# Tensor Operations on free sheaves. Is sheafification necessary?

I was trying to show that on a ringed space $$(X, \mathcal{O})$$, that $$T^n(\mathscr{F})$$ is free of finite rank if $$\mathscr{F}$$ is a free $$\mathcal{O}$$-module of finite rank. We recall that $$T^r(\mathscr{F})$$ is the sheafification of the presheaf $$t^r(\mathscr{F})$$ defined by $$U \mapsto \mathscr{F}(U)^{\otimes r }$$ for open sets $$U$$.

For simplicity assume that $$\mathscr{F} = \mathcal{O}^{\oplus n}$$. Then, if $$U \subseteq X$$ is open and $$e_1, \dots, e_n$$ is the canonical basis of $$\mathcal{O}(U)^{\oplus n}$$, then the usual argument for modules shows that $$t^n( \mathscr{F})(U)$$ has a basis given by $$e_{j_1} \otimes e_{j_2} \otimes \cdots \otimes e_{j_r}$$ for all $$r$$-tuples formed from $$\{1, \dots, n\}$$. This yields a canonical isomorphism $$t^n(\mathscr{F})(U) \to \mathcal{O}(U)^{\oplus n^r}$$. Since basis elements restrict to basis elements, this map is compatible with restrictions to form an isomorphism $$t^n(\mathscr{F}) \to \mathcal{O}^{\oplus n^r}$$.

This would then imply that $$t^n(\mathscr{F})$$ is already a sheaf, which seems off since the tensor product(as presheaves) of sheaves is not in general a sheaf.

Is it always a sheaf in the free situation?

Thanks!

What you want to show is that $$F$$ and $$G$$ are two (quasi-?)coherent sheaves of $$\mathcal O_X$$-modules, then for each open affine $$U \subset X$$ we have $$(F \otimes_{\mathcal O_X} G)(U) = F(U) \otimes_{\mathcal O(U)} G(U).$$
In general, you will certainly have to sheafify. For example, $$\dim \Gamma(\mathbb P^n, \mathcal O(1)) = n+1$$ and $$\mathcal O(1)^{\otimes 2} = \mathcal O(2)$$, but $$\dim \Gamma(\mathbb P^n, \mathcal O(2)) = \binom{n+2}{n} = \frac{(n+2)(n+1)}{2}\neq (n+1)^2.$$