Locally sheaf, does not imply globally sheaf I am searching for a presheaf $P$ on $X$, such that $P$ restricted to $U_i$  is a sheaf on $U_i$, where ${U_i}$'s form an open cover of $X$, but $P$ is not a sheaf on $X$. Please help.
 A: Let $P$ be the presheaf of bounded continuous functions on $\mathbb{R}$.  $P$ is a sheaf on each of the sets $U_n= (n,n+2)$ which cover $\mathbb{R}$.  It is not a sheaf because $f_n(x) = x$ on each of the $U_n$ do not patch together to form a bounded function.  You may find it instructive to show that the sheafification of $P$ is just the sheaf of (potentially unbounded) continuous functions.
A: The constant presheaf on a discrete space.
A: Consider either of the functions Logz or $\sqrt z$ defined on a region (open, connected set) $R$ that winds around the origin (but, in the case of $Logz$ , the region cannot contain the origin). Local expressions for both exist, but these local expressions cannot be patched to give a global expression in neither case, since neither can be defined globally in such a region. There is a result that  a holomorphic  $Logz$ can be defined in any simply-connected region that does not wind around $0$ , so that you can always define such a branch locally. Maybe you can use the fact that $Logz$ has non-trivial monodromy group (group is $\mathbb Z$), to show this. Notice that the argument increases by $2\pi$ everytime you wind around $0$.
EDIT: It seems like I misread the question again. For a presheaf that is not a sheaf, consider , in $\mathbb R$ , and for $W \subset \mathbb R$ , $F(W):=${$f:W \rightarrow \mathbb R , f(x)=c  $ for $x$ in $U$}.
