Morphism of Sheaves Given Morphism of Sheaves on a Base This is from Vakil's FOAG: exercise 2.5 C, part b. I understand how objects in the extension sheaf from a sheaf on a base $\mathcal B$ of a topology are created, but I am having trouble understanding how to produce a morphism of sheaves given a morphism of sheaves on a base. 
Assume we have topological space $X$. Supposing we have two sheaves on our base $\mathcal B$, say $F$ and $G$, and maps $F(B_i) \to G(B_i)$ for all $B_i \in \mathcal B$, these induce maps between any stalks $F_x \to G_x$ we like, and we know also for any $x \in X$, that $F_x \simeq F^{ext}_x$, where $F^{ext}$ is our extended sheaf (likewise for $G^{ext}$). After this, I do not know how to proceed, nor do I know if I needed all of that information.
 A: Here is my best attempt at an answer. Realize $\mathcal F^{ext}(U)$ instead as the limit of the $F(B_i)$ over all possible coverings of $U$ by basic open sets. This has by definition maps to each of the $F(B_i)$ commuting with known restriction morphisms between them. Since we have maps $F(B_i) \to G(B_i)$ for all $i$ with commuting squares (a condition imposed on our restriction functions initially), this gives us maps from $\varprojlim F(B_i)$ to each of the $G(B_i)$ which commute with restrictions. By the universal property of the limit applied to $\varprojlim G(B_i)$, there must be a unique map from $F^{ext}(U) = \varprojlim F(B_i) \to \varprojlim G(B_i) = G^{ext}(U)$ for all $U$, as desired. This creates a morphism of sheaves because it plays well with restrictions, again by the universal property of limits.
A: As indicated by the KReiser hint, you can do as follows:
We see the following diagram :

The first and second rows are exact , due to sheaf property, the second square commute do to compactible with restriction. and the universal property of the kernel gives the morphism we want.
