I am suppose to give en example of a variety $X$ where the constant presheaf $\mathcal{F}$ is not a sheaf, this is my attempt, is it ok?
Pick the constant abelian presheaf $\mathcal{F}$ with $\mathcal{F}(U)=\mathbb{Z}$ for every $U \subset X$ and $\mathcal{F}(\varnothing)=0$ for the variety $X$ and pick $X$ such that we can write $X=X_1\cup X_2$ with $X_1\cap X_2=\emptyset$. We set $U$ as an open subset of $X$ such that it includes elements from both $X_1$ and $X_2$ call these sets $U_1$ and $U_2$ respectively. Now we look at two sections $s,t$ with $\mathcal{F}(U_1)\ni s\neq t\in\mathcal{F}(U_2)$. Under these assumptions $\mathcal{F}$ satisfies all the sheaf conditions but the glueing property since the condition $s_i\vert_{U_i\cap U_j}=s_j\vert_{U_j\cap U_i}$ for all $i,j$ is satisfied since $\rho_{U\emptyset}(x)=0$ for all $x$. But this should imply that we have now a section $r$ over $U$ such that $r\vert_{U_i}=s_i$ and this is impossible since $\mathcal{F}(U_1)\ni s\neq t\in\mathcal{F}(U_2)$ by assumption.
is this ok?
Best Answer
The constant presheaf fails to be a sheaf on a variety consisting of two points. For the constant presheaf the group of global sections is isomorphic to one copy of the constant group, whereas in the associated sheaf the group of global sections is isomorphic to the direct sum of two copies of the constant group.
This is essentially the same as your proof, which is correct.