Sheafication of a sheaf restricted to a open set Let $X$ be a topological space and $U$ be open in $X$. Let $\mathcal F$ be a presheaf of rings on $X$. Let  $\mathcal F_u$ denote the presheaf restricted to the open set $U$. $\mathcal F^+$ denote the sheafication of $\mathcal F$. I want to show that $(\mathcal F_u)^+$ is  $(\mathcal F^+)_u$ using the universal property of sheafication. Now there is an obvious map $(\mathcal F_u)^+$ to $(\mathcal F^+)_u$. What to get an inverse map and prove they are isomorphic?
 A: Why not just compute the stalks and see that they are isomorphic? Let $i : U \hookrightarrow X$ denote the inclusion map and let $\mathcal{F}|_{U} := i^{-1}\mathcal{F}$ be the restriction of the presheaf $\mathcal{F}$ to $U$. From the canonical map $\theta : \mathcal{F} \to \mathcal{F}^+$ we get by applying the restriction functor a map (which for simplicity I will just call $\varphi$) $\varphi : \mathcal{F}|_U \to (\mathcal{F}^+)|_U$. The codomain is a sheaf and so the universal property of the sheafification gives a map $\overline{\varphi} : (\mathcal{F}|_U)^+ \to (\mathcal{F}^+)|_U$ such that the diagram
$\hspace{2.in}$

commutes. Now for any $p\in U$ let us pass to the stalk at $p$. The functored vertical column we know is an isomorphism, so we just need to prove that $\varphi_p$ is an isomorphism. To do this, for every $V \subseteq U$ open we have a square
$\hspace{2.5in}$
where now I consider $\mathcal{F}$ as a presheaf on $U$. Then when we pass to the direct limit over all open sets $V \subseteq U$ that contain $p$, the functored left column is an isomorphism, and the functored top and bottom rows are isomorphisms as well. Thus $\varphi_p$ is an isomorphism for all $p$ and so $\overline{\varphi}$ is an isomorphism.
A: Let $j : U \to X$ denote the open inclusion. The restriction functor $\mathrm{Sh}(X) \to \mathrm{Sh}(U), F \mapsto F|_U$ is isomorphic to $j^*$ and therefore left adjoint to the direct image functor $j_*$. But these functors are also defined on presheaves with the same formulas and there also satisfy this adjunction. It follows, that for every sheaf $G$ on $U$, we have
$\hom((F|_U)^+,G) = \hom(F|_U,G) = \hom(F,j_* G) = \hom(F^+,j_* G) =  \hom((F^+)|_U,G).$
The Yoneda Lemma does the rest.
