I'm trying to prove whether or not this is true:
Let $\mathcal{F}$ be a presheaf and $\mathcal{G}$ a sheaf (of rings, let's say) on a space $X$. If $\{U_i\}_i$ is a basis for the topology of $X$ and $\mathcal{F}\big|_{U_i}\simeq \mathcal{G}\big|_{U_i}$ for all $i$, then the sheafification $\mathcal{F}^{sh}$ is isomorphic to $\mathcal{G}$.
If this is true, I imagine it could be proven by universal property.
I can see how to build the morphism of pre-sheaves $\mathcal{F}\to\mathcal{G}$: if $f\in\mathcal{F}(U)$ we define it's image in $\mathcal{G}(U)$ as the gluing of the images $f\big|_{U\cap U_i}$ in $\mathcal{G}(U\cap U_i)$.
But don't know how to build the morphisms of sheaves $\mathcal{F}^{sh}\to\mathcal{G}$.
Any advice?
I think the assertion is true. Note however that I am learning algebraic geometry at the moment, so the following might be wrong. Please correct me, if this is the case and I will happily edit/delete it...
If you have a morphism of presheaves $\varphi:\mathcal{F} \rightarrow \mathcal{G}$ with the property that $$\varphi \vert_{U_i}: \mathcal{F}\vert_{U_i} \rightarrow \mathcal{G}\vert_{U_i}$$ is an isomorphism for all $U_i$, then functoriality of sheafification (by the universal property) yields a morphism of sheaves $$\varphi^+:\mathcal{F}^+ \rightarrow \mathcal{G}^+$$ with the property that $$\varphi^+\vert_{U_i}:\mathcal{F}^+\vert_{U_i} \cong \mathcal{G}^+\vert_{U_i}.$$ Hence for any $p\in X$ there is some $i$ with $p\in U_i$ and we get that on stalks $$(\varphi^+)_p:(\mathcal F^+)_p \rightarrow (\mathcal G^+)_p = (\varphi^+\vert_{U_i})_p:(\mathcal F^+\vert_{U_i})_p \rightarrow (\mathcal G^+\vert_{U_i})_p$$ is an isomorphism. Thus $\varphi^+$ is an isomorphism and we get $$\mathcal F^+ \cong \mathcal G^+ \cong \mathcal G$$