Prove that the presheaf of holomorphic functions that admit a square root does not form a sheaf.
Why is this? I can guess that this is because the square root of such a function contains two branches. However, I don't understand how this will lead to a contradiction (presumably in the "identity" axiom).
The problem is the "gluing axiom": Let $U^{-}$ and $U^{+}$ denote the complements of the non-negative and non-positive real axis, respectively. The function $f(z) = z$ admits a square root in each of $U^{\pm}$, but does not admit a square root on their union.