Possible Duplicate:
Failure of isomorphisms on stalks to arise from an isomorphism of sheaves
Let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves over a topological space X.
If we have that $\mathcal{F}_{x} \cong \mathcal{G}_{x}$ for all $x \in X$. (Without having a morphism between the sheaves)
Can we conclude that $\mathcal{F} \cong \mathcal{G}$ ?. How can we show the isomorphism explicitly?
Since $\mathcal{F} \cong \mathcal{F}^{+}$ (the sheafification), I was thinking about defining the morphism stalk by stalk. But the isomorphism between the stalks could be pretty different for $x, y\in X$ and the local condition in the sections in $\mathcal{G}$ would not be satisfied.