tensor product of sheaves is not a sheaf

Question

I was looking for an example of two sheaves that satisfy the gluing and uniqueness axioms but whose tensor product is a presheaf, but not a sheaf (doesn't satisfy uniqueness, for example).

Thank you very much.

Answer 1

Consider the sheaves $\mathcal{F}=\mathcal{G}=\underline{\mathbb{Z}}$ of locally constant functions with values in $\mathbb{Z}$. Almost by definition : $\mathcal{F}(U)=\mathbb{Z}^{\pi_0(U)}$.

Now consider an open subset $U$ with two connected components. You have $$\mathcal{F}(U)\otimes\mathcal{G}(U)=\mathbb{Z}^{\oplus 2}\otimes\mathbb{Z}^{\oplus 2}=\mathbb{Z}^{\oplus 4}.$$

However $\mathcal{P}:V\mapsto\mathcal{F}(V)\otimes\mathcal{G}(V)$ is not a sheaf (in fact its associated sheaf is again $\underline{\mathbb{Z}}$). To see this, consider the open covering of $U$ by its two components $U_1$ and $U_2$. If $\mathcal{P}$ was a sheaf, we would have the equality $\mathcal{P}(U)=\mathcal{P}(U_1)\oplus\mathcal{P}(U_2)$. But it doesn't hold.