We call Zariski sheaf a contravariant functor $F: (\operatorname{Aff \ Sch}/k) \to (\operatorname{Grp})$ such that $F_X: (\operatorname{Op}(X) \cap (\operatorname{Aff \ Sch}/k)) \to (\operatorname{Grp})$ is a sheaf for each affine scheme $X$ (here $\operatorname{Op}(X)$ is the category of open subsets of $X$)
Fix $n>0$ and consider the functor $F: (\operatorname{Aff \ Sch}/k) \to (\operatorname{Grp})$ such that $F(\operatorname{Spec}R) = \operatorname{GL}_n(R) / \operatorname{id_n}\cdot R^*$. I want to show that $F$ is not a Zariski sheaf.
I know that in general quotient of sheaves are not sheaves. In this case it seems to be easier to find a proof of the fact that $F$ is not a Zariski sheaf (since we ask for a quotient sheaf for each affine scheme) but I can't find a way.
Any hint or ideas? Thank you in advance