Let $X$ be a topological space containing two closed points $x,y$ and let $i : \{x,y\} \to X$ denote the inclusion map. Notice that $\{x,y\}$ carries the discrete topology. Let $F$ be a sheaf on $X$. Then $i^{-1} F$ is a presheaf on $\{x,y\}$ which is given by $(i^{-1} F)(\emptyset)=1$ (the terminal set), $(i^{-1} F)(\{x\}) = F_x$ (the stalk at $x$), $(i^{-1} F)(\{y\})=F_y$ (the stalk at $y$) and $$(i^{-1} F)(\{x,y\}) = \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U).$$ Why is, $i^{-1} F$ a sheaf if and only if the canonical map $$\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y.$$ is an isomorphism ?.
Thanks a lot to all of you.
A presheaf is a sheaf iff it is mono and conjunctive (maybe there is other terminology) For mono: Say that $f \in i^{-1}F(X)$ and if $r_x(f) \in i^{-1}F(\{x\})$ and $r_y(f) \in i^{-1}F(\{y\})$ are both zero then $f$ must be zero. this means that the map given by restrictions $$i^{-1}F(X) \rightarrow i^{-1}F(\{x\})\times i^{-1}F(\{y\})$$ is injective.
For conjuctive we get that the same map is surjective for if $f_x \in i^{-1}F(\{x\})$ and $f_y \in i^{-1}F(\{y\})$ then they agree on the common domain $\{x\} \cap \{y\}$ trivially so and there must be $f\in i^{-1}F(X)$ which resticts to $f_x$ and $f_y$.