are great. Here is my proposal: The dashed arrows arrise from the the UP of the tensor products of (pre)sheaves, and the dotted arrows arrise from the UP of sheafification. The labels $(1)$ and $(2)$ indicate w.r.t. which arrow we applied the universal property. Commutativity of this diagram can be argued by invoking the unicity of the arrows that arrise from the UP (and some subdiagrams are commutative by construction).

To see that our map $$ (F\otimes_{p,\mathcal O}G)^+\to F^+\otimes_{\mathcal O^+}G^+ $$ is an isomorphism, we can look at stalks, invoking four properties:

1. The stalk of the sheafification of a presheaf is canonically isomorphic to the stalk of the presheaf (such that the corresponding diagram is commutative);

2. a morphism between (pre)sheaves induces a unique morphism between stalks (such that the corresponding diagram is commutative);

3. the stalk of the tensor product is the tensor product of the stalks (see Lemma 17.16.1 (Stacks Project)) (and again, there are relevant diagrams where this isomorphism fits in);

4. a morphism between sheaves that induces an isomorphism between stalks is an isomorphism between sheaves.

Since the induced maps on the stalks is the identity map under these considerations, we are done.