# Constant presheaf is not a sheaf

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?

• They may want something more explicit. Can you write down the equations for a particular $X, X_1, X_2$? Hopefully "variety" doesn't mean irreducible here. – Hoot Jan 14 '15 at 22:26
• @Hoot I tried by I didnt manage to, can you give me a hint? – user117449 Jan 14 '15 at 22:29
• Well, it can't be connected. Look at closed subsets of $\mathbf A^1$. – Hoot Jan 14 '15 at 23:01
• @Hoot I took $X=V(J)$ where $J=((x-1)(x-2))$ in $\mathbb{R}[x]$, then $X= \{1, 2 \}$. Do you think this is good? – user117449 Jan 14 '15 at 23:06
• Looks good to me. – Hoot Jan 14 '15 at 23:17

• @Sodan your example is only missing the $X_1$ and $X_2$, which you can take to be single points. – Matt Samuel Jan 14 '15 at 23:16
• Yes, I thought i can take $X=X_1 \cup X_2$ where $X_1=\{1 \}$ and $X_2= \{ 2 \}$ – user117449 Jan 14 '15 at 23:18