I'm trying to learn algebraic geometry and am trying to think about what kinds of things are presheafs but not sheafs.
One exercise I had was to show that bounded holomorphic functions on open subsets of $\mathbb{C}$ do not form a sheaf. This is pretty obvious: $D_1(0), D_2(0), ...$ form an open cover of $\mathbb{C}$, and $f_n: D_n(0) \rightarrow \mathbb{C}$ given by $f_n(z) = z$ is holomorphic on $D_n(0)$ and bounded, and the restriction of any $f_n, f_m$ to $D_n(0) \cap D_m(0)$ gives you the same thing. If we had a sheaf structure, then there would be a bounded holomorphic function $g$ on $\mathbb{C}$ (necessarily $g(z) = z$) which restricts to $f_n$ on $D_n(0)$ for each $n$, which is absurd.
What about the presheaf of holomorphic functions on open subsets of $\mathbb{C}$ which have holomorphic square roots? This shouldn't be a sheaf, but I'm having trouble seeing why, mostly because I'm not really comfortable with saying with certainty that a given function has no holomorphic square root.
Every non-zero holomorphic function has a holomorphic square root on any simply connected open set. Pick any open $\Omega$ with a function $f:\Omega\to\mathbb C$ which does not has a square root and consider the covering of $\Omega$ by the discs it contains.