Let $\mathcal{F}$ be a presheaf on $\mathbb{R}$ such that $\mathcal{F}(U)$ is the abelian group of continuous functions with bounded support on $U$. Then what is the sheafification of $\mathcal{F}$?
I guess the sheafification should be the abelian group of continuous functions, but how should I prove it rigorously?
Thanks in advance!
Let $F$ be our presheaf of continuous functions with bounded support, $G$ be the sheaf of continuous functions, $F^+$ the sheafification of $F$.
Every continuous function with bounded support is a continuous function, so we have an injective morphism $F\to G$, which induces a map $F^+\to G$ by definition of the sheafification. To show that this is an isomorphism, we just need to prove that it is an isomorphism on stalks. However, $F^+$ has the same stalks as $F$, so it suffices to prove that the map $F\to G$ induces an isomorphism on stalks.
However, this is fairly clear. We already know injectivity, so it just remains to check surjectivity. Pick $x\in\Bbb{R}$. Choose $U$ some bounded open neighborhood of $\Bbb{R}$. Then for any $[(f,W)]\in G_x$, we have that the class $[(f|_{W\cap U},W\cap U)]\in F_x$ maps to $[(f,W)]$.